Basic Linear algebra c °A. Baker
Andrew Baker [08/12/2009]
Department of Mathematics, University of Glasgow. E-mail a...

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Basic Linear algebra c °A. Baker

Andrew Baker [08/12/2009]

Department of Mathematics, University of Glasgow. E-mail address: [email protected] URL: http://www.maths.gla.ac.uk/∼ajb

Linear Algebra is one of the most important basic areas in Mathematics, having at least as great an impact as Calculus, and indeed it provides a significant part of the machinery required to generalise Calculus to vector-valued functions of many variables. Unlike many algebraic systems studied in Mathematics or applied within or outwith it, many of the problems studied in Linear Algebra are amenable to systematic and even algorithmic solutions, and this makes them implementable on computers – this explains why so much calculational use of computers involves this kind of algebra and why it is so widely used. Many geometric topics are studied making use of concepts from Linear Algebra, and the idea of a linear transformation is an algebraic version of geometric transformation. Finally, much of modern abstract algebra builds on Linear Algebra and often provides concrete examples of general ideas. These notes were originally written for a course at the University of Glasgow in the years 2006–7. They cover basic ideas and techniques of Linear Algebra that are applicable in many subjects including the physical and chemical sciences, statistics as well as other parts of mathematics. Two central topics are: the basic theory of vector spaces and the concept of a linear transformation, with emphasis on the use of matrices to represent linear maps. Using these, a geometric notion of dimension can be made mathematically rigorous leading its widespread appearance in physics, geometry, and many parts of mathematics. The notes end by discussing eigenvalues and eigenvectors which play a rˆole in the theory of diagonalisation of square matrices, as well as many applications of linear algebra such as in geometry, differential equations and physics. There are some assumptions that the reader will already have met vectors in 2 and 3dimensional contexts, and has familiarity with their algebraic and geometric aspects. Basic algebraic theory of matrices is also assumed, as well as the solution of systems of linear equations using Gaussian elimination and row reduction of matrices. Thus the notes are suitable for a secondary course on the subject, building on existing foundations. There are very many books on Linear Algebra. The Bibliography lists some at a similar level to these notes. University libraries contain many other books that may be useful and there are some helpful Internet sites discussing aspects of the subject.

Contents Chapter 1. Vector spaces and subspaces 1.1. Fields of scalars 1.2. Vector spaces and subspaces

1 1 3

Chapter 2. Spanning sequences, linear independence and bases 2.1. Linear combinations and spanning sequences 2.2. Linear independence and bases 2.3. Coordinates with respect to bases 2.4. Sums of subspaces

11 11 13 22 23

Chapter 3. Linear transformations 3.1. Functions 3.2. Linear transformations 3.3. Working with bases and coordinates 3.4. Application to matrices and systems of linear equations 3.5. Geometric linear transformations

29 29 30 37 41 43

Chapter 4. Determinants 4.1. Definition and properties of determinants 4.2. Determinants of linear transformations 4.3. Characteristic polynomials and the Cayley-Hamilton theorem

45 45 50 51

Chapter 5. Eigenvalues and eigenvectors 5.1. Eigenvalues and eigenvectors for matrices 5.2. Some useful facts about roots of polynomials 5.3. Eigenspaces and multiplicity of eigenvalues 5.4. Diagonalisability of square matrices

55 55 56 58 63

Appendix A.

67

Appendix.

Complex solutions of linear ordinary differential equations

Bibliography

69

i

CHAPTER 1

Vector spaces and subspaces 1.1. Fields of scalars Before discussing vectors, first we explain what is meant by scalars. These are ‘numbers’ of various types together with algebraic operations for combining them. The main examples we will consider are the rational numbers Q, the real numbers R and the complex numbers C. But mathematicians routinely work with other fields such as the finite fields (also known as Galois fields) Fpn which are important in coding theory, cryptography and other modern applications. Definition 1.1. A field of scalars (or just a field ) consists of a set F whose elements are called scalars, together with two algebraic operations, addition + and multiplication ×, for combining every pair of scalars x, y ∈ F to give new scalars x + y ∈ F and x × y ∈ F . These operations are required to satisfy the following rules which are sometimes known as the field axioms. Associativity: For x, y, z ∈ F , (x + y) + z = x + (y + z), (x × y) × z = x × (y × z). Zero and unity: There are unique and distinct elements 0, 1 ∈ F such that for x ∈ F , x + 0 = x = 0 + x, x × 1 = x = 1 × x. Distributivity: For x, y, z ∈ F , (x + y) × z = x × z + y × z, z × (x + y) = z × x + z × y. Commutativity: For x, y ∈ F , x + y = y + x, x × y = y × x. Additive and multiplicative inverses: For x ∈ F there is a unique element −x ∈ F (the additive inverse of x) for which x + (−x) = 0 = (−x) + x. For each non-zero y ∈ F there is a unique element y −1 ∈ F (the multiplicative inverse of y) for which y × (y −1 ) = 1 = (y −1 ) × y. Remark 1.2. • Usually we just write xy instead of x × y, and then we always have xy = yx. 1

2

1. VECTOR SPACES AND SUBSPACES

• Because of commutativity, some of the above rules are redundant in the sense that they are consequences of others. • When working with vectors we will always have a specific field of scalars in mind and will make use of all of these rules. It is possible to remove commutativity or multiplicative inverses and still obtain mathematically interesting structures but in this course we definitely always assume the full strength of these rules. • The most important examples are R and C and it is worthwhile noting that the above rules are obeyed by these as well as Q. However, other examples of number systems such as N and Z do not obey all of these rules. Proposition 1.3. Let F be a field of scalars. For any x ∈ F , (a)

0x = 0,

(b)

− x = (−1)x.

Proof. Consider the following calculations which use many of the rules in Definition 1.1. For x ∈ F , 0x = (0 + 0)x = 0x + 0x, hence 0 = −(0x) + 0x = −(0x) + (0x + 0x) = (−(0x) + 0x) + 0x = 0 + 0x = 0x. This means that 0x = 0 as required for (a). Using (a) we also have x + (−1)x = 1x + (−1)x = (1 + (−1))x = 0x = 0, thus establishing (b).

¤

Example 1.4. Let F be a field. Let a, b ∈ F and assume that a 6= 0. Show that the equation ax = b has a unique solution for x ∈ F . Challenge: Now suppose that aij , b1 , b2 ∈ F for i, j = 1, 2. With the aid of the ‘usual’ method of solving a pair of simultaneous linear equations show that the system ) ( a11 x1 + a12 x2 = b1 a21 x1 + a22 x2 = b2 has a unique solution for x1 , x2 ∈ F if a11 a22 − a12 a21 6= 0. What can be said about solutions when a11 a22 − a12 a21 = 0? Solution. As a 6= 0 there is an inverse a−1 , hence the equation implies that x = 1x = (a−1 a)x = a−1 (ax) = a−1 b, so if x is a solution then it must equal a−1 b. But it is also clear that a(a−1 b) = (aa−1 )b = 1b = b, so this scalar does satisfy the equation. Notice that if a = 0, then the equation 0x = b can only have a solution if b = 0 and in that case any x ∈ F will work so the solution is not unique. Challenge: For this you will need to recall things about 2 × 2 linear systems. The upshot is that the system can have either no or infinitely many solutions. ¤

1.2. VECTOR SPACES AND SUBSPACES

3

1.2. Vector spaces and subspaces We now come to the key idea of a vector space. This involves an abstraction of properties already met with in special cases when dealing with vectors and systems of linear equations. Before meeting the general definition, here are some examples to have in mind as motivation. The plane and 3-dimensional space are usually modelled with coordinates and their points correspond to elements of R2 and R3 which we write in the form (x, y) or (x, y, z). These are usually called vectors and they are added by adding corresponding coordinates. Scalar multiplication is also defined by coordinate-wise multiplication with scalars. These operations have geometric interpretations in terms of the parallelogram rule and dilation of vectors. n It is usual to identify the set of all column vectors of length n with R , where (x1 , . . . , xn ) real x1 . . corresponds to the column vector . . Note that the use of the different types of brackets is xn very important here and will be used in this way throughout the course. Matrix addition and scalar multiplication correspond to coordinate-wise addition and scalar multiplication in Rn . More generally, the set of all m × n real matrices has an addition and scalar multiplication. In the above, R can be replaced by C and all the algebraic properties are still available. Now we give the very important definition of a vector space. Definition 1.5. A vector space over a field of scalars F consists of a set V whose elements are called vectors together with two algebraic operations, + (addition of vectors) and · (multiplication by scalars). Vectors will usually be denoted with boldface symbols such as v which is hand written as v . The operations + and · are required to satisfy the following rules, which are ∼ sometimes known as the vector space axioms. Associativity: For u, v, w ∈ V and s, t ∈ F , (u + v) + w = u + (v + w), (st) · v = s · (t · v). Zero and unity: There is a unique element 0 ∈ V such that for v ∈ V , v + 0 = v = 0 + v, and multiplication by 1 ∈ F satisfies 1 · v = v. Distributivity: For s, t ∈ F and u, v ∈ V , (s + t) · v = s · v + t · v, s · (u + v) = s · u + s · u. Commutativity: For u, v ∈ V , u + v = v + u. Additive inverses: For v ∈ V there is a unique element −v ∈ V for which v + (−v) = 0 = (−v) + v. Again it is normal to write sv in place of s · v when the meaning is clear. Care may need to be exercised in correctly interpreting expressions such as (st)v = (s × t) · v

4

1. VECTOR SPACES AND SUBSPACES

for s, t ∈ F and v ∈ V ; here the product (st) = (s × t) is calculated in F , while (st)v = (st) · v is calculated in V . Note that we do not usually multiply vectors together, although there is a special situation with the vector (or cross) product defined on R3 , but we will not often consider that in this course. From now on, let V be a vector space over a field of scalars F . If F = R we refer to V as a real vector space, while if F = C we refer to it as a complex vector space. Proposition 1.6. For s ∈ F and v ∈ V , the following identities are valid: (a)

0v = 0,

(b)

s0 = 0,

(c)

(−s)v = − (sv) = s(−v),

and in particular, taking s = 1 we have (d)

−v = (−1)v.

Proof. The proofs are similar to those of Proposition 1.3, but need care in distinguishing multiplication of scalars and scalar multiplication of vectors. ¤ Here are some important examples of vector spaces which will be met throughout the course. Example 1.7. Let F be any field of scalars. For n = 1, 2, 3, . . ., let F n denote the set of all n-tuples t = (t1 , . . . , tn ) of elements of F . For s ∈ F and u, v ∈ F n , we define (u1 , . . . , un ) + (v1 , . . . , vn ) =(u1 + v1 , . . . , un + vn ), s · (v1 , . . . , vn ) =s(v1 , . . . , vn ) = (sv1 , . . . , svn ). The zero vector is 0 = (0, . . . , 0). The cases F = R and F = C will be the most important in these notes. Example 1.8. Let F be a field. The set Mm×n (F ) of all m × n matrices with entries in F forms a vector space over F with addition and multiplication by scalars defined in the usual way. We identify the set Mn×1 (F ) with F n , to obtain the vector space of Example 1.7. We also set Mn (F ) = Mn×n (F ). In all cases, the zero vector of Mm×n (F ) is the zero matrix Om×n . Example 1.9. The complex numbers C can be viewed as a vector space over R with the usual addition and scalar multiplication being the usual multiplication. More generally, if V is a complex vector space then we can view it as a real vector space by only allowing real numbers as scalars. For example, Mm×n (C) becomes a real vector space in this way as does Cn . We then refer to the real vector space V as the underlying real vector space of the complex vector space V . Example 1.10. Let F be a field. Consider the set F [X] consisting of all polynomials in X with coefficients in F . We define addition and multiplication by scalars in F by (a0 + a1 X + · · · + ar X r ) + (b0 + b1 X + · · · + br X r ) =(a0 + b0 ) + (a1 + b1 )X + · · · + (ar + br )X r , s · (a0 + a1 X + · · · + ar X r ) =(sa0 ) + (sa1 )X + · · · + (sar )X r . The zero vector is the zero polynomial 0 = 0 + 0X + · · · + 0X r = 0.

1.2. VECTOR SPACES AND SUBSPACES

5

Example 1.11. Take R to be the field of scalars and consider the set D(R) consisting of all infinitely differentiable functions R −→ R, i.e., functions f : R −→ R for which all possible derivatives f (n) (x) exist for n > 1 and x ∈ R. We define addition and multiplication by scalars as follows. Let t ∈ R and f, g ∈ D(R), then f + g and s · f are the functions given by the following rules: for x ∈ R, (f + g)(x) = f (x) + g(x), (s · f )(x) = sf (x). Of course we really ought check that all the derivatives of f + g and s · f exist so that these function are in D(R), but this is an exercise in Calculus. The zero vector here is the constant function which sends every real number to 0, and this is usually written 0 in Calculus, although this might seem confusing in our context! More generally, for a given real number a, it is also standard to write a for the constant function which sends every real number to a. Note that Example 1.10 can also be viewed as a vector space consisting of functions since every polynomial over a field can be thought of as the function obtained by evaluation of the variable at values in F . When F = R, R[X] ⊆ D(R) and we will see that this is an example of a vector subspace which will be defined soon after some motivating examples. Example 1.12. Let V be a vector space over a field of scalars F . Suppose that W ⊆ V contains 0 and is closed under addition and multiplication by scalars, i.e., for s ∈ F and u, v ∈ W , we have u + v ∈ W, su ∈ W. Then W is also a vector space over F . Example 1.13. Consider the case where F = R and V = R2 . Then the line L with equation 2x − y = 0 consists of all vectors of the form (t, 2t) with t ∈ R, i.e., L = {(t, 2t) ∈ R2 : t ∈ R}. Clearly this passes through the origin so it contains 0. It is also closed under addition and scalar multiplication: if (t, 2t), (t1 , 2t1 ), (t2 , 2t2 ) ∈ L and s ∈ R then (t1 , 2t1 ) + (t2 , 2t2 ) = (t1 + t2 , 2t1 + 2t2 ) = (t1 + t2 , 2(t1 + t2 )) ∈ L, s(t, 2t) = (st, 2(st)) ∈ L. More generally, any line with equation of form ax + by = 0 with (a, b) 6= 0 = (0, 0) is similarly closed under addition and multiplication by scalars. Example 1.14. Consider the case where F = R and V = R3 . If (a, b, c) ∈ R3 is non-zero, then consider the plane P with equation ax + by + cz = 0, thus as a set P is given by P = {(x, y, z) ∈ R3 : ax + by + cz = 0}.

6

1. VECTOR SPACES AND SUBSPACES

Then P contains 0 and is closed under addition and scalar multiplication since whenever (x, y, z), (x1 , y1 , z1 ), (x2 , y2 , z2 ) ∈ P and s ∈ R, the sum (x1 , y1 , z1 ) + (x2 , y2 , z2 ) = (x1 + x2 , y1 + y2 , z1 + z2 ) satisfies a(x1 + x2 ) + b(y1 + y2 ) + c(z1 + z2 ) = (ax1 + by1 + cz1 ) + (ax2 + by2 + cz2 ) = 0, while the scalar multiple s(x, y, z) = (sx, sy, sz) satisfies a(sx) + b(sy) + c(sz) = s(ax + by + cz) = 0, showing that (x1 , y1 , z1 ) + (x2 , y2 , z2 ), s(x, y, z) ∈ P. R3

A line in through the origin consists of all vectors of the form tu for t ∈ R, where u is some non-zero direction vector. This is also closed under addition and scalar multiplication. Definition 1.15. Let V be a vector space over a field of scalars F . Suppose that the subset W ⊆ V is non-empty and is closed under addition and multiplication by scalars, thus it forms a vector space over F . Then W is called a (vector ) subspace of V . Remark 1.16. • In order to verify that W is non-empty it is usually easiest to show that it contains the zero vector 0. • The conditions for W to be closed under addition and scalar multiplication are equivalent to the single condition that for all s, t ∈ F and u, v ∈ W , su + tv ∈ W. • If W ⊆ V is a subspace, then for any w ∈ W , all of its scalar multiples (including 0w = 0 and (−1)w = −w) are also in W . Example 1.17. Let V be a vector space over a field F . • The sets {0} and V are both subspaces of V , usually thought of as its uninteresting subspaces. • If v ∈ V is a non-zero vector, then {v} is never a subspace of V since (for example) 0v = 0 is not an element of {v}. • The smallest subspace of V containing a given non-zero vector v is the line spanned by v which is the set of all scalar multiplies of v, {tv : t ∈ F }. Before giving more examples, here are some non-examples. Example 1.18. Consider the real vector space R2 and the following subsets with the same addition and scalar multiplication. (a) V1 = {(x, y) ∈ R2 : x > 0} is not a real vector space: for example, given any vector (x, y) ∈ V1 with x > 0, there is no additive inverse −(x, y) ∈ V1 since this would have to be (−x, −y) which has −x < 0. So V1 is not closed under addition. (b) V2 = {(x, y) ∈ R2 : y = x2 } is not a real vector space: for example, (1, 1) ∈ V2 but 2(1, 1) = (2, 2) does not satisfy y = x2 since x2 = 4 and y = 2. So V2 is not closed under scalar multiplication. (c) V3 = {(x, y) ∈ R2 : x = 2} is not a real vector space: for example, (2, 0), (2, 1) ∈ V3 but (2, 0) + (2, 1) = (4, 1) ∈ / V3 , so V3 is not closed under addition.

1.2. VECTOR SPACES AND SUBSPACES

7

Here are some further examples of subspaces. Example 1.19. Suppose that F is a equations a11 x1 + .. (1.1) . a x + m1 1

field and that we have a system of homogenous linear ··· .. .

+

a1n xn .. .

···

+ amn xn

= 0 .. . = 0

x1 . n . for scalars aij ∈ F . Then the set W of solutions (x1 , . . . , xn ) = . in F of the system (1.1) xn is a subspace of F n . This is most easily checked by recalling that the system is equivalent to a single matrix equation

(1.2)

Ax = 0,

where A = [aij ] and then verifying that given solutions x, y ∈ W and s ∈ F , A(x + y) = Ax + Ay = 0 + 0 = 0, and A(sx) = s(Ax) = s(A0) = s0 = 0. This example indicates an extremely important relationship between vector space ideas and linear equations, and this is one of the main themes in the subject of Linear Algebra. Example 1.20. Recall the vector space F [X] of Example 1.10. Consider the following subsets of F [X]: W0 = F [X], W1 = {f (X) ∈ F [X] : f (0) = 0}, W2 = {f (X) ∈ F [X] : f (0) = f 0 (0) = 0}, and in general, for n > 1, Wn = {f (X) ∈ F [X] : for k = 0, . . . , n − 1, f (k) (0) = 0}. Show that each Wn is a subspace of F [X] and that Wn is a subspace of Wn−1 . Solution. Let n > 1. Suppose that s, t ∈ F and f (X), g(X) ∈ Wn . Then for each k = 0, . . . , n − 1 we have f (k) (0) = 0 = g (k) (0), hence

dk (sf (X) + tg(X)) = sf (k) (X) + tg (k) (X), dX k and evaluating at X = 0 gives dk (sf + tg)(0) = sf (k) (0) + tg (k) (0) = 0. dX k This shows that Wn is closed under addition and scalar multiplication, therefore it is a subspace of F [X]. Clearly Wn ⊆ Wn−1 and as it is closed under addition and scalar multiplication, it is a subspace of Wn−1 . ¤

8

1. VECTOR SPACES AND SUBSPACES

We can combine subspaces to form new subspaces. Given subsets Y1 , . . . , Yn ⊆ X of a set X, their intersection is the subset Y1 ∩ Y2 ∩ · · · ∩ Yn = {x ∈ X : for k = 1, . . . , n, x ∈ Yk } ⊆ X, while their union is the subset Y1 ∪ Y2 ∪ · · · ∪ Yn = {x ∈ X : there is a k = 1, . . . , n such that x ∈ Yk } ⊆ X, Proposition 1.21. Let V be a vector space. If W1 , . . . , Wn ⊆ V are subspaces of V , then W1 ∩ · · · ∩ Wn is a subspace of V . Proof. Let s, t ∈ F and u, v ∈ W1 ∩ · · · ∩ Wn . Then for each k = 1, . . . , n we have u, v ∈ Wk , and since Wk is a subspace, su + tv ∈ Wk . But this means that su + tv ∈ W1 ∩ · · · ∩ Wn , i.e., W1 ∩ · · · ∩ Wn is a subspace of V .

¤

On the other hand, the union of subspace does not always behave so well. Example 1.22. Consider the real vector space R2 and the subspaces W 0 = {(s, 0) : s ∈ R},

W 00 = {(0, t) : t ∈ R}.

Show that W 0 ∪ W 00 is not a subspace of R2 and determine W 0 ∩ W 00 . Solution. The union of these subspaces is W 0 ∪ W 00 = {v ∈ R2 : v = (s, 0) or v = (0, s) for some s ∈ R}. Then (1, 0) ∈ W 0 and (0, 1) ∈ W 00 , but (1, 1) = (1, 0) + (0, 1) is visibly not in W 0 ∪ W 00 , so this is not closed under addition. (However, it is closed under scalar multiplication, so care is required when checking this sort of example.) If x ∈ W 0 ∩ W 00 then x = (s, 0) and x = (0, t) for some s, t ∈ R, which is only possible if s = t = 0 and so x = (0, 0). Therefore W 0 ∩ W 00 = {0}. ¤ Here is another way to think about Example 1.19. Example 1.23. Suppose that F is a equations a11 x1 + .. . a x + m1 1

field and that we have a system of homogenous linear ··· .. .

+

a1n xn .. .

···

+ amn xn

= 0 .. . = 0

with the aij ∈ F . For each r = 1, . . . , m, let Wr = {(x1 , . . . , xn ) : ar1 x1 + · · · + arn xn = 0} ⊆ F n , which is a subspace of F n .

1.2. VECTOR SPACES AND SUBSPACES

9

Then the set W of solutions (x1 , . . . , xn ) of the above linear system is given by the intersection W = W1 ∩ · · · ∩ Wm , therefore W is a subspace of F n . Example 1.24. In the real vector space R4 , determine the intersection U ∩V of the subspaces U = {(w, x, y, z) ∈ R4 : x + 2y = 0 = w + 2x − z},

V = {(w, x, y, z) ∈ R4 : w + x + 2y = 0}.

Solution. Before doing this we first use elementary row operations to find the general solutions of the systems of equations used to define U and V . For U we consider the system ) ( x + 2y = 0 w + 2x − z = 0 with augmented matrix # " # " # " 1 2 0 −1 0 1 0 −4 −1 0 0 1 2 0 0 ∼ ∼ , 2 0 0 1 2 0 −1 0 R1 ↔R2 0 1 2 0 0 R1 →R1 −2R2 0 1 where the last matrix is a reduced echelon matrix. So the general solution vector of this system is (4s + t, −2s, s, t) where s, t ∈ R. Thus we have U = {(4s + t, −2s, s, t) ∈ R4 : s, t ∈ R}. Similarly, the general solution vector of the equation w + x + 2y = 0 is (−r − 2s, r, s, t) where r, s, t ∈ R and we have V = {(−r − 2s, r, s, t) ∈ R4 : r, s, t ∈ R}. Now to find U ∩ V we have to solve the system x + 2y = 0 w + 2x − z = 0 w + x + 2y = 0 with augmented matrix 1 0 1 2 0 0 1 2 0 −1 0 ∼ ∼ 0 1 2 0 0 1 2 0 −1 0 0 R3 →R3 −R1 R1 ↔R2 0 1 1 2 0 0 1 1 2 0 0 1 0 −4 −1 0 1 ∼ ∼ 1 2 0 0 0 0 R1 →R1 −2R2 R3 →(1/4)R3 R3 →R3 +R1 0 0 4 1 0 0

2 0 −1 0 1 2 0 0 −1 2 1 0 0 −4 1 2 0 1

−1 0 0 0 1/4 0

1 0 0 0 0 ∼ 0 1 0 −1/2 0 , R1 →R1 +4R3 R2 →R2 −2R3 0 0 1 1/4 0

where the last matrix is reduced echelon. The general solution vector is (0, t/2, −t/4, t) where t ∈ R, thus we have U ∩ V = {(0, t/2, −t/4, t) : t ∈ R} = {(0, 2s, −s, 4s) : s ∈ R}.

¤

CHAPTER 2

Spanning sequences, linear independence and bases When working with R2 or R3 we often use the fact that every vector v can be expressed in terms of the standard basis vectors e1 , e2 or e1 , e2 , e3 i.e., v = xe1 + ye2

or v = xe1 + ye2 + ze3 ,

where v = (x, y) or v = (x, y, z). Something similar is true for every vector space and involves the concepts of spanning sequence and basis. 2.1. Linear combinations and spanning sequences We will always assume that V is a vector space over a field F . The first notion we will require is that of a linear combination. Definition 2.1. If t1 , . . . , tr ∈ F and v1 , . . . , vr is a sequence of vectors in V (some of which may be 0 or may be repeated), then the vector t1 v1 + · · · + tr vr ∈ V is said to be a linear combination of the sequence v1 , . . . , vr (or of the vectors vi ). Sometimes it is useful to allow r = 0 and view the 0 as a linear combination of the elements of the empty sequence ∅. A sequence is often denoted (v1 , . . . , vr ), but we will not use that notation to avoid confusion with the notation for elements of F n . We will sometimes refer to a sequence v1 , . . . , vr of vectors in V as a sequence in V . There are various operations which can be performing on sequences such as deleting terms, reordering terms, multiplying terms by non-zero scalars, combining two sequences into a new one by concatinating them in either order. Notice that 0 can be expressed as a linear combination in other ways. For example, when F = R and V = R2 , 3(1, −1) + (−3)(2, 0) + 3(1, 1) = 0,

1(0, 0) = 0,

1(5, 0) + 0(0, 0) + (−5)(1, 0) = 0,

so 0 is a linear combination of each of the sequences (1, −1), (2, 0), (1, 1),

0,

(5, 0), 0, (1, 0).

For any sequence v1 , . . . , vr we can always write 0 = 0v1 + · · · + 0vr , so 0 is a linear combination of any sequence (even the empty sequence of length 0). Sometimes it is useful to allow infinite sequences as in the next example. 11

12

2. SPANNING SEQUENCES, LINEAR INDEPENDENCE AND BASES

Example 2.2. Let F be a field and consider the vector space of polynomials F [X]. Then every f (X) ∈ F [X] has a unique expression of the form f (X) = a0 + a1 X + · · · + ar X r + · · · , where ai ∈ F and for large i, ai = 0. Thus every polynomial in F [X] can be uniquely written as a linear combination of the sequence 1, X, X 2 , . . . , X r , . . .. Such a special set of vectors is called a basis and soon we will develop the theory of bases. Bases are extremely important and essential for working in vector spaces. In applications such as to Fourier series, there are bases consisting of trigonometric functions, while in other context orthogonal polynomials play this rˆole. Definition 2.3. Let S : w1 , . . . be a sequence (possibly infinite) of elements of V . Then S is a spanning sequence for V (or is said to span V ) if for every element v ∈ V there is an expression of the form v = s1 w1 + · · · + sk wk , where s1 , . . . , sk ∈ F , i.e., every vector in V is a linear combination of the sequence S. Notice that in this definition we do not require any sort of uniqueness of the expansion. Definition 2.4. Let S : w1 , . . . be a sequence of elements of V . Then the linear span of S, Span(S) ⊆ V , is the subset consisting of all linear combinations of S, i.e., Span(S) = {t1 w1 + · · · + tr wr : r = 0, 1, 2, . . . , ti ∈ F }. Remark 2.5. • By definition, S is a spanning sequence for Span(S). • We always have 0 ∈ Span(S). • If S = ∅ is the empty sequence, then we make the definition Span(S) = {0}. Lemma 2.6. Let S : w1 , . . . be a sequence of elements of V . Then the following hold. (a) Span(S) ⊆ V is a subspace and S is a spanning sequence of Span(S). (b) Every subspace of V containing the elements of S contains Span(S) as a subspace, therefore Span(S) is the smallest such subspace. (c) If S 0 is a subsequence of S, then Span(S 0 ) ⊆ Span(S). In (c), we use the idea that S 0 is a subsequence of S if it is obtained from S by removing some terms and renumbering. For example, if S : u, v, w is a sequence then the following are subsequences: S 0 : u, v,

S 00 : u, w,

S 000 : v.

Proof. (a) It is easy to see that any sum of linear combinations of elements for S is also such a linear combination, similarly for scalar multiples. (b) If a subspace W ⊆ V contains all the elements of S, then all linear combinations of S are in W , therefore Span(S) ⊆ W . (c) This follows from (b) since the elements of S 0 are all in Span(S). ¤ Definition 2.7. The vector space V is called finite dimensional if it has a spanning sequence S : w1 , . . . , w` of finite length `.

2.2. LINEAR INDEPENDENCE AND BASES

13

Much of the theory we will develop will be about finite dimensional vector spaces although many important examples are infinite dimensional. Here are some further useful properties of sequences and the subspaces they span. Proposition 2.8. Let S : w1 , . . . be a sequence in V . Then the following hold. (a) If S 0 : w10 , . . . is a sequence obtained by permuting (i.e., reordering) the terms of S, then Span(S 0 ) = Span(S). Hence the span of a sequence is unchanged by permuting its terms. (b) If r > 1 and wr ∈ Span(w1 , . . . , wr−1 ), then for the sequence S 00 : w1 , . . . , wr−1 , wr+1 , . . . , we have Span(S 00 ) = Span(S). In particular, we can remove 0’s and any repetitions of vectors without changing the span of a sequence. (c) If s > 1 and S 000 is obtained from S by replacing ws by a vector of the form ws000 = (x1 w1 + · · · + xs−1 ws−1 ) + ws , where x1 , . . . , xs−1 ∈ F , then Span(S 000 ) = Span(S). (d) If t1 , . . . is sequence of non-zero scalars, then for the sequence T : t1 w1 , t2 w2 , . . . we have Span(T ) = Span(S). Proof. (a),(b),(d) are simple consequences of the commutativity, associativity and distributivity of addition and scalar multiplication of vectors. (c) follows from the obvious fact that Span(w1 , . . . , ws−1 , ws000 ) = Span(w1 , . . . , ws−1 , ws ). ¤ 2.2. Linear independence and bases Now we introduce a notion which is related to the uniqueness issue. Definition 2.9. Let S : w1 , . . . be a sequence in V . • S is linearly dependent if for some r > 1 there are scalars s1 , . . . , sr ∈ F not all zero and for which s1 w1 + · · · + sr wr = 0. • S is linearly independent if it is not linearly dependent. This means that whenever s1 , . . . , sr ∈ F satisfy s1 w1 + · · · + sr wr = 0, then s1 = · · · = sr = 0. Example 2.10. • Any sequence S containing 0 is linearly dependent since 1 · 0 = 0. • If a vector v occurs twice in S then 1v + (−1)v = 0, so again S is linearly dependent. • If a vector in S can be expressed as a linear combination of the list obtained by deleting it, then S is linearly dependent. • In the real vector space R2 , the sequence (1, 2), (3, −1), (0, 7) is linearly dependent since 3(1, 2) + (−1)(3, −1) + (−1)(0, 7) = (0, 0), while its subsequence (1, 2), (3, −1) is linearly independent.

14

2. SPANNING SEQUENCES, LINEAR INDEPENDENCE AND BASES

Lemma 2.11. Suppose that S : w1 , . . . is a linearly dependent sequence in V . Then there is some k > 1 and scalars t1 , . . . , tk−1 ∈ F for which wk = t1 w1 + · · · + tk−1 wk−1 . Hence if S 0 is the sequence obtained by deleting wk then Span(S 0 ) = Span(S). Proof. Since S is linearly dependence, there must be an r > 1 and scalars s1 , . . . , sr ∈ F where not all of the si are zero and they satisfy s1 w1 + · · · + sr wr = 0. Let k be the largest value of i for which si 6= 0. Then s1 w1 + · · · + sk wk = 0. with sk 6= 0. Multiplying by s−1 k gives −1 s−1 k (s1 w1 + · · · + sk wk ) = sk 0 = 0,

whence −1 −1 −1 −1 −1 s−1 k s1 w1 + sk s2 w2 + · · · + sk sk wk = sk s1 w1 + sk s2 w2 + · · · + sk sk−1 wk−1 + wk = 0.

Thus we have −1 wk = (−s−1 k s1 )w1 + · · · + (−sk sk−1 )wk−1 .

The equality of the two spans is clear.

¤

The next result is analogous to Proposition 2.8 for spanning sequences. Proposition 2.12. Let S : w1 , . . . be a sequence in V . Then the following hold. (a) If S 0 : w10 , . . . is a sequence obtained by permuting the terms of S, then S 0 is linearly dependent (respectively linearly independent) if S is. (b) If s > 1 and S 00 is obtained from S by replacing ws by a vector of the form ws00 = (x1 w1 + · · · + xs−1 ws−1 ) + ws , where x1 , . . . , xs−1 ∈ F , then S 00 is linearly dependent (respectively linearly independent) if S is. (c) If t1 , . . . is sequence of non-zero scalars, then for the sequence T : t1 w1 , t2 w2 , . . . , we have T linearly dependent (resp. linearly independent) if S is. Proof. Again these are simple consequences of properties of addition and scalar multiplication. ¤ Proposition 2.13. Suppose that S : w1 , . . . be a linearly independent sequence in V . If for some k > 1, the scalars s1 , . . . , sk , t1 , . . . , tk ∈ F satisfy s1 w1 + · · · + sk wk = t1 w1 + · · · + tk wk , then s1 = t1 , . . . , sk = tk . Hence every vector has at most one expression as a linear combination of S.

2.2. LINEAR INDEPENDENCE AND BASES

15

Proof. The equation can be rewritten as (s1 − t1 )w1 + · · · + (sk − tk )wk = 0. Now linear independence shows that each coefficient satisfies (si − ti ) = 0.

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Proposition 2.14. Let S : w1 , . . . be a spanning sequence for V . Then there is a subsequence T of S which is a spanning sequence for V and is also linearly independent. Proof. We indicate a proof for the case when S is finite, the general case is slightly more involved. Notice that Span(S) = V and for some n > 1, the sequence is S : w1 , . . . , wn . If S is already linearly independent then we can take T = S. Otherwise, by Lemma 2.11, one of the wi can be expressed as a linear combination of the others; let k1 be the largest such i. Then wk1 can be expressed as a linear combination of the list S1 : w1 , . . . wk1 −1 , wk1 +1 , . . . , wn and we have Span(S1 ) = Span(S) = V , so S1 is a spanning sequence for V . The length of S1 is n − 1 < n, so this sequence is shorter than S. Clearly we can repeat this obtaining new spanning sequences S1 , S2 , . . . for V , where the lengths of Sr is n − r. Eventually we must get to a spanning sequence Sk which cannot be reduced further, but this must be linearly independent. ¤ Example 2.15. Taking F = R and V = R2 , show that the sequence S : (1, 2), (1, 1), (−1, 0), (1, 3) spans R2 and find a linearly independent spanning subsequence of S. Solution. Following the idea in the proof of Proposition 2.14, first observe that (1, 3) = 3(1, 1) + 2(−1, 0), so we can replace S by the sequence S 0 : (1, 2), (1, 1), (−1, 0) for which Span(S 0 ) = Span(S). Next notice that (−1, 0) = (1, 2) + (−2)(1, 1), so we can replace S 0 by the sequence S 00 : (1, 2), (1, 1) for which Span(S 00 ) = Span(S 0 ) = Span(S). For every (a, b) ∈ R2 , the equation x(1, 2) + y(1, 1) = (a, b) corresponds to the matrix equation

" #" # " # 1 1 x a = , 2 1 y b

" # 1 1 and since det = −1 6= 0, this has the unique solution 2 1 " # " #−1 " # " #" # x 1 1 a −1 1 a = = . y 2 1 b 2 −1 b Thus S 00 spans R2 and is linearly independent. The next definition is very important.

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16

2. SPANNING SEQUENCES, LINEAR INDEPENDENCE AND BASES

Definition 2.16. A linearly independent spanning sequence of a vector space V is called a basis of V . The next result shows why bases are useful and leads to the notion of coordinates with respect to a basis which will be introduced in Section 2.3. Proposition 2.17. Let V be a vector space over F . A sequence S in V is a basis of V if and only if every vector of V has a unique expression as a linear combination of S. Proof. This follows from Proposition 2.13 together with the fact that a basis is a spanning sequence. ¤ Theorem 2.18. If S is a spanning sequence of V , then there is a subsequence of S which is a basis. In particular, if V is finite dimensional then it has a finite basis. Proof. The first part follows from Proposition 2.14. For the second part, start with a finite spanning sequence then find a subsequence which is a basis, and note that this sequence is finite. ¤ Example 2.19. Consider the vector space V = R3 over R. Show that the sequence S : (1, 0, 1), (1, 1, 1), (1, 2, 1), (1, −1, 2) is a spanning sequence of V and find a subsequence which is a basis. Solution. The following method for the first part uses the general approach to solving systems of linear equations. Given any numbers h1 , h2 , h3 ∈ R, consider the equation x1 (1, 0, 1) + x2 (1, 1, 1) + x3 (1, 2, 1) + x4 (1, −1, 2) = (h1 , h2 , h3 ) = h. This is equivalent to the system x + x + x + x = h 1 2 3 4 1 x2 + 2x3 − x4 = h2 x1 + x2 + x3 + 2x4 = h3 which has augmented matrix

Performing elementary row 1 1 [A | h] ∼ 0 1 R3 →R3 −R1 0 0

1 1 1 1 h1 [A | h] = 0 1 2 −1 h2 . 1 1 1 2 h3

operations we find that 1 1 h1 1 ∼ 2 −1 h2 0 R1 →R1 −R2 0 1 h3 − h1 0 1 ∼ 0 R1 →R1 −2R3 R2 →R2 +R3 0

0 −1 2 h1 − h2 1 2 −1 h2 0 0 1 h3 − h1

0 −1 0 3h1 − h2 − 2h3 1 2 0 −h1 + h2 + h3 , 0 0 1 −h1 + h3

where the last matrix is reduced echelon. So the system is consistent with general solution x1 = s + 3h1 − h2 − 2h3 ,

x2 = −2s − h1 + h2 + h3 ,

x3 = s,

x4 = −h1 + h3

So every element of R3 can indeed be expressed as a linear combination of S.

(s ∈ R).

2.2. LINEAR INDEPENDENCE AND BASES

17

Notice that when h = 0 we have the solution x1 = s,

x2 = −2s,

x3 = s,

x4 = 0

(s ∈ R).

Thus shows that S is linearly dependent. Taking for example s = 1, we see that (1, 0, 1) − 2(1, 1, 1) + (1, 2, 1) + 0(1, −1, 2) = 0, hence (1, 2, 1) = (−1)(1, 0, 1) + 2(1, 1, 1) + 0(1, −1, 2), so the sequence T : (1, 0, 1), (1, 1, 1), (1, −1, 2) also spans R3 . Notice that there is exactly one solution of the above system of equations for h = 0 with s = 0, so T is linearly independent and therefore it is a basis for R3 . ¤ The approach of the last example leads to a general method. We state it over any field F since the method works in general. For this we need to recall the idea of a reduced echelon matrix and the way this is used to determine the general solution of a system of linear equations. Theorem 2.20. Suppose that a1 , . . . , an is a sequence of vectors in F m . Let A = [a1 · · · an ] be the m × n matrix with aj as its j-th column and let A0 be its reduced echelon matrix. (a) a1 , . . . , an is a spanning sequence of F m if and only if the number of non-zero rows of A0 is m. (b) If the j-th column of A0 does not contain a pivot, then the corresponding vector aj is expressible as a linear combination of those associated with columns which do contain a pivot. Such a linear combination can found by setting the free variable for the j-th column equal to 1 and all the other free variables equal to 0. (c) The vectors ai for those i where the i-th column of A0 contains a pivot, form a basis for Span(a1 , . . . , an ). Here is another kind of situation. Example 2.21. Consider the vector space R3 over R. Find a basis for the plane W with equation x − 2y + 3z = 0. Solution. This h i time we need to start with a system of one equation and its associated matrix 1 −2 3 which is reduced echelon, hence we have the general solution x = 2s − 3t,

y = s,

z=t

(s, t ∈ R).

So every vector (x, y, z) in W is a linear combination (x, y, z) = s(2, 1, 0) + t(−3, 0, 1) of the sequence (2, 1, 0), (−3, 0, 1) obtained by setting s = 1, t = 0 and s = 0, t = 1 respectively. In fact this expression is unique since these two vectors are linearly independent. ¤ Example 2.22. Consider the vector space D(R) over R discussed in Example 1.11. Let W ⊆ D(R) be the set of all solutions of the differential equation dy + 3y = 0. dx Show that W is a subspace of D(R) and find a basis for it.

18

2. SPANNING SEQUENCES, LINEAR INDEPENDENCE AND BASES

Solution. It is easy to see that W is a subspace. Notice that if y1 , y2 are two solutions then d(y1 + y2 ) dy1 dy2 + 3(y1 + y2 ) = + + 3y1 + 3y2 dx dx ¶ µ µdx ¶ dy1 dy2 = + 3y1 + + 3y2 = 0, dx dx hence y1 + y2 is a solution. Also, if y is a solution and s ∈ R, then d(sy) dy + 3(sy) = s + s(3y) dx dx µ ¶ dy =s + 3y = 0, dx so sy is a solution. Clearly the constant function 0 is a solution. The general solution is y = ae−3x

(a ∈ R),

so the function e−3x spans W . Furthermore, if for some t ∈ R, se−3x = 0, where the right hand side means the constant function, this has to hold for every x ∈ R. Taking x = 0 we find that t = te0 = 0, so t = 0. Hence the sequence e−3x is a basis of W .

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A good exercise is to try to generalise this, for example by considering the solutions of the differential equation d2 y − 6y = 0. dx2 There is a brief review of the solution of ordinary differential equations in Appendix A. Here is another example. Example 2.23. Let W ⊆ D(R) be the set of all solutions of the differential equation d2 y + 4y = 0. dx2 Show that W is a subspace of D(R) and find a basis for it. Solution. Checking W is a subspace is similar to Example 2.22 The general solution has the form y = a cos 2x + b sin 2x

(a, b ∈ R),

so the functions cos 2x, b sin 2x span W . Suppose that for some s, t ∈ R, s cos 2x + t sin 2x = 0, where the right hand side means the constant function. This has to hold whatever the value of x ∈ R, so try taking values x = 0, π/4. These give the equations s cos 0 + t sin 0 = 0,

s cos π/2 + t sin π/2 = 0,

i.e., s = 0, t = 0. So cos 2x, sin 2x are linearly independent and hence cos 2x, sin 2x is a basis. Notice that this is a finite dimensional vector space. ¤

2.2. LINEAR INDEPENDENCE AND BASES

19

Example 2.24. Let R∞ denote the set consisting of all sequences (an ) of real numbers which satisfy an = 0 for all large enough n, i.e., (an ) = (a1 , a2 , . . . , ak , 0, 0, 0, . . .). This can be viewed as a vector space over R with addition and scalar multiplication given by (an ) + (bn ) = (an + bn ), t · (cn ) = (tcn ). Then the sequences er = (0, . . . , 0, 1, 0, . . .) with a single 1 in the r-th place, form a basis for R∞ . Solution. First note that (xn ) = (x1 , x2 , . . . , xk , 0, 0, 0, . . .) can be expressed as (xn ) = x1 e1 + · · · + xk ek , so the ei span R∞ . If t1 e1 + · · · + tk ek = (0), then (tn ) = (t1 , . . . , tk , 0, 0, 0, . . .) = (0, . . . , 0, . . .), and so tk = 0 for all k. Thus the ei are also linearly independent.

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Here is another important method for constructing bases. Theorem 2.25. Let V be a finite dimensional vector space and let S be a linearly independent sequence in V . Then there is a basis S 0 of V in which S is a subsequence. In particular, S and S 0 are both finite. Proof. We will indicate a proof for the case where S is a finite sequence. Choose a finite spanning sequence, say T : w1 , . . . , wk . By Proposition 2.14 we might as well assume that T is linearly independent. If S does not span V then certainly S, T (the sequence obtained by joining S and T together) spans (since T does). Now choose the largest value of ` = 1, . . . , k for which the sequence S` = S, w1 , . . . , w` is linearly independent. Then S, T is linearly dependent and if ` < k, we can write wk as a linear combination of S, w1 , . . . , wk−1 , so this is still a spanning set for V . Similarly we can keep discarding the vectors wk−j for j = 0, . . . , (k − ` − 1) and at each stage we have a spanning sequence S, w1 , . . . , wk−j−1 . Eventually we are left with just S` which is still a spanning set for V as well as being linearly independent, and this is the required basis. ¤ Theorem 2.26. Let V be a finite dimensional vector space and suppose that it has a basis with n elements. Then any other basis is also finite and has n elements. This result allows us to make the following important definition.

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2. SPANNING SEQUENCES, LINEAR INDEPENDENCE AND BASES

Definition 2.27. Let V be a finite dimensional vector space. Then the number of elements in any basis of V is finite and is called the dimension of the vector space V and is denoted dimF V or just dim V if the field is clear. Example 2.28. Let F be any field. Then the vector space F n over F has dimension n, i.e., dimF F n = n. Proof. For each k = 1, 2, . . . , n let ek = (0, . . . , 0, 1, 0, . . . , 0) (with the 1 in the k-th place). Then the sequence e1 , . . . , en spans F n since for any vector (x1 , . . . , xn ) ∈ F n , (x1 , . . . , xn ) = x1 e1 + · · · + xn en . Also, if the right hand side is 0, then (x1 , . . . , xn ) = (0, . . . , 0) and so x1 = · · · = xn = 0, showing that this sequence is linearly independent. Therefore, e1 , . . . , en is a basis for F n and dimF F n = n. The basis e1 , . . . , en of F n is often called the standard basis of F n . ¤ Proof of Theorem 2.26. Let B : b1 , . . . , bn be a basis and suppose that S : w1 , . . . is another basis for V . We will assume that S is finite, say S : w1 , . . . , wm , the more general case is similar. The sequence w1 , b1 , . . . , bn must be linearly dependent since B spans V . As B is linearly independent, there is a number k1 = 1, . . . , n for which bk1 = s0 w1 + s1 b1 + · · · + sk1 −1 bk1 −1 for some scalars s0 , s1 . . . , sk1 −1 . By reordering B, we can assume that k1 = n, so bn = s0 w1 + s1 b1 + · · · + sn−1 bn−1 , where s0 6= 0 since B is linearly independent. Thus the sequence B1 : w1 , b1 , . . . , bn−1 spans V . If it were linearly dependent then there would be an equation of the form t0 w1 + t1 b1 + · · · + tn−1 bn−1 = 0 in which not all the ti are zero; in fact t0 6= 0 since B is linearly independent. From this we obtain a non-trivial linear combination of the form t01 b1 + · · · + t0n bn = 0, which is impossible. Thus B1 is a basis. Now we repeat this argument with w2 , w3 , and so on, at each stage forming (possibly after reordering and renumbering the bi ’s) a sequence Bk : w1 , . . . , wk , b1 , . . . , bn−k which is linearly independent and a spanning sequence for V , i.e., it is a basis. If the length of S were finite (say equal to `) and less than the length of B (i.e., ` < n), then the sequence B` : w1 , . . . , w` , b1 , . . . , bn−` would be a basis. But then B` = S and so we could not have any bi at the right hand end since S is a spanning sequence, so B` would be linearly dependent. Thus we must be able to find at least n terms of S, and so eventually we obtain the sequence Bn : w1 , . . . , wn which is a basis. This shows that S must have finite length which equals the length of B, i.e., any two bases have finite and equal length . ¤ The next result is obtained from our earlier results.

2.2. LINEAR INDEPENDENCE AND BASES

21

Proposition 2.29. Let V be a finite dimensional vector space of dimension n. Suppose that R : v1 , . . . , vr is a linearly independent sequence in V and that S : w1 , . . . , ws is a spanning sequence in V . Then the following inequalities are satisfied: r 6 n 6 s. Furthermore, R is a basis if and only if r = n, while S is a basis if and only if s = n. Theorem 2.30. Let V be a finite dimensional vector space and suppose that W ⊆ V is a vector subspace. Then W is finite dimensional and dim W 6 dim V, with equality precisely when W = V . Proof. Choose a basis S for W . Then by Theorem 2.25, this extends to a basis S 0 for and these are both finite sets with dim W and dim V elements respectively. As S is subsequence of S 0 , we have dim W 6 dim V, with equality exactly when S = S 0 , which can only happen when W = V .

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A similar inequality holds when the vector space V is infinite dimensional. Example 2.31. Recall from Example 2.23 the infinite dimensional real vector space D(R) and the subspace W ⊆ D(R). Then dim W = 2 since the vectors cos 2x, sin 2x were shown to form a basis. The next result gives are some other characterisations of bases in a vector space V . Let X and Y be sets; then X is a proper subset of Y (indicated by X ( Y ) if X ⊆ Y and X = 6 Y. Proposition 2.32. Let S be a sequence in V . • Let S be a spanning sequence of V . Then S is a basis if and only if it is a minimal spanning sequence, i.e., no proper subsequence of S is a spanning sequence. • Let S be a linearly independent sequence. Then is a basis if and only if it is a maximal linearly independent set, i.e., no sequence in V containing S as a proper subsequence is linearly independent. Proof. Again this follows from things we already know.

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Example 2.33. Recall Example 1.8: given a field of scalars F , for each pair m, n > 1 we have the vector space over F consisting of all m × n matrices with entries in F , namely Mm×n (F ). For r = 1, . . . , m and s = 1, . . . , n let E rs be the m × n matrix all of whose entries is 0 except for the (r, s) entry which is 1. Now any matrix [aij ] ∈ Mm×n (F ) can be expressed in the form n m X X

ars E rs ,

r=1 s=1

E rs

so the matrices span Mm×n (F ). But the only way that the expression on the right hand side can be the zero matrix is if [aij ] = Om×n , i.e., if aij = 0 for every entry aij . Thus the E rs form a basis and so dimF Mm×n (F ) = mn.

22

2. SPANNING SEQUENCES, LINEAR INDEPENDENCE AND BASES

In particular, dimF Mn (F ) = n2 and dimF Mn×1 (F ) = n = dimF M1×n (F ). The next result provides another useful way to find a basis for a subspace of F n . Proposition 2.34. Suppose that S : w1 , . . . , wm is a sequence of vectors in F n and that W = Span(S). Arrange the coordinates of each wr as the r-th row of an m × n matrix A. Let A0 be the reduced echelon form of A, and use the coordinates of the r-th row of A0 to form the vector wr0 . If A0 has k non-zero rows, then S 0 : w10 , . . . , wk0 is a basis for W . Proof. To see this, notice that we can recover A from A0 by reversing the elementary row operations required to obtain A0 from A, hence the rows of A0 are all in Span(A) and vice versa. This shows that Span(A0 ) = Span(A) = W . Furthermore, it is easy to see that the non-zero rows of A0 are linearly independent. ¤ Example 2.35. Consider the real vector space R4 and let W be the subspace spanned by the vectors (1, 2, 3, 0), (2, 0, 1, −2), (0, 1, 0, 4), (2, 1, 1, 2). Find a basis for W . Solution. Let

1 2 A= 0 2

2 0 1 1

3 0 1 −2 . 0 4 1 2

Then 1 2 3 0 1 2 3 0 0 −4 −5 −2 0 1 0 4 A ∼ ∼ R2 →R2 −2R1 0 1 0 4 R2 ↔R3 0 −4 −5 −2 R4 →R4 −2R1 0 −3 −5 2 0 −3 −5 2 1 0 3 −8 1 0 0 1 0 4 0 1 ∼ ∼ R1 →R1 −2R2 0 0 −5 14 R4 →R4 −R3 0 0 R3 →R3 +4R2 R4 →R4 +3R2 0 0 −5 14 0 0 1 1 0 3 −8 0 1 0 4 0 ∼ ∼ R3 →(−1/5)R4 0 0 1 −14/5 R1 →R1 −3R3 0 0 0 0 0 0

3 0 −5 0 0 1 0 0

−8 4 14 0

0 2/5 0 0 . 1 −14/5 0 0

Since the last matrix is reduced echelon, the sequence (1, 0, 0, 2/5), (0, 1, 0, 0), (0, 0, 1, −14/5) is a basis of W . An alternative basis is (5, 0, 0, 2), (0, 1, 0, 0), (0, 0, 5, −14). ¤ 2.3. Coordinates with respect to bases Let V be a finite dimensional vector space over a field F and let n = dim V . Suppose that v1 , . . . , vn form a basis of V . Then by Proposition 2.17, every vector x ∈ V has a unique expression of the form x = x1 v1 + · · · + xn vn ,

2.4. SUMS OF SUBSPACES

23

where x1 , . . . , xn ∈ F are the coordinates of x with respect to this basis. Of given the course, x1 . vectors v1 , . . . , vn , each vector x is determined by the coordinate vector .. with respect xn to this basis. If v10 , . . . , vn0 is another basis and if the coordinates of x with respect it are x01 , . . . , x0n ∈ F , then x = x01 v10 + · · · + x0n vn0 , If we write vk0 = a1k v1 + · · · + ank vn for ars ∈ F , we obtain x = x01

n X

ar1 vr + · · · + x0n

n X

r=1 n X

r=1 n X

s=1

s=1

a1s x0s )v1 + · · · + (

=(

arn vr ,

ans x0s )vn .

Thus we obtain the equation n n X X x1 v1 + · · · + xn vn = ( a1s x0s )v1 + · · · + ( ans x0s )vn . s=1

s=1

and so for each r = 1, . . . , n we have (2.1a)

xr =

n X

ars x0s .

s=1

This is more efficiently expressed in the matrix form x1 a11 · · · a1n x01 . . .. .. .. .. = .. (2.1b) . . . . xn an1 · · · ann x0n Clearly we could also express the coordinates x0r in terms of the xs . Thus implies that the matrix [aij ] is invertible and in fact we have −1 x01 a11 · · · a1n x1 . . . . . .. = .. .. .. .. . (2.1c) x0n an1 · · · ann xn 2.4. Sums of subspaces Let V be a vector space over a field F . In this section we will study another way to combine subspaces of a vector space by taking a kind of sum. Definition 2.36. Let V 0 , V 00 be two vector subspaces of V . Then their sum is the subset V 0 + V 00 = {v0 + v00 : v0 ∈ V 0 , v00 ∈ V 00 }. More generally, for subspaces V1 , . . . , Vk of V , V1 + · · · + Vk = {v1 + · · · + vk : for j = 1, . . . , k, vj ∈ Vj }. Notice that V1 + · · · + Vk contains each Vj as a subset.

24

2. SPANNING SEQUENCES, LINEAR INDEPENDENCE AND BASES

Proposition 2.37. For subspaces V1 , . . . , Vk of V , the sum V1 + · · · + Vk is also a subspace of V which is the smallest subspace of V containing each of the Vj as a subset. Proof. Suppose that x, y ∈ V1 +· · ·+Vk . Then for j = 1, . . . , k there are vectors xj , yj ∈ Vj for which x = x1 + · · · + xk , y = y1 + · · · + yk . We have x + y = (x1 + · · · + xk ) + (y1 + · · · + yk ) = (x1 + y1 ) + · · · + (xk + yk ). Since each Vj is a subspace it is closed under addition, hence xj + yj ∈ Vj , therefore x + y ∈ V1 + · · · + Vk . Also, for s ∈ F , sx = s(x1 + · · · + xk ) = sx1 + · · · + sxk , and each Vj is closed under multiplication by scalars, hence sx ∈ V1 + · · · + Vk . Therefore V1 + · · · + Vk is a subspace of V . Any subspace W ⊆ V which contains each of the Vj as a subset has to also contain every element of the form v1 + · · · + vk (vj ∈ Vj ) and hence V1 + · · · + Vk ⊆ W . Therefore V1 + · · · + Vk is the smallest such subspace.

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Definition 2.38. Let V 0 , V 00 be two vector subspaces of V . Then V 0 + V 00 is a direct sum if V 0 ∩ V 00 = {0}. More generally, for subspaces V1 , . . . , Vk of V , V1 +· · ·+Vk is a direct sum if for each r = 1, . . . , k, Vr ∩ (V1 + · · · + Vr−1 + Vr+1 + · · · + Vk ) = {0}. Such direct sums are sometimes denoted V 0 ⊕ V 00 and V1 ⊕ · · · ⊕ Vk . Theorem 2.39. Let V 0 , V 00 be subspaces of V for which V 0 + V 00 is a direct sum V 0 ⊕ V 00 . Then every vector v ∈ V 0 ⊕ V 00 has a unique expression v = v0 + v00

(v0 ∈ V 0 , v00 ∈ V 00 ).

More generally, if V1 , . . . , Vk are subspaces of V for which V1 +· · ·+Vk is a direct sum V1 ⊕· · ·⊕Vk , then every vector v ∈ V1 ⊕ · · · ⊕ Vk has a unique expression v = v1 + · · · + vk

(vj ∈ Vj , j = 1, . . . , k).

Proof. We will give the proof for the first case, the general case is more involved but similar. Let v ∈ V 0 ⊕ V 00 . Then certainly there are vectors v0 ∈ V 0 , v00 ∈ V 00 for which v = v0 + v00 . Suppose there is a second pair of vectors w0 ∈ V 0 , w00 ∈ V 00 for which v = w0 + w00 . Then (w0 − v0 ) + (w00 − v00 ) = 0, and w0 − v0 ∈ V 0 , w00 − v00 ∈ V 00 . But this means that w0 − v0 = v00 − w00 where the left hand side is in V 0 and the right hand side is in V 00 . But the equality here means that each of these terms is in V 0 ∩ V 00 = {0}, hence w0 = v0 and w00 = v00 . Thus such an expression for v must be unique. ¤

2.4. SUMS OF SUBSPACES

25

Proposition 2.40. Let V 0 , V 00 be subspaces of V . If these are finite dimensional, then so is V 0 + V 00 and dim(V 0 + V 00 ) = dim V 0 + dim V 00 − dim(V 0 ∩ V 00 ). Proof. Notice that V 0 ∩ V 00 is a subspace of each of V 0 and V 00 and of V 0 + V 00 . Choose a basis for V 0 ∩ V 00 , say v1 , . . . , va . This extends to bases for V 0 and V 00 , say v1 , . . . , va , v10 , . . . , vb0 and v1 , . . . , va , v100 , . . . , vc00 . Every element of V 0 + V 00 is a linear combination of the vectors v1 , . . . , va , v10 , . . . , vb0 , v100 , . . . , vc00 , we will show that they are linearly independent. Suppose that for scalars s1 , . . . , sa , s01 , . . . , s0b , s001 , . . . , s00c , the equation s1 v1 + · · · + sa va + s01 v10 + · · · + s0b vb0 + s001 v100 + · · · + s00c vc00 = 0 holds. Then we have s001 v100 + · · · + s00c vc00 = −(s1 v1 + · · · + sa va + s01 v10 + · · · + s0b vb0 ) ∈ V 0 , but the left hand side is also in V 00 , hence both sides are in V 0 ∩ V 00 . By our choice of basis for V 0 ∩ V 00 , we find that s001 v100 + · · · + s00c vc00 is a linear combination of the vectors v1 , . . . , va and since v1 , . . . , va , v100 , . . . , vc00 is linearly independent the only way that can happen is if all the coefficients are zero, hence s001 = · · · = s00c = 0. A similar argument also shows that s01 = · · · = s0b = 0. We are left with the equation s1 v1 + · · · + sa va = 0 which also implies that s1 = · · · = sa = 0 since the vi form a basis for V 0 ∩ V 00 . Now we have dim(V 0 ∩ V 00 ) = a,

dim V 0 = a + b,

dim V 00 = a + c,

and dim(V 0 + V 00 ) = a + b + c = dim V 0 + dim V 00 − dim(V 0 ∩ V 00 ).

¤

Definition 2.41. Let V be a vector space and W ⊆ V be a subspace. A subspace W 0 ⊆ V for which W + W 0 = V and W ∩ W 0 = {0} is called a linear complement of W in V . For such a linear complement we have V = W ⊕ W 0 . We remark that although linear complements always exist they are by no means unique! The next result summarises the situation. Proposition 2.42. Let W ⊆ V be subspace. (a) There is a linear complement W 0 ⊆ V of W in V . (b) If V is finite dimensional, then any subspace W 00 ⊆ V satisfying the conditions • W ∩ W 00 = {0}, • dim W 00 = dim V − dim W is a linear complement of W in V . (c) Any subspace W 000 ⊆ V satisfying the conditions • W + W 000 = V , • dim W 000 = dim V − dim W is a linear complement of W in V .

26

2. SPANNING SEQUENCES, LINEAR INDEPENDENCE AND BASES

Proof. (a) We show how to find a linear complement in the case where V is finite dimensional with n = dim V . By Theorem 2.30, we know that dim W 6 dim V . So let W have a basis w1 , . . . , wr , where r = dim W . Now by Theorem 2.25 there is an extension of this to a basis of V , say 0 w1 , . . . , wr , w10 , . . . , wn−r . 0 Now let W 0 = Span(w10 , . . . , wn−r ). Then by Lemma 2.6, W 0 is a subspace of V and it has 0 w10 , . . . , wn−r as a basis, so dim W 0 = n − r. Now it is clear that W + W 0 = V . Suppose that v ∈ W ∩ W 0 . Then there are expressions

v = s1 w1 + · · · + sr wr , 0 v = s01 w10 + · · · + s0n−r wn−r ,

hence there is an equation 0 s1 w1 + · · · + sr wr + (−s01 )w10 + · · · + (−s0n−r )wn−r = 0.

But the vectors wi , wj0 form a basis so are linearly independent. Therefore s1 = · · · = sr = s01 = · · · = s0n−r = 0. This means that v = 0, hence W ∩ W 0 = {0} and the sum W + W 0 is direct. So we have shown that W 0 is a linear complement of W in V . (b) and (c) can be verified in similar fashion. ¤ Example 2.43. Consider the vector space R2 over R. If V 0 ⊆ R2 and V 00 ⊆ R2 are two distinct lines through the origin, show that V 0 + V 00 is a direct sum and that V 0 ⊕ V 00 = R2 . Solution. As these lines are distinct and V 0 ∩ V 00 is subspace of each of V 0 and V 00 , we have dim(V 0 ∩ V 00 ) < dim V 0 = dim V 00 = 1, hence dim(V 0 ∩ V 00 ) = 0. This means that V 0 ∩ V 00 = {0} which is geometrically obvious. Now notice that dim(V 0 + V 00 ) = dim V 0 + dim V 00 − 0 = 1 + 1 = 2, hence dim(V 0 + V 00 ) = dim R2 and so V 0 + V 00 = R2 .

¤

Example 2.44. Consider the vector space R3 over R. If V 0 ⊆ R3 is a line through the origin and V 00 ⊆ R2 is a plane through the origin which does not contain V 0 , show that V 0 + V 00 is a direct sum and that V 0 ⊕ V 00 = R3 . Solution. V 0 ∩ V 00 is subspace of each of V 0 and V 00 , so we have dim(V 0 ∩ V 00 ) 6 dim V 0 = 1 < 2 = dim V 00 . In fact the inequality must be strict since otherwise V 0 ⊆ V 00 , hence dim(V 0 ∩ V 00 ) = 0. This means that V 0 ∩ V 00 = {0} which is also geometrically obvious. Now notice that dim(V 0 + V 00 ) = dim V 0 + dim V 00 − 0 = 1 + 2 = 3, therefore dim(V 0 + V 00 ) = dim R3 and so V 0 + V 00 = R3 .

¤

2.4. SUMS OF SUBSPACES

27

Example 2.45. Consider the vector space R4 over R and the subspaces V 0 = {(s + t, s, −s, −t) : s, t ∈ R},

V 00 = {(u, 2v, −v, −u) : u, v ∈ R}.

Show that dim(V 0 + V 00 ) = 3. Solution. V 0 is spanned by the vectors (1, 1, −1, 0), (1, 0, 0, −1), which are linearly independent and hence form a basis. V 00 is spanned by (1, 0, 0, −1), (0, 2, −1, 0) which are linearly independent and hence form a basis. Thus we find that dim V 0 = 2 = dim V 00 . Now suppose that (w, x, y, z) ∈ V 0 ∩ V 00 . Then there are s, t, u, v ∈ R for which (w, x, y, z) = (s + t, s, −s, −t) = (u, 2v, −v, −u). By comparing coefficients, this leads to the system of four linear equations s + t − u = 0 s − 2v = 0 (2.2) − s + v = 0 − t + u = 0 which has associated augmented matrix 1 1 −1 0 0 1 1 0 0 −2 0 0 ∼ −1 0 0 1 0 R2 →R2 −R1 0 R3 →R3 +R1 0 −1 1 0 0 0 1 0 0 0 1 −1 ∼ R1 →R1 −R2 0 0 0 R3 →R3 +R2 R4 →R4 +R2 0 0 0

1 −1 0 0 1 1 −1 0 0 0 −1 1 −2 0 1 −1 1 0 ∼ 1 −1 1 0 R2 ↔R3 0 −1 1 −2 0 −1 1 0 0 0 −1 1 0 0 −1 0 1 0 0 −1 0 1 0 1 0 0 1 −1 ∼ R ↔R −1 0 3 4 0 0 0 1 0 1 0 0 0 0 −1 0 1 0 0 0 0 0 1 −1 0 0 ∼ , R1 →R1 +R3 0 0 0 1 0 R2 →R2 −R3 R4 →R4 +R3 0 0 0 0 0

where the last matrix is reduced echelon. The general solution of the system (2.2) is s = 0,

t = u,

u ∈ R,

v = 0.

hence we have V 0 ∩ V 00 = {(u, 0, 0, −u) : u ∈ R}, and this has basis (1, 0, 0, −1), hence dim(V 0 ∩ V 00 ) = 1. This means that these two planes intersect in a line in R4 . Now we see that dim(V 0 + V 00 ) = dim V 0 + dim V 00 − dim(V 0 ∩ V 00 ) = 2 + 2 − 1 = 3.

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CHAPTER 3

Linear transformations 3.1. Functions Before introducing linear transformations, we will review some basic ideas about functions, including properties of composition and inverse functions which will be required later. Let X, Y and Z be sets. Definition 3.1. A function (or mapping) f : X −→ Y from X to Y involves a rule which associates to each x ∈ X a unique element f (x) ∈ Y . X is called the domain of f , denoted dom f , while Y is called the codomain of f , denoted codom f . Sometimes the rule is indicated by writing x 7→ f (x). Given functions f : X −→ Y and g : Y −→ Z, we can form their composition g ◦ f : X −→ Z which has the rule g ◦ f (x) = g(f (x)). Note the order of composition here! g◦f

X

f

/Y

g

Ã /Z

We often write gf for g ◦ f when no confusion will arise. If X, Y , Z are different it may not be possible to define f ◦ g since dom f = X may not be the same as codom g = Z. Example 3.2. For any set X, the identity function IdX −→ X has rule x 7→ IdX (x) = x. Example 3.3. Let X = Y = R. Then the following rules define functions R −→ R: x 7→ x + 1,

x 7→ x2 ,

x 7→

x2

1 , +1

3

x 7→ sin x, x 7→ e5x .

Example 3.4. Let X = Y = R+ , the set of all positive real numbers. Then the following rules define two functions R+ −→ R: √ √ x 7→ x, x → 7 − x. Example 3.5. Let X and Y be sets and suppose that w ∈ Y is an element. The constant function cw : X −→ Y taking value w has the rule x 7→ cw (x) = w. For example, if X = Y = R then c0 is the function which returns the value c0 (x) = 0 for every real number x ∈ R. Definition 3.6. A function f : X −→ Y is • injective (or an injection or one-to-one) if for x1 , x2 ∈ X, 29

30

3. LINEAR TRANSFORMATIONS

f (x1 ) = f (x2 ) implies x1 = x2 , • surjective (or a surjection or onto) if for every y ∈ Y there is an x ∈ X for which y = f (x), • bijective (or injective or a one-to-one correspondence) if it is both injective and surjective. We will use the following basic fact. Proposition 3.7. The function f : X −→ Y is a bijection if and only if there is an inverse function Y −→ X which is the unique function h : Y −→ X satisfying h ◦ f = IdX ,

f ◦ h = IdY .

If such an inverse exists it is usual to denote it by f −1 : Y −→ X, and then we have f −1 ◦ f = IdX ,

f ◦ f −1 = IdY .

Later we will see examples of all these notions in the context of vector spaces. 3.2. Linear transformations In this section, let F be a field of scalars. Definition 3.8. Let V and W be two vector spaces over F and let f : V −→ W be a function. Then f is called a linear transformation or linear mapping if it satisfies the two conditions • for all vectors v1 , v2 ∈ V , f (v1 + v2 ) = f (v1 ) + f (v2 ), • for all vectors v ∈ V and scalars t ∈ F , f (tv) = tf (v). These two conditions together are equivalent to the single condition • for all vectors v1 , v2 ∈ V , and scalars t1 , t2 ∈ F , f (t1 v1 + t2 v1 ) = t1 f (v1 ) + t2 f (v1 ). Remark 3.9. For a linear transformation f : V −→ W , we always have f (0) = 0, where the left hand 0 means the zero vector in V and the right hand 0 means the zero vector in W . The next example introduces an important kind of linear transformation associated with a matrix. Example 3.10. Let A ∈ Mm×n be an m × n matrix. Define fA : F n −→ F m by the rule fA (x) = Ax. Then for s, t ∈ F and u, v ∈ F n , fA (su + tv) = A(su + tv) = sAu + tAv = sfA (u) + tfA (v). So fA is a linear transformation. Observe that the standard basis vectors e1 , . . . , en (viewed as column vectors) satisfy fA (ej ) = j-th column of A,

3.2. LINEAR TRANSFORMATIONS

so

h A = fA (e1 ) · · ·

i h fA (en ) = Ae1 · · ·

31

i Aen .

Now it easily follows that fA (e1 ), . . . , fA (en ) spans Im fA since every element of the latter is a linear combination of this sequence. Theorem 3.11. Let f : V −→ W and g : U −→ V be two linear transformations between vector spaces over F . Then the composition f ◦ g : U −→ W is also a linear transformation. Proof. For s1 , s2 ∈ F and u1 , u2 ∈ V , we have to show that f ◦ g(s1 u1 + s2 u2 ) = s1 f ◦ g(u1 ) + s2 f ◦ g(u2 ). Using the fact that both f and g are linear transformations we have f ◦ g(s1 v1 + s2 v2 ) = f (s1 g(u1 ) + s2 g(u2 )) = s1 f (g(u1 )) + s2 f (g(u2 )) = s1 f ◦ g(u1 ) + s2 f ◦ g(u2 ), as required.

¤

Theorem 3.12. Let V and W be vector spaces over F . Suppose that f, g : V −→ W are linear transformations and s, t ∈ F . Then the function sf + tg : V −→ W ;

(sf + tg)(v) = sf (v) + tg(v)

is a linear transformation. Definition 3.13. Let f : V −→ W be a linear transformation. • The kernel (or nullspace) of f is the following subset of V : Ker f = {v ∈ V : f (v) = 0} ⊆ V. • The image (or range) of f is the following subset of W : Im f = {w ∈ W : there is an v ∈ V s.t. w = f (v)} ⊆ W. Theorem 3.14. Let f : V −→ W be a linear transformation. Then (a) Ker f is a subspace of V , (b) Im f is a subspace of W . Proof. (a) Let s1 , s2 ∈ F and v1 , v2 ∈ Ker f . We must show that s1 v1 + s2 v2 ∈ Ker f , i.e., that f (s1 v1 + s2 v2 ) = 0. Since f is a linear transformation and f (v1 ) = 0 = f (v2 ), we have f (s1 v1 + s2 v2 ) = s1 f (v1 ) + s2 f (v2 ) = s1 0 + s2 0 = 0, hence f (s1 v1 + s2 v2 ) ∈ Ker f . Therefore Ker f is a subspace of V . Now suppose that t1 , t2 ∈ F and w1 , w2 ∈ Im f . We must show that t1 w1 + t2 w2 ∈ Im f , i.e., that t1 w1 + t2 w2 = f (v) for some v ∈ V . Since w1 , w2 ∈ Im f , there are vectors v1 , v2 ∈ V for which f (v1 ) = w1 ,

f (v2 ) = w2 .

32

3. LINEAR TRANSFORMATIONS

As f is a linear transformation, t1 w1 + t2 w2 = t1 f (v1 ) + t2 f (v2 ) = f (t1 v1 + t2 v2 ), so we can take v = t1 v1 + t2 v2 to get t1 w1 + t2 w2 = f (v). Hence Im f is a subspace of W .

¤

For linear transformations the following result holds, where we make use of notions introduced in Section 3.1. Theorem 3.15. Let f : V −→ W be a linear transformation. (a) f is injective if and only if Ker f = {0}. (b) f is surjective if and only if Im f = W . (c) f is bijective if and only if Ker f = {0} and Im f = W . Proof. (a) Notice that f (0) = 0. Hence if f is injective then f (v) = 0 implies v = 0, i.e., Ker f = {0}. We need to show the converse, i.e., if Ker f = {0} then f is injective. So suppose that Ker f = {0}, v1 , v2 ∈ V , and f (v1 ) = f (v2 ). Since f is a linear transformation, f (v1 − v2 ) = f (v1 + (−1)v2 ) = f (v1 ) + (−1)f (v2 ) = f (v1 ) − f (v2 ) = 0, hence f (v1 − v2 ) = 0 and therefore v1 − v2 = 0, i.e., v1 = v2 . This shows that f is injective. (b) This is immediate from the definition of surjectivity. (c) This comes by combining (a) and (b). ¤ Theorem 3.16. Let f : V −→ W be a bijective linear transformation. Then the inverse function f −1 : W −→ V is a linear transformation. Proof. Let t1 , t2 ∈ F and w1 , w2 ∈ W . We must show that f −1 (t1 w1 + t2 w2 ) = t1 f −1 (w1 ) + t2 f −1 (w2 ).

3.2. LINEAR TRANSFORMATIONS

33

The following calculation does this: ¡ ¢ f f −1 (t1 w1 + t2 w2 ) − t1 f −1 (w1 ) + t2 f −1 (w2 ) =f ◦ f −1 (t1 w1 + t2 w2 ) − f (t1 f −1 (w1 ) + t2 f −1 (w2 )) = IdW (t1 w1 + t2 w2 ) − f (t1 f −1 (w1 ) + t2 f −1 (w2 )) =(t1 w1 + t2 w2 ) − (t1 f ◦ f −1 (w1 ) + t2 f ◦ f −1 (w2 )) =(t1 w1 + t2 w2 ) − (t1 IdW (w1 ) + t2 IdW (w2 )) =(t1 w1 + t2 w2 ) − (t1 w1 + t2 w2 ) = 0. Since f is injective this means that f −1 (t1 w1 + t2 w2 ) − (t1 f −1 (w1 ) + t2 f −1 (w2 )) = 0, giving the desired formula.

¤

Definition 3.17. A linear transformation f : V −→ W which is a bijective function is called an isomorphism from V to W , and these are said to be isomorphic vector spaces. Notice that for an isomorphism f : V −→ W , the inverse f −1 : W −→ V is also an isomorphism. Definition 3.18. Let f : V −→ W be a linear transformation. Then the rank of f is rank f = dim Im f, and the nullity of f is null f = dim Ker f, whenever these are finite. Theorem 3.19 (The rank-nullity theorem). Let f : V −→ W be a linear transformation with V finite dimensional. Then dim V = rank f + null f. Proof. Start by choosing a basis for Ker f , say u1 , . . . , uk . Notice that k = null f . By Theorem 2.25, this can be extended to a basis of V , say u1 , . . . , uk , uk+1 , . . . , un , where n = dim V . Now for s1 , . . . , sn ∈ F , f (s1 u1 + · · · + sn un ) = s1 f (u1 ) + · · · + sn f (un ) = s1 0 + · · · + sk 0 + sk+1 f (uk+1 ) + · · · + sn f (un ) = sk+1 f (uk+1 ) + · · · + sn f (un ). This shows that Im f is spanned by the vectors f (uk+1 ), . . . , f (un ). In fact, this last expression is only zero when s1 u1 + · · · + sn un lies in Ker f which can only happen if sk+1 = · · · = sn = 0. So the vectors f (uk+1 ), . . . , f (un ) are linearly independent and therefore form a basis for Im f . Thus rank f = n − k and the result follows. ¤ We can also reformulate the results of Theorem 3.15.

34

3. LINEAR TRANSFORMATIONS

Theorem 3.20. Let f : V −→ W be a linear transformation. (a) f is injective if and only if null f = 0. (b) f is surjective if and only if rank f = dim W . (c) f is bijective if and only if f = 0 and rank f = dim W . Corollary 3.21. Let f : V −→ W be a linear transformation with V and W finite dimensional vector spaces. The following hold: (a) if f is an injection, then dimF V 6 dimF W ; (b) if f is a surjection, then dimF V > dimF W ; (c) if f is a bijection, then dimF V = dimF W . Proof. This makes use of the rank-nullity Theorem 3.19 and Theorem 3.20. (a) Since Im f is a subspace of W , we have dim Im f 6 dim W , hence if f is injective, dim V = rank f 6 dim W. (b) This time, if f is surjective we have dim V > rank f = dim W. Combining (a) and (b) gives (c).

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Example 3.22. Let D(R) be the vector space of differentiable real functions considered in Example 1.11. (a) For n > 1, define the function Dn : D(R) −→ D(R);

Dn (f ) =

dn f . dxn

Show that this is a linear transformation and that Dn = D ◦ Dn−1 = Dn−1 ◦ D and so Dn is the n-fold composition D ◦ · · · ◦ D. Identify the kernel and image of Dn . (b) More generally, for real numbers a0 , a1 , . . . , an , define the function a0 + a1 D + · · · + an Dn : D(R) −→ D(R); (a0 + a1 D + · · · + an Dn )(f ) = a0 f + a1 D(f ) + · · · + an Dn (f ), and show that this is a linear transformation and identify its kernel and image. Solution. (a) When n = 1, D has the effect of differentiation on functions. For scalars s, t and functions f, g in D(R), d(sf + tg) df dg D(sf + tg) = =s +t = sD(f ) + tD(g). dx dx dx Hence D is a linear transformation. The formulae for Dn are clear and compositions of linear transformations are also linear transformations by Theorem 3.11. To understand the kernel of Dn , notice that f ∈ Ker Dn if and only if Dn (f ) = 0, so we are looking for the general solution of the differential equation dn f =0 dxn which has general solution f (x) = c0 + c1 x + · · · cn−1 xn−1

(c0 , . . . , cn−1 ∈ R).

3.2. LINEAR TRANSFORMATIONS

35

To find the image of Dn , first notice that for any real function g we have µZ x ¶ Z x d g(t) dt = g(x). D g(t) dt = dx 0 0 This shows that Im D = D(R), i.e., D is surjective. More generally, we find that Im Dn = D(R). (b) This uses Theorem 3.12. The point is that a0 + a1 D + · · · + an Dn = a0 IdD(R) +a1 D + · · · an Dn . We have f ∈ Ker(a0 + a1 D + · · · + an Dn ) if and only if df dn f + · · · + an n = 0, dx dx so the kernel is the same as the set of solutions of the latter differential equation. Again we find that this linear transformation is surjective and we leave this as an exercise. ¤ a0 f + a1

Example 3.23. Let F be a field of scalars. For n > 1, consider the vector space Mn (F ). Define the trace function tr : Mn (F ) −→ F by tr([aij ]) =

n X

arr = a11 + · · · + ann .

r=1

Show that tr is linear transformation and find bases for its kernel and image. Solution. For [aij ], [bij ] ∈ Mn (F ) and s, t ∈ F we have tr(s[aij ] + t[bij ]) = tr([saij + tbij ]) =

n X (sarr + tbrr ) r=1

=

n X

sarr +

r=1 n X

=s

n X

tbrr

r=1 n X

arr + t

r=1

brr

r=1

= s tr([aij ]) + t tr([bij ]), hence tr is a linear transformation. Notice that dimF Mn (F ) = n2 and dimF F = 1. Also, for t ∈ F , t 0 ··· 0 . . . . . . . ... = t. tr . . 0 0 ··· 0 This shows that Im tr = F , i.e., tr is surjective and Im tr = F has basis 1. By the rank-nullity Theorem 3.19, null tr = dimF Ker tr = n2 − 1. Now the matrices E rs (r, s = 1, . . . , n − 1, r 6= s) and E rr − E (r+1)(r+1) (r = 1, . . . , n − 1) lie in Ker tr and are linearly independent. There are n(n − 1) + (n − 1) = (n + 1)(n − 1) = n2 − 1 of these, so they form a basis of Ker tr.

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36

3. LINEAR TRANSFORMATIONS

Example 3.24. Let F be a field and let m, n > 1. Consider the transpose function (−)T : Mm×n (F ) −→ Mn×m (F ) defined by [aij ]T = [aji ], i.e., the (i, j)-th entry of [aij ]T is the (j, i)-th entry of [aij ]. Show that (−)T is a linear transformation which is an isomorphism. What is its inverse? Solution. For [aij ], [bij ] ∈ Mm×n (F ), ([aij ] + [bij ])T = [aij + bij ]T = [aji + bji ] = [aji ] + [bji ] = [aij ]T + [bij ]T , and if s ∈ F , (s[aij ])T = [saij ]T = [saji ] = s[aji ] = s[aji ]T . Thus (−)T is a linear transformation. If [aij ] ∈ Ker(−)T then [aji ] = On×m (the n × m zero matrix), hence aij = 0 for i = 1, . . . , m and j = 1, . . . , n and so [aji ] = Om×n . Thus null(−)T = 0 and by the rank-nullity theorem, rank(−)T = dim Mm×n (F ) = mn = dim Mn×m (F ). This shows that Im(−)T = Mn×m (F ). Hence (−)T is an isomorphism. Its inverse is (−)T : Mn×m (F ) −→ Mm×n (F ).

¤

Similar arguments apply to the next example. Example 3.25. Let m, n > 1 and consider the Hermitian conjugate function (−)∗ : Mm×n (C) −→ Mn×m (C) defined by [aij ]∗ = [aji ], i.e., the (i, j)-th entry [aij ]∗ is the complex conjugate of the (j, i)-th entry of [aij ]. If Mm×n (C) and Mn×m (C) are viewed as real vector spaces, then (−)∗ is a linear transformation which is an isomorphism. There are some useful rules for dealing with transpose and Hermitian conjugates of products of matrices. Proposition 3.26 (Reversal rules). (a) If F is any field, the transpose operation satisfies (AB)T = B T AT for all matrices A ∈ M`×m (F ) and B ∈ Mm×n (F ). (b) For all complex matrices P ∈ M`×m (C) and Q ∈ Mm×n (C), (P Q)∗ = Q∗ P ∗ .

3.3. WORKING WITH BASES AND COORDINATES

37

Proof. (a) Writing A = [aij ] and B = [bij ], we obtain "m #T " m # X X (AB)T = ([aij ][bij ])T = air brj = ajr bri r=1

r=1

=

"m X

# bri ajr

r=1

= [bji ][aji ] = [bij ]T [aij ]T = B T AT . (b) is proved in a similar way but with the appearance of complex conjugates in the formulae. ¤ Definition 3.27. For an arbitrary field F , an matrix A ∈ Mn (F ) is called • symmetric if AT = A, • skew-symmetric if AT = −A. A complex matrix B ∈ Mn (C) is called • Hermitian if B ∗ = A, • skew-Hermitian if B ∗ = −A. 3.3. Working with bases and coordinates When performing calculations with a linear transformation it is often useful to use coordinates with respect to bases of the domain and codomain. In fact, this reduces every linear transformation to one associated to a matrix as described in Example 3.10. Lemma 3.28. Let f, g : V −→ W be linear transformations and let S : v1 , . . . , vn be a basis for V . Then f = g if and only if the values of f and g agree on the elements of S. Hence a linear transformation is completely determined by its values on the elements of a basis for its domain. Proof. Every element v ∈ V can be uniquely expressed as a linear combination v = s1 v1 + · · · + sn vn Now f (v) = f (s1 v1 + · · · + sn vn ) = s1 f (v1 ) + · · · + sn f (vn ) and similarly g(v) = g(s1 v1 + · · · + sn vn ) = s1 g(v1 ) + · · · + sn g(vn ). So the functions f and g agree on v if and only if for each i, f (vi ) = g(vi ). We conclude that f and g agree on V if and only if they agree on S. ¤ Here is another useful result involving bases. Lemma 3.29. Let V and W be finite dimensional vector spaces over F , and let S : v1 , . . . , vn be a basis for V . Suppose that f : V −→ W is a linear transformation. Then (a) if f (v1 ), . . . , f (vn ) is linearly independent in W , then f is an injection; (b) if f (v1 ), . . . , f (vn ) spans W , then f is surjective; (b) if f (v1 ), . . . , f (vn ) forms a basis for W , then f is bijective.

38

3. LINEAR TRANSFORMATIONS

Proof. (a) The equation f (s1 v1 + · · · + sn vn ) = 0 is equivalent to s1 f (v1 ) + · · · + sn f (vn ) = 0 and so if the vectors f (v1 ), . . . , f (vn ) are linearly independent, then this equation has unique solution s1 = · · · = sn = 0. This shows that Ker f = {0} and hence by Theorem 3.15 we see that f is injective. (b) Im f ⊆ W contains every vector of the form f (t1 v1 + · · · + tn vn ) = t1 f (v1 ) + · · · + tn f (vn ) (t1 , . . . , tn ∈ F ). If the vectors f (v1 ), . . . , f (vn ) span W , this shows that W ⊆ Im f and so f is surjective. (c) This follows by combining (a) and (b).

¤

Theorem 3.30. Let S : v1 , . . . , vn be a basis for the vector space V . Let W be a second vector space and let T : w1 , . . . , wn be any sequence of vectors in W . Then there is a unique linear transformation f : V −→ W satisfying f (vi ) = wk

(i = 1, . . . , n).

Proof. For every vector v ∈ V , there is a unique expression v = s1 v1 + · · · + sk vk , where si ∈ F . Then we can set f (v) = s1 w1 + · · · + sk wk since the right hand side is clearly well defined in terms of v. In particular, if v = vi , then f (vi ) = wi and that f is a linear transformation f : V −→ W . Uniqueness follows from Lemma 3.28. ¤ A useful consequence of this result is the following. Theorem 3.31. Suppose that the vector space V is a direct sum V = V1 ⊕V2 of two subspaces and let f1 : V1 −→ W and f2 : V2 −→ W be two linear transformations. Then there is a unique linear transformation f : V −→ W extending f1 , f2 , i.e., satisfying f (u) if u ∈ V , 1 1 f (u) = f2 (u) if u ∈ V2 . Proof. Use bases of V1 and V2 to form a basis of V , then use Theorem 3.30.

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Here is another useful result. Theorem 3.32. Let V and W be finite dimensional vector spaces over F with n = dimF V and m = dimF W . Suppose that S : v1 , . . . , vn and T : w1 , . . . , wm are bases for V and W respectively. Then there is a unique transformation f : V −→ W with the following properties: (a) if n 6 m then f (vj ) = wj for j = 1, . . . , n; (b) if n > m then

w j f (vj ) = 0

when j = 1, . . . , m, when j > m.

3.3. WORKING WITH BASES AND COORDINATES

39

Furthermore, f also satisfies: (c) if n 6 m then f is injective; (d) if n > m then f is surjective; (e) if m = n then f is an isomorphism. Proof. (a) Starting with the sequence w1 , . . . , wn , we can apply Theorem 3.30 to show that there is a unique linear transformation f : V −→ W with the stated properties. (b) Starting with the sequence w1 , . . . , wm , 0, . . . , 0 of length n, we can apply Theorem 3.30 to show that there is a unique linear transformation f : V −→ W with the stated properties. (c), (d) and (e) now follow from Lemma 3.29. ¤ The next result provides a convenient way to decide if two vector spaces are isomorphic: simply show that they have the same dimension. Theorem 3.33. Let V and W be finite dimensional vector spaces over the field F . Then there is an isomorphism V −→ W if and only if dimF V = dimF W . Proof. If there is an isomorphism f : V −→ W , then dimF V = dimF W by Corollary 3.21(c). For the converse, if dimF V = dimF W then by Theorem 3.32(e) there is an isomorphism V −→ W . ¤ Now we will see how to make use of bases for the domain and codomain to work with a linear transformation by using matrices. Let V and W be two finite dimensional vector spaces. Let S : v1 , . . . , vn and T : w1 , . . . , wm be bases for V and W respectively. Suppose that f : V −→ W is a linear transformation. Then there are unique scalars tij ∈ F (i = 1, . . . , m, j = 1, . . . , n) for which (3.1)

f (vj ) = t1j w1 + · · · + tmj wm .

We can form the m × n matrix

(3.2)

t11 · · · . .. . T [f ]S = [tij ] = .. tm1 · · ·

t1n .. . . tmn

Remark 3.34. (a) Notice that this matrix depends on the bases S and T as well as f . (b) The coefficients are easily found: the j-th column is made up from the coefficients that occur when expressing f (vj ) in terms of the wi . 0 In the next result we will also assume that there are bases S 0 : v10 , . . . , vn0 and T 0 : w10 , . . . , wm for V and W respectively, for which

vj0 = a1j v1 + · · · + anj vn =

n X

aij vi

i=1 n X

wk0 = b1k v1 + · · · + bmk wm =

i=1

bik wi

(j = 1, . . . , n), (k = 1, . . . , m).

40

3. LINEAR TRANSFORMATIONS

Working with these bases we obtain the matrix

t011 · · · .. .. = [t0ij ] = . . t0m1 · · ·

T 0 [f ]S 0

We will set

a11 · · · . .. A = [aij ] = .. . an1 · · ·

a1n .. . , ann

t01n .. . . t0mn

b11 · · · . .. B = [bij ] = .. . bm1 · · ·

b1m .. . . bmm

Theorem 3.35. (a) Let x ∈ V have coordinates x1 , . . . , xn with respect to the basis S : v1 , . . . , vn of V and let f (x) ∈ W have coordinates y1 , . . . , ym with respect to the basis T : w1 , . . . , wm of W . Then x1 t11 · · · t1n x1 y1 . . . . . . . . .. .. . . (3.3a) . . . = T [f ]S . = . xn tm1 · · · tmn xn ym (b) The matrices T [f ]S and

T 0 [f ]S 0

are related by the formula

(3.3b)

T 0 [f ]S 0

= B −1 T [f ]S A.

Proof. These follow from calculations with the above formulae.

¤

The formulae of (3.3), allow us to reduce calculations to ones with coordinates and matrices. Example 3.36. Consider the linear transformation f : R3 −→ R2 ;

f (x, y, z) = (2x − 3y + 7z, x − z)

between the real vector spaces R3 and R2 . (a) Write down the matrix of f with respect to the standard bases of R3 and R2 . (b) Determine the matrix T [f ]S , where S and T are the bases S : (1, 0, 1), (0, 1, 1), (0, 0, 1),

T : (3, 1), (2, 1).

Solution. (a) In terms of the standard basis vectors (1, 0, 0), (0, 1, 0), (0, 0, 1) and (1, 0), (0, 1) we have f (1, 0, 0) = (2, 1) = 2(1, 0) + 1(0, 1), f (0, 1, 0) = (−3, 0) = −3(1, 0) + 0(0, 1), f (0, 0, 1) = (7, −1) = 7(1, 0) + (−1)(0, 1), " # 2 −3 7 hence the matrix is . 1 0 −1 (b) We have (1, 0, 1) = 1(1, 0, 0) + 0(0, 1, 0) + 1(0, 0, 1), (0, 1, 1) = 0(1, 0, 0) + 1(0, 1, 0) + 1(0, 0, 1), (0, 0, 1) = 0(1, 0, 0) + 0(0, 1, 0) + 1(0, 0, 1),

3.4. APPLICATION TO MATRICES AND SYSTEMS OF LINEAR EQUATIONS

so we take

41

1 0 0 A = 0 1 0 . 1 1 1

We also have (3, 1) = 3(1, 0) + 1(0, 1), and so we take

(2, 1) = 2(1, 0) + 1(0, 1),

" # 3 2 B= . 1 1

The inverse of B is B −1

# # " " 1 −2 1 −2 1 , = = 3 − 2 −1 −1 3 3

so we have T [f ]S

=B

−1

" # 2 −3 7 A 1 0 −1

#" # 1 0 0 1 −2 2 −3 7 = 0 1 0 −1 3 1 0 −1 1 1 1 " # 9 6 9 = . −9 −7 −10 "

¤

Theorem 3.37. Suppose that f : V −→ W and g : U −→ V are linear transformations and that R, S, T are bases for U, V, W respectively. Then the matrices T [f ]S , S [g]R , T [f ◦ g]R are related by the matrix equation T [f ◦ g]R = T [f ]S S [g]R . Proof. This follows by a calculation using the formula (3.1). Notice that the right hand matrix product does make sense because T [f ]S is dim W × dim V , S [g]R and is dim V × dim U , while T [f ◦ g]R is dim W × dim U . ¤ 3.4. Application to matrices and systems of linear equations Let F be a field and consider an m × n matrix A ∈ Mm×n (F ). Recall that by performing a suitable sequence of elementary row operations we may obtain a reduced echelon matrix A0 which is row equivalent to A, i.e., A0 ∼ A. This can be used to solve a linear system of the form Ax = b. It is also possible to perform elementary column operations on A, or equivalently, to form T A then perform elementary row operations until we get a reduced echelon matrix A00 , then transpose back to get another column reduced echelon matrix (A00 )T . Definition 3.38. The number of non-zero rows in A0 is called the row rank of A. The number of non-zero columns in (A00 )T is called the column rank of A. Example 3.39. Taking F = R, find the row and column ranks of the matrix 1 2 A = 0 3 . 5 1

42

3. LINEAR TRANSFORMATIONS

Solution. For the row rank we have 1 2 1 2 1 2 1 0 A = 0 3 ∼ 0 3 ∼ 0 1 ∼ 0 1 = A0 , 5 1 0 −9 0 −9 0 0 so the row rank is 2. For the column rank,

" # " # " # 1 0 5 1 0 5 1 0 5 A = ∼ ∼ = A00 , 2 3 1 0 3 −9 0 1 −3 T

which gives

1 0 (A00 )T = 0 1 5 −3

and the column rank is 2.

¤

Theorem 3.40. Let A ∈ Mm×n (F ). Then row rank of A = column rank of A. Proof of Theorem 3.40. We will make use of ideas from Example 3.10. Recall the linear transformation fA : F n −→ F m determined by fA (x) = Ax (x ∈ F n ). First we identify the value of the row rank of A in other terms. Let r be the row rank of A. Then Ker fA = {x ∈ F n : Ax = 0} is the set of solution of the associated system of homogeneous linear equations. Recall that once the reduced echelon matrix A0 has been found, the general solution can be expressed in terms of n − r basic solutions each obtained by setting one of the parameters equal to 1 and the rest equal to 0. These basic solutions form a spanning sequence for Ker fA which is linearly independent and hence is a basis for Ker fA . So we have (3.4)

row rank of A = r = n − dim Ker fA .

On the other hand, if c is the column rank of A, we find that the columns of (A00 )T are a spanning sequence for the subspace of F m spanned by the columns of A, and in fact, they are linearly independent. As the columns of A span Im fA , (3.5)

column rank of A = c = dim Im fA .

But now we can use the rank-nullity Theorem 3.19 to deduce that n = dim Ker fA + dim Im fA = (n − r) + c = n + (c − r), giving c = r as claimed.

¤

Remark 3.41. In this proof we saw that the columns of A span Im fA and also how to find a basis for Ker fA by the method of solution of linear systems using elementary row operations.

3.5. GEOMETRIC LINEAR TRANSFORMATIONS

43

Definition 3.42. The rank of A is the number rank A = rank fA = dim Im fA = column rank of A = row rank of A, where rank fA was defined in Definition 3.18. It agrees with the number of non-zero rows in the reduced echelon matrix similar to A. The kernel of A is Ker A = Ker fA and the image of A is Im A = Im fA . 3.5. Geometric linear transformations When considering a linear transformation f : Rn −→ Rn on the real vector space Rn it is often important or useful to understand its geometric content. Example 3.43. For θ ∈ R, consider the linear transformation ρθ : R2 −→ R2 given by ρθ (x, y) = (x cos θ − y sin θ, x sin θ + y cos θ). Investigate the geometric effect of ρθ on the plane. Solution. The effect on the standard basis vectors is ρθ (1, 0) = (cos θ, sin θ),

ρθ (0, 1) = (− sin θ, cos θ).

Using the dot product of vectors it is possible to check that the angle between ρθ (x, y) and (x, y) is always θ if (x, y) 6= (0, 0), and that the lengths of ρθ (x, y) and (x, y) are equal. So ρθ has the effect of rotating the plane about the origin through the angle θ measured in the anti-clockwise direction. Notice that the matrix of ρθ with respect to the standard basis of R2 is the rotation matrix " # cos θ − sin θ . ¤ sin θ cos θ Example 3.44. For θ ∈ R, consider the linear transformation σθ : R2 −→ R2 given by σθ (x, y) = (x cos θ + y sin θ, x sin θ − y cos θ). Investigate the geometric effect of σθ on the plane. Solution. The effect on the standard basis vectors is σθ (1, 0) = (cos θ, sin θ),

σθ (0, 1) = (sin θ, − cos θ).

The angle between every pair of vectors is preserved by σθ , as is the length of a vector. We also find that σθ (cos(θ/2), sin(θ/2)) = (cos(θ/2), sin(θ/2)), σθ (− sin(θ/2), cos(θ/2)) = (sin(θ/2), − cos(θ/2)) = −(− sin(θ/2), cos(θ/2)), from which it follows that σθ has the effect of reflecting the plane in the line through the origin and parallel to (cos(θ/2), sin(θ/2)). Notice that the matrix of σθ with respect to the standard basis of R2 is the reflection matrix " # cos θ sin θ . ¤ sin θ − cos θ

44

3. LINEAR TRANSFORMATIONS

# " 1 t Example 3.45. Let S = and let fS : R2 −→ R2 be the real linear transformation 0 1 given by fS (x) = Sx. Then fS has the geometric effect of a shearing parallel to the x-axis. Solution. This can be seen by plotting the effect of fS on points. For example, points on the x-axis are fixed by fS , while points on the y-axis get moved parallel to the x-axis into the line of slope t. ¤ " # d 0 Example 3.46. Let D = where d > 0 and let fD : R2 −→ R2 be the real linear 0 1 transformation defined by fD (x) = Dx. Then fD has the geometric effect of a dilation parallel to the x-axis. Solution. When applying fD , vectors along the x-axis get dilated by a factor d, while those on the y-axis are unchanged. ¤ Example 3.47. Let V a finite dimensional vector space over a field F . Suppose that V = V 0 ⊕ V 00 is the direct sum of two subspaces V 0 , V 00 (see Definition 2.38). Then there is a linear transformation pV 0 ,V 00 : V −→ V given by pV 0 ,V 00 (v0 + v00 ) = v0

(v0 ∈ V 0 , v00 ∈ V 00 ).

Then Ker pV 0 ,V 00 = V 00 and Im pV 0 ,V 00 = V 0 . pV 0 ,V 00 is sometimes called the projection of V onto V 0 with kernel V 00 . The case where V 00 is a line (i.e., dim V 00 = 1) is particularly important. Example 3.48. Find a formula for the projection of R3 onto the plane P through the origin and perpendicular to (1, 0, 1) with kernel equal to the line L spanned by (1, 0, 1). Solution. Let this projection be p : R3 −→ R3 . Then every vector x ∈ R3 can be expressed as

³ √ √ √ √ ´ √ √ √ √ x = x − [(1/ 2, 0, 1/ 2) · x](1/ 2, 0, 1/ 2) + [(1/ 2, 0, 1/ 2) · x](1/ 2, 0, 1/ 2),

where and

√ √ √ √ x − [(1/ 2, 0, 1/ 2) · x](1/ 2, 0, 1/ 2) ∈ P √ √ √ √ [(1/ 2, 0, 1/ 2) · x](1/ 2, 0, 1/ 2) ∈ L.

This corresponds to the direct sum decomposition R3 = P ⊕ L. The projection p is given by √ √ √ √ p(x) = x − [(1/ 2, 0, 1/ 2) · x](1/ 2, 0, 1/ 2).

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CHAPTER 4

Determinants For 2 × 2 and 3 × 3 matrices determinants are defined by the formulae ¯ # ¯ " ¯ ¯a a11 a12 ¯ 11 a12 ¯ (4.1) det =¯ ¯ = a11 a22 − a12 a21 ¯a21 a22 ¯ a21 a22 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯a a a a11 a12 a13 11 12 13 ¯ ¯ ¯a ¯ ¯a ¯ ¯a ¯ ¯ ¯ 11 a12 ¯ ¯ 11 a12 ¯ ¯ 11 a12 ¯ ¯ (4.2) det a21 a22 a23 = ¯a21 a22 a23 ¯ = a11 ¯ ¯ − a12 ¯ ¯ + a13 ¯ ¯. ¯ ¯a21 a22 ¯ ¯a21 a22 ¯ ¯a21 a22 ¯ ¯ ¯ ¯ a31 a32 a33 a31 a32 a33 We will introduce determinants for all n × n matrices over any field of scalars and these will have properties analogous to that of (4.2) which allows the calculation of 3 × 3 determinants from 2 × 2 ones. Definition 4.1. For the n×n matrix A = [aij ] with entries in F , for each pair r, s = 1, . . . , n, the cofactor matrix Ars is the (n − 1) × (n − 1) matrix obtained by deleting the r-th row and s-th column of A. Warning. When working with the determinant of a matrix [aij ] we will often write |A| for det(A) and display it as an array with vertical lines rather than square brackets as in (4.1) and (4.2). But beware, the two expressions a11 a12 a13 a21 a22 a23 a31 a32 a33 and

¯ ¯ ¯a ¯ a a a a a 11 12 13 11 12 13 ¯ ¯ ¯ ¯ ¯a21 a22 a23 ¯ = det a21 a22 a23 ¯ ¯ ¯a31 a32 a33 ¯ a31 a32 a33

are very different things: the first is a matrix, whereas the second is a scalar. They should not be confused! 4.1. Definition and properties of determinants Let F be a field of scalars. Theorem 4.2. For each n > 1, and every n × n matrix A = [aij ] with entries in F , there is a uniquely determined scalar det A = det(A) = |A| ∈ F which satisfies the following properties. (A) For the n × n identity matrix In , det(In ) = 1. 45

46

4. DETERMINANTS

(B) If A, B are two n × n matrices then det(AB) = det(A) det(B). (C) Let A be an n × n matrix. Then we have Expansion along the r-th row: n X (−1)r+j arj det(Arj )

det(A) =

j=1

for any r = 1, . . . , n. Expansion along the s-th column: det(A) =

n X

(−1)s+i ais det(Ais )

i=1

for any s = 1, . . . , n. The signs in these ¯ ¯+ ¯ ¯ ¯− ¯ ¯+ ¯ ¯ .. ¯.

expansions can be obtained from the pattern ¯ − + − · · · ¯¯ ¯ + − + ··· ¯ ¯ − + − · · · ¯¯ .. ¯ .. . . ¯

(D) For an n × n matrix A, det(AT ) = det(A). Notice that the formula det(A) =

n X

(−1)j−1 a1j det(A1j )

j=1

from (C) is a direct generalisation of (4.2). Here are some important results on determinants of invertible matrices that follow from the above properties. Proposition 4.3. Let A be an invertible n × n matrix. (i) det(A) 6= 0 and 1 . det(A−1 ) = det(A)−1 = det(A) (ii) If B is any n × n matrix then det(ABA−1 ) = det(B). Proof. (i) By (B), det(In ) = det(AA−1 ) = det(A) det(A−1 ), hence using (A) we find that det(A) det(A−1 ) = 1, so det(A−1 ) 6= 0 and det(A−1 ) = det(A)−1 =

1 . det(A)

4.1. DEFINITION AND PROPERTIES OF DETERMINANTS

47

(ii) By (B) and (i), det(ABA−1 ) = det(AB) det(A−1 ) = det(A) det(B) det(A−1 ) = det(B) det(A) det(A−1 ) = det(B).

¤

Let us relate this to elementary row operations and elementary matrices, whose basic properties we will assume. Suppose that R is an elementary row operation and E(R) is the corresponding elementary matrix. Then the matrix A0 obtained from A by applying R satisfies A0 = E(R)A, hence (4.3)

det(A0 ) = det(E(R)A) = det(E(R)) det(A).

Proposition 4.4. If A0 = E(R)A is the to the n × n matrix A, then − det(A) det(A0 ) = λ det(A) det(A) In particular,

−1 det(E(R)) = λ 1

result of applying the elementary row operation R

if R = (Rr ↔ Rs ), if R = (Rr → λRr ), if R = (Rr → Rr + λRs ).

if R = (Rr ↔ Rs ), if R = (Rr → λRr ), if R = (Rr → Rr + λRs ).

Proof. If R = Rr ↔ Rs , then it is possible to prove by induction on the size n that det(E(Rr ↔ Rs )) = −1. The initial case is n = 2 where

¯ ¯ ¯0 1¯ ¯ ¯ det(E(R1 ↔ R2 )) = ¯ ¯ = −1. ¯1 0¯

If R = Rr → λRr for λ 6= 0, then expanding along the r-th row gives det(E(Rr → λRr )) = λ det(In ) = λ. Finally, if R = Rr → Rr + λRs with r 6= s, then expanding along the r-th row gives det(E(Rr → Rr + λRs )) = det(In−1 ) + λ det(Inrs ) = 1 + λ × 0 = 1.

¤

Thus to calculate a determinant, first find any sequence R1 , . . . Rk of elementary row operations so that the combined effect of applying these successively to A is an upper triangular matrix L. Then L = E(Rk ) · · · E(R1 )A and so det(L) = det(E(Rk )) · · · det(E(R1 )) det(A), giving det(A) = det(E(Rk ))−1 · · · det(E(R1 ))−1 det(L),

48

4. DETERMINANTS

To calculate det(L) is easy, since

¯ ¯`1 ∗ ∗ ¯ ¯0 ` ∗ ¯ 2 ¯ ¯ 0 0 ` 3 det(L) = ¯ ¯ .. .. .. ¯. . . ¯ ¯0 0 0

∗ ∗ ∗

∗ ∗ ∗ .. .

0 ···

¯ ¯ ∗ ¯¯ ¯` ¯ ¯ 2 ∗ ∗ ∗ ∗¯ ¯ ¯ ¯ 0 `3 ∗ ∗ ∗ ¯ = `1 ¯ . . .. ¯ ¯. . . .. ¯ ¯. . ¯ . ¯¯ ¯0 0 0 ··· 0 `n ¯ ∗ ∗ ∗

¯ ∗ ¯¯ ¯ ∗¯ .. ¯¯ .¯ ¯ 0 `n ¯ ∗ ∗

by definition and repeating this gives det(L) = `1 `2 · · · `n . Using this we can obtain a useful criterion for invertibility of a square matrix.

Proposition 4.5. Let A be an n × n matrix with entries in a field of scalars F . Then det(A) = 0 if and only if A is singular. Equivalently, det(A) 6= 0 if and only if A is invertible.

Proof. By Proposition 4.3 we know that if A is invertible, then det(A) 6= 0. On the other hand, if det(A) 6= 0 then suppose that the reduced echelon form of A is A0 . If there is a zero row in A0 then det(A0 ) = 0. But if A0 = E(Rk ) · · · E(R1 )A, then

0 = det(A0 ) = det(E(Rk )) · · · det(E(R1 )) det(A),

where the scalars det(E(Rk )), . . . , det(E(R1 )), det(A) are all non-zero. Hence this cannot happen and so A0 = In , showing that A is invertible. ¤

There is an explicit formula, Cramer’s Rule, for finding the inverse of a matrix using determinants. However, it is not much use for computations with large matrices and methods based on row and column operations are usually much more efficient. In practise, evaluation of determinants is often simplified by using elementary row operations to create zero’s or otherwise reduced calculations.

Example 4.6. Evaluate the following determinants:

¯ ¯4 ¯ ¯0 ¯ ¯ ¯3 ¯ ¯2

5 3 0 7

1 2 1 0

¯ 2¯¯ 1¯¯ ¯, 3¯ ¯ 3¯

¯ ¯ ¯ x y 2 1 ¯¯ ¯ ¯ ¯ 2 x¯ . ¯ y ¯ ¯ ¯x + y 0 x¯

4.1. DEFINITION AND PROPERTIES OF DETERMINANTS

Solution. ¯ ¯4 5 ¯ ¯0 3 ¯ ¯ ¯3 0 ¯ ¯2 7

1 2 1 0

¯ ¯ x y2 ¯ ¯ 2 ¯ y ¯ ¯x + y 0

49

¯ ¯ ¯ 2¯¯ ¯¯4 0 3 2¯¯ 1¯¯ ¯¯5 3 0 7¯¯ ¯=¯ ¯ [transposing] 3¯ ¯1 2 1 0¯ ¯ ¯ ¯ 3¯ ¯2 1 3 3¯ ¯ ¯ ¯ 4 0 3 2¯¯ ¯ ¯−1 0 −9 −2¯ ¯ ¯ =¯ ¯ [R2 → R2 − 3R4 , R3 → R3 − 2R4 ] ¯−3 0 −5 −6¯ ¯ ¯ ¯ 2 1 3 3¯ ¯ ¯ ¯ 4 3 2¯¯ ¯ ¯ ¯ = ¯−1 −9 −2¯ [expanding along column 2] ¯ ¯ ¯−3 −5 −6¯ ¯ ¯ ¯4 3 2¯ ¯ ¯ ¯ ¯ = ¯1 9 2¯ [multiplying rows 2 and 3 by (−1)] ¯ ¯ ¯3 5 6¯ ¯ ¯ ¯ 4 3 2¯¯ ¯ ¯ ¯ = ¯−3 6 0¯ [R2 → R2 − R1 , R3 → R3 − 3R1 ] ¯ ¯ ¯ 0 −22 0¯ ¯ ¯ ¯ 4 2¯ ¯ ¯ = 22 ¯ ¯ = 22 × 6 = 132, [expanding along row 2] ¯−3 0¯ ¯ ¯ ¯ ¯ ¯ 1 ¯¯ ¯y 2 1 ¯ ¯x y 2 ¯ ¯ ¯ ¯ ¯ ¯ x¯ = (x + y) ¯ ¯ + x¯ ¯ [expanding along row 3] ¯ ¯ 2 x¯ ¯y 2 ¯ ¯ x = (x + y)(xy 2 − 2) + x(2x − y 3 ) = x2 y 2 − 2x + xy 3 − 2y + 2x2 − xy 3 = x2 y 2 − 2x − 2y + 2x2 .

Example 4.7. Solve the equation ¯ ¯1 2 ¯ ¯ ¯x x + 1 ¯ ¯1 1

¤

¯ −1 ¯¯ ¯ x2 ¯ = 0. ¯ 5¯

Solution. Expanding the determinant gives ¯ ¯ ¯ ¯ ¯1 2 −1 ¯¯ 2 −1 ¯¯ ¯¯1 ¯ ¯ ¯ ¯ ¯ x2 + x¯ [R2 → R2 − xR1 , R3 → R3 − R1 ] x2 ¯ = ¯0 1 − x ¯x x + 1 ¯ ¯ ¯ ¯ ¯1 6 ¯ 1 5 ¯ ¯0 −1 ¯ ¯ ¯1 − x x2 + x¯¯ ¯ =¯ ¯ [expanding along column 1] ¯ −1 6 ¯ = x2 + x + 6 − 6x = x2 − 5x + 6 = (x − 2)(x − 3). So the solutions are x = 2, 3.

¤

50

4. DETERMINANTS

Example 1 x matrix 1 y 1 z

¯ ¯ ¯1 x x2 ¯ ¯ ¯ ¯ ¯ 4.8. Evaluate the Vandermonde determinant ¯1 y y 2 ¯. Determine when the ¯ ¯ ¯1 z z 2 ¯ x2 y 2 is singular. z2

Solution.

¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯1 x x2 ¯ ¯1 x x2 x x2 ¯¯ ¯¯1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯1 y y 2 ¯ = ¯0 y − x y 2 − x2 ¯ = ¯0 y − x (y − x)(y + x)¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯1 z z 2 ¯ ¯0 z − x z 2 − x2 ¯ ¯0 z − x (z − x)(z + x)¯ ¯ ¯ 2 ¯ ¯1 x x ¯ ¯ ¯ ¯ = (y − x)(z − x) ¯0 1 y + x¯ ¯ ¯ ¯0 1 z + x ¯ ¯ ¯ ¯1 y + x¯ ¯ ¯ = (y − x)(z − x) ¯ ¯ ¯0 z − y ¯ = (y − x)(z − x)(z − y).

The matrix is singular precisely when its determinant vanishes, and this happens when at least one of (y − x), (z − x), (z − y) is zero, i.e., when two of the scalars x, y, z are equal. ¤ Remark 4.9. For n > 1, ¯ ¯1 ¯ ¯ ¯1 ¯. ¯. ¯. ¯ ¯1

there is a general formula for an n × n Vandermonde determinant, ¯ ¯ x1 x21 · · · xn−1 ¯ 1 ¯ Y x2 x22 · · · xn−1 ¯ 2 (xs − xr ). .. ¯¯ = .. . . ¯ 16r<s6n ¯ ¯ xn x2n · · · xn−1 n

The proof is similar to that for the case n = 3. Again we can see when the matrix 1 x1 x21 · · · xn−1 1 1 x2 x22 · · · xn−1 2 . .. .. . . . . 1 xn x2n · · · xn−1 n is singular, namely when xr = xs for some pair of distinct indices r, s. 4.2. Determinants of linear transformations Let V be a finite dimensional vector space over the field of scalars F and suppose that f : V −→ V is a linear transformation. Write n = dim V . If we choose a basis S : v1 , . . . vn for V , then there is a matrix S [f ]S and we can find its determinant det(S [f ]S ). A second basis T : w1 , . . . wn leads to another determinant det(T [f ]T ). Appealing to Theorem 3.35(b) we find that for a suitable invertible matrix P , T [f ]T

= P −1 S [f ]S P,

hence by Proposition 4.3, det(T [f ]T ) = det(P −1 S [f ]S P ) = det(S [f ]S ).

4.3. CHARACTERISTIC POLYNOMIALS AND THE CAYLEY-HAMILTON THEOREM

51

This shows that det(S [f ]S ) does not depend on the choice of basis, only on f , hence it is an invariant of f . Definition 4.10. Let f : V −→ V is a linear transformation on a finite dimensional vector space. Then the determinant of f is det f = det(f ) = det(S [f ]S ), where S is any basis of V . This value only depends on f and not on the basis S. Remark 4.11. Suppose that V = Rn . Then there is a geometric interpretation of det f (at least for n = 2, 3, but it makes sense in the general case if we suitably interpret volume in Rn ). The standard basis vectors e1 , . . . , en form the edges of a unit cube based at the origin, with volume 1. The vectors f (e1 ), . . . , f (en ) form edges of a parallellipiped based at the origin. The volume of this parallellipiped is | det f | (note that volumes are non-negative!). This is used in change of variable formulae for multivariable integrals. The value of det f (or more accurately of | det f |) indicates how much volumes are changed by applying f . The sign of det f indicates whether or not ‘orientations’ change and this is also encountered in multivariable integration formulae. 4.3. Characteristic polynomials and the Cayley-Hamilton theorem For a field F , let A be an n × n matrix with entries in F . Definition 4.12. The characteristic polynomial of A is the polynomial χA (t) ∈ F [t] (often denoted charA (t)) defined by χA (t) = det(tIn − A). Proposition 4.13. The polynomial χA (t) has degree n and is monic, i.e., χA (t) = tn + cn−1 tn−1 + · · · + c1 t + c0 , where c0 , c1 , . . . , cn−1 ∈ F . Furthermore, cn−1 = − tr A,

c0 = (−1)n det(A).

Proof. To see the first part, write

¯ ¯t − a ¯ 11 · · · ¯ . .. ¯ det(tIn − A) = ¯ .. . ¯ ¯ −an1 · · ·

¯ ¯ ¯ ¯ ¯ ¯ ¯ t − ann ¯ −a1n .. .

and then apply property (C) repeatedly. The formula for cn−1 follows from the fact that the terms contributing to the tn−1 term are obtained by multiplying n − 1 of the diagonal terms and then multiplying by the constant term in the remaining one. The result is then cn−1 = −(a11 + · · · + ann ) = − tr(A). For the constant term c0 , taking t = 0 and using Proposition 4.4 we obtain c0 = det(−A) = (−1)n det(A). Proposition 4.14. Let B be an invertible n × n matrix, then χBAB −1 (t) = χA (t).

¤

52

4. DETERMINANTS

Proof. This follows from the fact that ¡ ¢ χBAB −1 (t) = det(tIn − BAB −1 ) = det B(tIn − A)B −1 = det(tIn − A) by Proposition 4.3(ii).

¤

This allows us to make the following definition. See Definition 4.10 for a similar idea. Let V be a finite dimensional vector space over a field F and let f : V −→ V be a linear transformation. Definition 4.15. The characteristic polynomial of f is χf (t) = χS [f ]S (t) = det(tIn − S [f ]S ), where S is any basis of V . We now come to an important result which is useful in connection with the notion of eigenvalues to be considered in Chapter 5. Let p(t) = ak tk + ak−1 tk−1 + · · · + a1 t + a0 ∈ F [t] be a polynomial (so a0 , a1 , . . . , ak ∈ F ). For an n × n matrix A we write p(A) = ak Ak + ak−1 Ak−1 + · · · + a1 A + a0 In , which is also an n×n matrix. If p(A) = On then we say that A satisfies this polynomial identity. Similarly, if f : V −→ V is a linear transformation, p(f ) = ak f k + ak−1 f k−1 + · · · + a1 f + a0 IdV . Theorem 4.16 (Cayley-Hamilton theorem). (i) Let A be an n × n matrix. Then χA (A) = On . (ii) Let V be a finite dimensional vector space of dimension n and let f : V −→ V be a linear transformation. Then χf (f ) = 0, where the right hand side is the constant function taking value 0. Informally, these results are sometimes summarised by saying that a square matrix A or a linear transformation f : V −→ V satisfies its own characteristic equation. 0 0 2 Example 4.17. Show that the matrix A = 1 0 0 satisfies a polynomial identity of 0 1 3 degree 3. Hence show that A is invertible and express A−1 as a polynomial in A. Solution. The characteristic polynomial of A is ¯ ¯ ¯ t 0 −2 ¯¯ ¯ ¯ ¯ χA (t) = ¯−1 t 0 ¯ = t3 − 3t2 − 2. ¯ ¯ ¯ 0 −1 t − 3¯ So det(A) = 2 6= 0 and A is invertible. Also, A3 − 3A2 − 2I3 = O3 so A(A2 − 3A) = (A2 − 3A)A = A3 − 3A2 = 2I3 ,

giving

4.3. CHARACTERISTIC POLYNOMIALS AND THE CAYLEY-HAMILTON THEOREM

53

1 A−1 = (A2 − 3A). 2

¤

CHAPTER 5

Eigenvalues and eigenvectors 5.1. Eigenvalues and eigenvectors for matrices Let F be a field and A ∈ Mn (F ) = Mn×n (F ) be a square matrix. We know that there is an associated linear transformation fA : F n −→ F n given by fA (x) = Ax. We can ask whether there are any vectors which are fixed by fA , i.e., which satisfy the equation fA (x) = x. Of course, 0 is always fixed so we should ask this question with the extra requirement that x be non-zero. Usually there will be no such vectors, but a slightly more general situation that might occur is that fA will dilate some non-zero vectors, i.e., fA (x) = λx for some λ ∈ F and non-zero vector x. This amounts to saying that fA sends all vectors in some line through the origin into that line. Of course, such a line is a 1-dimensional subspace of F n . " # 2 1 . Example 5.1. Let F = R and consider the 2 × 2 matrix A = 0 3 (i) Show that the linear transformation fA : R2 −→ R2 does not fix any non-zero vectors. (ii) Show thatfA dilates all vectors on the x-axis by a factor of 2. (iii) Find all the non-zero vectors which fA dilates by a factor of 3. Solution. (i) Suppose that a vector x ∈ R2 satisfies Ax = x, then it also satisfies the homogenous equation # " −1 −1 (I2 − A)x = 0, i.e., x = 0. 0 −2 But the only solution of this is x = 0. (ii) We have " #" # " # " # 2 1 x 2x x = =2 , 0 3 0 0 0 so vectors on the x-axis get dilated under fA by a factor of 2. (iii) Suppose that Ax = 3x, then " # 1 −1 (3I2 − A)x = 0, i.e., x = 0. 0 0 The set of all solution vectors of this is the line L = {(t, t) : t ∈ R}.

¤

Definition 5.2. Let F be a field and A ∈ Mn (F ). Then λ ∈ F is an eigenvalue for A if there is a non-zero vector v ∈ F n for which Av = λv; such a vector v is called an eigenvalue associated with λ. Definition 5.2 crucially depends on the field F as the following example shows. 55

56

5. EIGENVALUES AND EIGENVECTORS

# " 0 2 Example 5.3. Let A = . 1 0 (i) Show that A has no eigenvalues in Q. (ii) Show that A has two eigenvalues in R. Solution. Notice that A2 − 2I2 = O2 , so if λ is a real eigenvalue with associated eigenvector v, then (A2 − 2I2 )v = 0, which gives (λ2 − 2)v = 0. √ √ Thus we must have λ2 = 2, hence λ = ± 2. However, neither of ± 2 is a rational number although both are real. Working in R2 we have "√ # "√ # " √ # " √ # √ √ − 2 2 2 − 2 A = 2 , A =− 2 . ¤ 1 1 1 1 " # 0 −2 A similar problem occurs with the real matrix which has the two imaginary eigen1 0 √ values ±i 2. Because of this it is preferable to work in an algebraically closed field such as the complex numbers. So from now on we will work with complex matrices and consider eigenvalues in C and eigenvectors in Cn . When discussing real matrices we need to take care to check for real eigenvalues. This is covered by the next result which can be proved easily from the definition. Lemma 5.4. Let A be an n × n real matrix which we can also view as a complex matrix. Suppose that λ ∈ R is an eigenvalue for which there is an associated eigenvector in Cn . Then there is an associated eigenvector in Rn . Theorem 5.5. Let A be an n × n complex matrix and let λ ∈ C. Then λ is an eigenvalue of A if and only if λ is a root of the characteristic polynomial χA (t), i.e., χA (λ) = 0. Proof. From Definition 5.2, λ is an eigenvalue for A if and only if λIn − A is singular, i.e., is not invertible. By Proposition 4.5, this is equivalent to requiring that χA (λ) = det(λIn − A) = 0.

¤

5.2. Some useful facts about roots of polynomials Recall that every complex (which includes real) polynomial p(t) ∈ C[t] of positive degree n admits a factorisation into linear polynomials p(t) = c(t − α1 ) · · · (t − αn ), where c 6= 0 and the roots αk are unique apart from the order in which they occur. It is often useful to write c(t − β1 )r1 · · · (t − βm )rm ,

5.2. SOME USEFUL FACTS ABOUT ROOTS OF POLYNOMIALS

57

where β1 , . . . , βm are the distinct complex roots of p(t) and rk > 1. Then rk is the (algebraic) multiplicity of the root βk . Again, this factorisation is unique apart form the order in which the distinct roots are listed. When p(t) is a real polynomial, i.e., p(t) ∈ R[t], the complex roots fall into two types: real roots and non-real roots which come in complex conjugate pairs. Thus for such a real polynomial we have p(t) = c(t − α1 ) · · · (t − αr )(t − γ1 )(t − γ 1 ) · · · (t − γs )(t − γ s ), where α1 , . . . , αr are the real roots (possibly with repetitions) and γ1 , γ 1 . . . , γs , γ s are the nonreal roots which occur in complex conjugate pairs γk , γ j . For a non-real complex number γ, (t − γ)(t − γ) = t2 − (γ + γ)t + γγ = t2 − (2 Re γ)t + |γ|2 , where Re γ is the real part of γ and |γ| is the modulus of γ. Proposition 5.6. Let p(t) = tn + an−1 tn−1 + · · · + a1 t + a0 ∈ C[t] be a monic complex polynomial which factors completely as p(t) = (t − α1 )(t − α2 ) · · · (t − αn ), where α1 , . . . , αn are the complex roots, occurring with multiplicities. The following formulae apply for the sum and product of these roots: n X

αi = α1 + · · · + αn = −an−1 ,

i=1

n Y

αi = α1 · · · αn = (−1)n a0 .

i=1

Example 5.7. Let p(t) be a complex polynomial of degree 3. Suppose that p(t) = t3 − 6t2 + at − 6, where a ∈ C, and also that 1 is a root of p(t). Find the other roots of p(t). Solution. Suppose that the other roots are α and β. Then by Proposition 5.6, −(α + β + 1) = −6,

i.e.,

β = 5 − α;

also, (−1) × 1 × αβ = −6,

i.e.,

αβ = 6.

Hence we find that (5 − α)α = 6,

i.e.,

α2 − 5α + 6 = 0.

The roots of this are 2, 3, therefore we obtain these as the remaining roots of p(t).

¤

Of course we can use this when working with the characteristic polynomial of an n×n matrix A when the roots are the eigenvalues λ1 , . . . , λn of A occurring with suitable multiplicities. If χA (t) = tn + cn−1 tn−1 + · · · + c1 t + c0 , then by Proposition 4.13, (5.1) (5.2)

λ1 + · · · + λn = tr A = −cn−1 , λ1 · · · λn = det A = (−1)n c0 .

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5. EIGENVALUES AND EIGENVECTORS

5.3. Eigenspaces and multiplicity of eigenvalues In this section we will work over the field of complex numbers C. Definition 5.8. Let A be an n × n complex matrix and let λ ∈ C be an eigenvalue of A. Then the λ-eigenspace of A is EigA (λ) = {v ∈ Cn : Av = λv} = {v ∈ Cn : (λIn − A)v = 0}, the set of all eigenvectors associated with λ together with the zero vector 0. If A is a real matrix and if λ ∈ R is a real eigenvalue, then the real λ-eigenspace of A is n n EigR A (λ) = {v ∈ R : Av = λv} = EigA (λ) ∩ R .

Proposition 5.9. Let A be an n × n complex matrix and let λ ∈ C be an eigenvalue of A. Then EigA (λ) is a subspace of Cn . If A is a real matrix and λ is a real eigenvalue, then EigR A (λ) is a subspace of the real vector n space R . Proof. By definition, 0 ∈ EigA (λ). Now notice that if u, v ∈ EigA (λ) and z, w ∈ C, then A(zu + wv) = A(zu) + A(wv) = zAu + wAv = zλu + wλv = λ(zu + wv). Hence EigA (λ) is a subspace of Cn . n The proof in the real case easily follows from the fact that EigR A (λ) = EigA (λ) ∩ R .

¤

Definition 5.10. Let A be an n × n complex matrix and let λ ∈ C be an eigenvalue of A. The geometric multiplicity of λ is dimC EigA (λ). Note that dimC EigA (λ) > 1. In fact there are some other constraints on geometric multiplicity which we will not prove. Theorem 5.11. Let A be an n × n complex matrix and let λ ∈ C be an eigenvalue of A. (i) Let the algebraic multiplicity of λ in the characteristic polynomial χA (t) be rλ . Then geometric multiplicity of λ = dimC EigA (λ) 6 rλ . (ii) If A is a real matrix, then dimC EigA (λ) = dimR EigR A (λ). 2 1 0 Example 5.12. For the real matrix A = −1 0 1, determine its complex eigenvalues. 2 0 0 For each eigenvalue λ, find the corresponding eigenspace EigA (λ). For each real eigenvalue λ, find EigR A (λ). Solution. We have ¯ ¯ ¯ ¯ ¯ ¯ ¯t − 2 −1 0¯¯ ¯ t −1¯ ¯ 1 −1¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ χA (t) = ¯ 1 t −1¯ = (t − 2) ¯ ¯ ¯+¯ ¯ 0 ¯ ¯ t¯ t ¯ ¯−2 ¯ −2 0 t¯ = (t − 2)t2 + (t − 2) = (t − 2)(t2 + 1) = (t − 2)(t − i)(t + i).

5.3. EIGENSPACES AND MULTIPLICITY OF EIGENVALUES

59

So the complex roots of χA (t) are 2, i, −i and there is one real root 2. Now we can find the three eigenspaces. EigA (2): Working over C, we need to solve 2 − 2 −1 0 (2I3 − A)z = 1 2 −1 z = 0, −2 0 2 i.e.,

0 −1 0 2 −1 z = 0. 1 −2 0 2

This is equivalent to the equation

1 0 −1 0 z = 0. 0 1 0 0 0

in which the matrix is reduced echelon, and the general solution is z = (z, 0, z) (z ∈ C). Hence, EigA (2) = {(z, 0, z) ∈ C3 : z ∈ C} and so dimC EigA (2) = 1. Since 2 is real, we can form 3 EigR A (2) = {(t, 0, t) ∈ R : t ∈ R}

which also has dimR EigR A (2) = 1. EigA (i): Working over C, we need to solve i − 2 −1 0 (iI3 − A)z = 1 i −1 z = 0, −2 0 i which is equivalent to the reduced echelon equation 1 0 −i/2 (1 + 2i)/2 z = 0. 0 1 0 0 0 This has general solution (iz, −(1 + 2i)z, 2z)

(z ∈ C),

so EigA (i) = {(iz, −(1 + 2i)z, 2z) ∈ C3 : z ∈ C} and dimC EigA (i) = 1. EigA (−i): Similarly, we have EigA (−i) = {(−iz, −(1 − 2i)z, 2z) ∈ C3 : z ∈ C} and dimC EigA (−i) = 1.

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Notice that in this example, each eigenvalue has geometric multiplicity equal to 1 which is also its algebraic multiplicity. Before giving a general result on this, here is an important observation.

60

5. EIGENVALUES AND EIGENVECTORS

Proposition 5.13. Let A be an n × n complex matrix. (i) Suppose that λ1 , . . . , λk are distinct eigenvalues for A with associated eigenvectors v1 , . . . , vk . Then v1 , . . . , vk is linearly independent. (ii) The sum of the eigenspaces EigA (λ1 ) + · · · + EigA (λk ) is a direct sum, i.e., EigA (λ1 ) + · · · + EigA (λk ) = EigA (λ1 ) ⊕ · · · ⊕ EigA (λk ). Proof. (i) Choose the largest r 6 k for which v1 , . . . , vr is linearly independent. Then if r < k, v1 , . . . , vr+1 is linearly dependent. Suppose that for some z1 , . . . , zr+1 ∈ C, not all zero, (5.3)

z1 v1 + · · · + zr+1 vr+1 = 0.

Multiplying by A we obtain z1 Av1 + · · · + zr+1 Avr+1 = 0 and hence (5.4)

z1 λ1 v1 + · · · + zr+1 λr+1 vr+1 = 0.

Now subtracting λr+1 × (5.3) from (5.4) we obtain z1 (λ1 − λr+1 )v1 + · · · + zr (λr − λr+1 )vr + 0vr+1 = 0. But since v1 , . . . , vr is linearly independent, this means that z1 (λ1 − λr+1 ) = · · · = zr (λr − λr+1 ) = 0, and so since the λj are distinct, z1 = · · · = zr = 0. hence we must have zr+1 vr+1 = 0 which implies that zr+1 = 0 since vr+1 6= 0. So we must have r = k and v1 , . . . , vk is linearly independent. (ii) This follows by a similar argument to (i). ¤ Theorem 5.14. Let A be an n × n complex matrix. Suppose that the distinct complex eigenvalues of A are λ1 , . . . , λ` and that for each j = 1, . . . , `, geometric multiplicity of λj = algebraic multiplicity of λj . Then Cn is the direct sum of the eigenspaces EigA (λj ), i.e., Cn = EigA (λ1 ) ⊕ · · · ⊕ EigA (λ` ). In particular, Cn has a basis consisting of eigenvectors of A. −1 2 2 Example 5.15. Let A = 2 2 2. Find the eigenvalues of A and show that C3 has −3 −6 −6 a basis consisting of eigenvectors of A. Show that the real vector space R3 also has a basis consisting of eigenvectors of A.

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61

Solution. The characteristic polynomial of A is ¯ ¯ ¯t + 1 −2 ¯ −2 ¯ ¯ ¯ ¯ χA (t) = ¯ −2 t − 2 −2 ¯ ¯ ¯ ¯ 3 6 t + 6¯ ¯ ¯ ¯t + 1 −2 ¯ 0 ¯ ¯ ¯ ¯ = ¯ −2 t − 2 −t ¯ ¯ ¯ ¯ 3 6 t¯ ¯ ¯ ¯ ¯ ¯t − 2 −t¯ ¯−2 −t¯ ¯ ¯ ¯ ¯ = (t + 1) ¯ ¯ + 2¯ ¯ ¯ 6 ¯ 3 t¯ t¯ ¯ ¯ ¯ ¯! Ã ¯t − 2 −1¯ ¯−2 −1¯ ¯ ¯ ¯ ¯ = t (t + 1) ¯ ¯ + 2¯ ¯ ¯ 6 ¯ 3 1¯ 1¯ ¡ ¢ = t (t + 1)(t + 4) + 2 = t(t2 + 5t + 6) = t(t + 2)(t + 3). Thus the roots of χA (t) are 0, −2, −3 which are all real numbers. Notice that these each have algebraic multiplicity 1. EigA (0): We need to find the solutions of u 1 −2 −2 u 0 (0I3 − A) v = −2 −2 −2 v = 0 w 3 6 6 w 0 which is equivalent to

1 0 0 u 0 0 1 1 v = 0 0 0 0 w 0

and the general solution is (u, v, w) = (0, z, −z) ∈ C3 . Thus a basis for EigA (0) is (0, 1, −1) which is also a basis for the real vector space EigR A (0). So dimC EigA (0) = dimR EigR A (0) = 1. EigA (−2): We need to find the solutions of u −1 −2 −2 u 0 ((−2)I3 − A) v = −2 −4 −2 v = 0 w 3 6 4 w 0 which is equivalent to

1 2 0 u 0 0 0 1 v = 0 0 0 0 w 0

and the general solution is (u, v, w) = (2z, −z, 0) ∈ C3 . Thus a basis for EigA (−2) is (2, −1, 0) which is also a basis for the real vector space EigR A (−2). So dimC EigA (−2) = dimR EigR A (−2) = 1. EigA (−3): We need to find the solutions of u −2 −2 −2 u 0 ((−3)I3 − A) v = −2 −5 −2 v = 0 w 3 6 3 w 0

62

5. EIGENVALUES AND EIGENVECTORS

which is equivalent to

1 0 1 u 0 = 0 1 0 v 0 0 0 0 w 0

and the general solution is (u, v, w) = (z, 0, −z) ∈ C3 . Thus a basis for EigA (−2) is (1, 0, −1) which is also a basis for the real vector space EigR A (−3). So dimC EigA (−3) = dimR EigR A (−3) = 1. Then (0, 1, −1), (2, −1, 0), (1, 0, −1) is a basis for the complex vector space C3 and for the real vector space R3 . ¤ 0 1 2 Example 5.16. Let B = −1 0 2. Find the eigenvalues of B and show that C3 has a −2 −2 0 basis consisting of eigenvectors of B. For each real eigenvalue λ determine EigR B (λ). Solution. The characteristic polynomial of B is ¯ ¯ ¯ t −1 −2¯ ¯ ¯ ¯ ¯ χB (t) = ¯1 t −2¯ = t3 + 9t = t(t2 + 9) = t(t − 3i)(t + 3i), ¯ ¯ ¯2 2 t¯ so the eigenvalues are 0, 3i, −3i. We find that EigB (0) = {(2z, −2z, z) : z ∈ C}, EigB (3i) = {((1 + 3i)z, (−1 + 3i)z, −4z) : z ∈ C}, EigB (−3i) = {((1 − 3i)z, (−1 − 3i)z, −4z)) : z ∈ C}. For the real eigenvalue 0, EigR B (0) = {(2t, −2t, t) : t ∈ R}. Note that vectors (2t, −2t, t) with t ∈ R and t 6= 0 are the only real eigenvectors of B. Then (2, −2, 1), (1 + 3i, −1 + 3i, −4), (1 − 3i, −1 − 3i, −4) is a basis for the complex vector space C3 . ¤ 0 0 −1 Example 5.17. Let A = 0 2 0. Find the eigenvalues of A and show that C3 has a 1 0 2 basis consisting of eigenvectors of A. For each real eigenvalue λ determine EigR A (λ). Show that 3 C has no basis consisting of eigenvectors for A. Solution. The characteristic polynomial of A is ¯ ¯ ¯ ¯ t 0 1 ¯ ¯ ¯ ¯ χA (t) = ¯ 0 t − 2 0 ¯ = (t − 1)2 (t − 2), ¯ ¯ ¯−1 0 t − 2¯ hence the eigenvalues are 1 with algebraic multiplicity 2, and 2 with algebraic multiplicity 1. EigA (1): We have to solve the equation 1 0 1 u 0 0 v = 0 , 0 −1 −1 0 −1 w 0

5.4. DIAGONALISABILITY OF SQUARE MATRICES

which is equivalent to

63

1 0 1 u 0 0 1 0 v = 0 . 0 0 0 w 0

This gives EigA (1) = {(z, 0, −z) : z ∈ C}. Notice that dim EigA (1) = 1 which is less than the algebraic multiplicity of this eigenvalue. EigA (2): We have to solve the equation 0 u 2 0 1 0 0 0 v = 0 , 0 −1 0 0 w which is equivalent to

1 0 0 u 0 0 0 1 v = 0 . 0 0 0 w 0

This gives EigA (2) = {(0, z, 0) : z ∈ C}. Thus we obtain dim(EigA (1) + EigA (2)) = dim EigA (1) + dim EigA (2) − dim EigA (1) ∩ EigA (2) = 1 + 1 − 0 = 2. So the eigenvectors cannot span C3 . Similarly, EigR A (1) = {(s, 0, −s) : s ∈ R},

EigR A (2) = {(0, t, 0) : t ∈ R},

and R R R R R dimR (EigR A (1) + EigA (2)) = dimR EigA (1) + dimR EigA (2) − dimR EigA (1) ∩ EigA (2)

= 1 + 1 − 0 = 2.

¤

This shows that eigenvalues cannot span C3 . These examples show that when the algebraic multiplicity of an eigenvalue is greater than 1, its geometric multiplicity need not be the same as its algebraic multiplicity. 5.4. Diagonalisability of square matrices Now we will consider the implications of the last section for diagonalisability of square matrices. Definition 5.18. Let A, B ∈ Mn (F ) be n × n matrices with entries in a field F . Then A is similar to B if there is an invertible n × n matrix P ∈ Mn (F ) for which B = P −1 AP.

64

5. EIGENVALUES AND EIGENVECTORS

Remark 5.19. Notice that if B = P −1 AP then A = P BP −1 = (P −1 )−1 BP −1 , so B is also similar to A. If we take P = In = In−1 , then A = In−1 AIn , so A is similar to itself. Finally, if A is similar to B (say B = P −1 AP ) and B is similar to C (say C = Q−1 BQ), then C = Q−1 (P −1 AP )Q = (Q−1 P −1 )A(P Q) = (P Q)−1 A(P Q), hence A is similar to C. These three observations show that the notion of similarity defines an equivalence relation on Mn (F ). Definition 5.20. Let λ1 , . . . , λn ∈ F be scalars. the n × n matrix λ1 . . diag(λ1 , . . . , λn ) = . 0

The diagonal matrix diag(λ1 , . . . , λn ) is ··· .. . ···

0 .. . λn

with the scalars λ1 , . . . , λn down the main diagonal and 0 everywhere else. An n × n matrix A ∈ Mn (F ) is diagonalisable (over F ) if it is similar to a diagonal matrix, i.e., if there is an invertible matrix P for which P −1 AP is a diagonal matrix. Now we can ask the question: When is a square matrix similar to a diagonal matrix? This is answered by our next result. Theorem 5.21. A matrix A ∈ Mn (F ) is diagonalisable over F if and only F n has a basis consisting of eigenvectors of A. Proof. If F n has a basis of eigenvectors of A, say v1 , . . . , vn with associated eigenvalues λ1 , . . . , λn , let P be the invertible matrix with these vectors as its columns. We have h i AP = Av1 · · · Avn h i = λ1 v1 · · · λn vn h i = v1 · · · vn diag(λ1 , . . . , λn ) = P diag(λ1 , . . . , λn ), from which we obtain P −1 AP = diag(λ1 , . . . , λn ), hence A is similar to the diagonal matrix diag(λ1 , . . . , λn ). On the other hand, if A is similar to a diagonal matrix, say P −1 AP = diag(λ1 , . . . , λn ), then AP = P diag(λ1 , . . . , λn ), and the columns of P are eigenvectors associated with the eigenvalues λ1 , . . . , λn and they also form a basis of F n . ¤

5.4. DIAGONALISABILITY OF SQUARE MATRICES

65

The process of finding a diagonal matrix similar to a given matrix is called diagonalisation. It usually works best over C, but sometimes a real matrix can be diagonalised over R if all its eigenvalues are real. Example 5.22. Diagonalise the real matrix 2 −2 3 A = 1 1 1 . 1 3 −1 Solution. First we ¯ ¯t − 2 ¯ ¯ χA (t) = ¯ −1 ¯ ¯ −1 ¯ ¯t − 2 ¯ ¯ =¯ 0 ¯ ¯ −1

find the complex eigenvalues of A. The characteristic polynomial is ¯ 2 −3 ¯¯ ¯ t − 1 −1 ¯ ¯ −3 t + 1¯ ¯ 2 −3 ¯¯ ¯ t + 2 −t − 2¯ ¯ −3 t + 1¯ ¯ ¯ ¯t − 2 2 −3 ¯¯ ¯ ¯ ¯ = (t + 2) ¯ 0 1 −1 ¯ ¯ ¯ ¯ −1 −3 t + 1¯ ¯ ¯ ¯ ¯! Ã ¯ 1 ¯ ¯ 2 −3¯ −1 ¯ ¯ ¯ ¯ = (t + 2) (t − 2) ¯ ¯−¯ ¯ ¯−3 t + 1¯ ¯ 1 −1¯ = (t + 2) ((t − 2)(t + 1 − 3) − (−2 + 3)) ¡ ¢ = (t + 2) (t − 2)2 − 1 = (t + 2)(t2 − 4t + 4 − 1) = (t + 2)(t2 − 4t + 3) = (t + 2)(t − 3)(t − 1).

Hence the roots of χA (t) are the real numbers −2, 1, 3 and these are the eigenvalues of A. Now solving the equation ((−2)I3 − A)x = 0 we find that the general real solution is x = (11t, t, −14t) with t ∈ R. So an eigenvector for the eigenvalue −2 is (11, 1, −14). Solving (I3 − A)x = 0 we find that the general real solution is x = (t, −t, −t) with t ∈ R. So an eigenvector for the eigenvalue 1 is (1, −1, −1). Solving (3I3 − A)x = 0 we find that the general real solution is x = (t, t, t) with t ∈ R. So an eigenvector for the eigenvalue 3 is (1, 1, 1). 3 The vectors (11, 1, −14), (1, −1, −1), (1, 1, 1) form a basis for the real vector space R and 11 1 1 the matrix P = 1 −1 1 satisfies −14 −1 1 11 1 1 −22 1 3 11 1 1 A 1 −1 1 = −2 −1 3 = 1 −1 1 diag(−2, 1, 3), −14 −1 1 28 −1 3 −14 −1 1 hence AP = P diag(−2, 1, 3). Therefore diag(−2, 1, 3) = P −1 AP , showing that A is diagonalisable.

¤

APPENDIX A

Complex solutions of linear ordinary differential equations It often makes sense to look for complex valued solutions of differential equations of the form dn f dn−1 f df (A.1) + a + · · · + a1 + a0 f = 0, n−1 n n−1 dx dx dx where the ar are real or complex numbers which do not depend on x. The solutions are functions defined on R and taking values in C. The number n is called the degree of the differential equation. When n = 1, for any complex number a, the differential equation df + af = 0 dx has as its solutions all the functions of the form te−ax , for t ∈ C. Here we recall that for a complex number z = x + yi with x, y ∈ R, ez = ex+yi = ex eyi = ex (cos y + i sin y) = ex cos y + iex sin y. When n = 2, for any complex numbers a, b, we consider the differential equation d2 f df + bf = 0, +a 2 dx dx where we assume that the polynomial z 2 + az + b has the complex roots α, β, which might be equal. (A.2)

• If α 6= β, then (A.2) has as its solutions all the functions of the form seαx + tseβx , for s, t ∈ C. • If α = β, then (A.2) has as its solutions all the functions of the form seαx + txeαx , for s, t ∈ C. When a, b are real, then either both roots are real or there are two complex conjugate roots α, α; suppose that α = u + vi where u, v ∈ R, hence the complex conjugate of α is α = u − vi. In this case we can write the solutions in the form • peux cos vx + qeux sin vx for p, q ∈ C if v 6= 0, • peux + qxeux for p, q ∈ C if v = 0. In either case, to get real solutions we have to take p, q ∈ R.

67

Bibliography [BLA] [FLA] [LADR]

T. S. Blyth & E. F. Robertson, Basic Linear Algebra, Springer-Verlag (1998), ISBN 3540761225. T. S. Blyth & E. F. Robertson, Further Linear Algebra, Springer-Verlag (2002), ISBN 1852334258. S. Axler, Linear Algebra Done Right, 2nd edition, corrected 7th printing, Springer-Verlag (2004), ISBN 0387982582. [Schaum] S. Lipschutz & M. Lipson, Schaum’s Outline of Linear algebra, McGraw-Hill (2000), ISBN 0071362002. [Strang] G. Strang, Linear Algebra and Its Applications, 4th edition, BrooksCole, ISBN 0534422004.

69

Andrew Baker [08/12/2009]

Department of Mathematics, University of Glasgow. E-mail address: [email protected] URL: http://www.maths.gla.ac.uk/∼ajb

Linear Algebra is one of the most important basic areas in Mathematics, having at least as great an impact as Calculus, and indeed it provides a significant part of the machinery required to generalise Calculus to vector-valued functions of many variables. Unlike many algebraic systems studied in Mathematics or applied within or outwith it, many of the problems studied in Linear Algebra are amenable to systematic and even algorithmic solutions, and this makes them implementable on computers – this explains why so much calculational use of computers involves this kind of algebra and why it is so widely used. Many geometric topics are studied making use of concepts from Linear Algebra, and the idea of a linear transformation is an algebraic version of geometric transformation. Finally, much of modern abstract algebra builds on Linear Algebra and often provides concrete examples of general ideas. These notes were originally written for a course at the University of Glasgow in the years 2006–7. They cover basic ideas and techniques of Linear Algebra that are applicable in many subjects including the physical and chemical sciences, statistics as well as other parts of mathematics. Two central topics are: the basic theory of vector spaces and the concept of a linear transformation, with emphasis on the use of matrices to represent linear maps. Using these, a geometric notion of dimension can be made mathematically rigorous leading its widespread appearance in physics, geometry, and many parts of mathematics. The notes end by discussing eigenvalues and eigenvectors which play a rˆole in the theory of diagonalisation of square matrices, as well as many applications of linear algebra such as in geometry, differential equations and physics. There are some assumptions that the reader will already have met vectors in 2 and 3dimensional contexts, and has familiarity with their algebraic and geometric aspects. Basic algebraic theory of matrices is also assumed, as well as the solution of systems of linear equations using Gaussian elimination and row reduction of matrices. Thus the notes are suitable for a secondary course on the subject, building on existing foundations. There are very many books on Linear Algebra. The Bibliography lists some at a similar level to these notes. University libraries contain many other books that may be useful and there are some helpful Internet sites discussing aspects of the subject.

Contents Chapter 1. Vector spaces and subspaces 1.1. Fields of scalars 1.2. Vector spaces and subspaces

1 1 3

Chapter 2. Spanning sequences, linear independence and bases 2.1. Linear combinations and spanning sequences 2.2. Linear independence and bases 2.3. Coordinates with respect to bases 2.4. Sums of subspaces

11 11 13 22 23

Chapter 3. Linear transformations 3.1. Functions 3.2. Linear transformations 3.3. Working with bases and coordinates 3.4. Application to matrices and systems of linear equations 3.5. Geometric linear transformations

29 29 30 37 41 43

Chapter 4. Determinants 4.1. Definition and properties of determinants 4.2. Determinants of linear transformations 4.3. Characteristic polynomials and the Cayley-Hamilton theorem

45 45 50 51

Chapter 5. Eigenvalues and eigenvectors 5.1. Eigenvalues and eigenvectors for matrices 5.2. Some useful facts about roots of polynomials 5.3. Eigenspaces and multiplicity of eigenvalues 5.4. Diagonalisability of square matrices

55 55 56 58 63

Appendix A.

67

Appendix.

Complex solutions of linear ordinary differential equations

Bibliography

69

i

CHAPTER 1

Vector spaces and subspaces 1.1. Fields of scalars Before discussing vectors, first we explain what is meant by scalars. These are ‘numbers’ of various types together with algebraic operations for combining them. The main examples we will consider are the rational numbers Q, the real numbers R and the complex numbers C. But mathematicians routinely work with other fields such as the finite fields (also known as Galois fields) Fpn which are important in coding theory, cryptography and other modern applications. Definition 1.1. A field of scalars (or just a field ) consists of a set F whose elements are called scalars, together with two algebraic operations, addition + and multiplication ×, for combining every pair of scalars x, y ∈ F to give new scalars x + y ∈ F and x × y ∈ F . These operations are required to satisfy the following rules which are sometimes known as the field axioms. Associativity: For x, y, z ∈ F , (x + y) + z = x + (y + z), (x × y) × z = x × (y × z). Zero and unity: There are unique and distinct elements 0, 1 ∈ F such that for x ∈ F , x + 0 = x = 0 + x, x × 1 = x = 1 × x. Distributivity: For x, y, z ∈ F , (x + y) × z = x × z + y × z, z × (x + y) = z × x + z × y. Commutativity: For x, y ∈ F , x + y = y + x, x × y = y × x. Additive and multiplicative inverses: For x ∈ F there is a unique element −x ∈ F (the additive inverse of x) for which x + (−x) = 0 = (−x) + x. For each non-zero y ∈ F there is a unique element y −1 ∈ F (the multiplicative inverse of y) for which y × (y −1 ) = 1 = (y −1 ) × y. Remark 1.2. • Usually we just write xy instead of x × y, and then we always have xy = yx. 1

2

1. VECTOR SPACES AND SUBSPACES

• Because of commutativity, some of the above rules are redundant in the sense that they are consequences of others. • When working with vectors we will always have a specific field of scalars in mind and will make use of all of these rules. It is possible to remove commutativity or multiplicative inverses and still obtain mathematically interesting structures but in this course we definitely always assume the full strength of these rules. • The most important examples are R and C and it is worthwhile noting that the above rules are obeyed by these as well as Q. However, other examples of number systems such as N and Z do not obey all of these rules. Proposition 1.3. Let F be a field of scalars. For any x ∈ F , (a)

0x = 0,

(b)

− x = (−1)x.

Proof. Consider the following calculations which use many of the rules in Definition 1.1. For x ∈ F , 0x = (0 + 0)x = 0x + 0x, hence 0 = −(0x) + 0x = −(0x) + (0x + 0x) = (−(0x) + 0x) + 0x = 0 + 0x = 0x. This means that 0x = 0 as required for (a). Using (a) we also have x + (−1)x = 1x + (−1)x = (1 + (−1))x = 0x = 0, thus establishing (b).

¤

Example 1.4. Let F be a field. Let a, b ∈ F and assume that a 6= 0. Show that the equation ax = b has a unique solution for x ∈ F . Challenge: Now suppose that aij , b1 , b2 ∈ F for i, j = 1, 2. With the aid of the ‘usual’ method of solving a pair of simultaneous linear equations show that the system ) ( a11 x1 + a12 x2 = b1 a21 x1 + a22 x2 = b2 has a unique solution for x1 , x2 ∈ F if a11 a22 − a12 a21 6= 0. What can be said about solutions when a11 a22 − a12 a21 = 0? Solution. As a 6= 0 there is an inverse a−1 , hence the equation implies that x = 1x = (a−1 a)x = a−1 (ax) = a−1 b, so if x is a solution then it must equal a−1 b. But it is also clear that a(a−1 b) = (aa−1 )b = 1b = b, so this scalar does satisfy the equation. Notice that if a = 0, then the equation 0x = b can only have a solution if b = 0 and in that case any x ∈ F will work so the solution is not unique. Challenge: For this you will need to recall things about 2 × 2 linear systems. The upshot is that the system can have either no or infinitely many solutions. ¤

1.2. VECTOR SPACES AND SUBSPACES

3

1.2. Vector spaces and subspaces We now come to the key idea of a vector space. This involves an abstraction of properties already met with in special cases when dealing with vectors and systems of linear equations. Before meeting the general definition, here are some examples to have in mind as motivation. The plane and 3-dimensional space are usually modelled with coordinates and their points correspond to elements of R2 and R3 which we write in the form (x, y) or (x, y, z). These are usually called vectors and they are added by adding corresponding coordinates. Scalar multiplication is also defined by coordinate-wise multiplication with scalars. These operations have geometric interpretations in terms of the parallelogram rule and dilation of vectors. n It is usual to identify the set of all column vectors of length n with R , where (x1 , . . . , xn ) real x1 . . corresponds to the column vector . . Note that the use of the different types of brackets is xn very important here and will be used in this way throughout the course. Matrix addition and scalar multiplication correspond to coordinate-wise addition and scalar multiplication in Rn . More generally, the set of all m × n real matrices has an addition and scalar multiplication. In the above, R can be replaced by C and all the algebraic properties are still available. Now we give the very important definition of a vector space. Definition 1.5. A vector space over a field of scalars F consists of a set V whose elements are called vectors together with two algebraic operations, + (addition of vectors) and · (multiplication by scalars). Vectors will usually be denoted with boldface symbols such as v which is hand written as v . The operations + and · are required to satisfy the following rules, which are ∼ sometimes known as the vector space axioms. Associativity: For u, v, w ∈ V and s, t ∈ F , (u + v) + w = u + (v + w), (st) · v = s · (t · v). Zero and unity: There is a unique element 0 ∈ V such that for v ∈ V , v + 0 = v = 0 + v, and multiplication by 1 ∈ F satisfies 1 · v = v. Distributivity: For s, t ∈ F and u, v ∈ V , (s + t) · v = s · v + t · v, s · (u + v) = s · u + s · u. Commutativity: For u, v ∈ V , u + v = v + u. Additive inverses: For v ∈ V there is a unique element −v ∈ V for which v + (−v) = 0 = (−v) + v. Again it is normal to write sv in place of s · v when the meaning is clear. Care may need to be exercised in correctly interpreting expressions such as (st)v = (s × t) · v

4

1. VECTOR SPACES AND SUBSPACES

for s, t ∈ F and v ∈ V ; here the product (st) = (s × t) is calculated in F , while (st)v = (st) · v is calculated in V . Note that we do not usually multiply vectors together, although there is a special situation with the vector (or cross) product defined on R3 , but we will not often consider that in this course. From now on, let V be a vector space over a field of scalars F . If F = R we refer to V as a real vector space, while if F = C we refer to it as a complex vector space. Proposition 1.6. For s ∈ F and v ∈ V , the following identities are valid: (a)

0v = 0,

(b)

s0 = 0,

(c)

(−s)v = − (sv) = s(−v),

and in particular, taking s = 1 we have (d)

−v = (−1)v.

Proof. The proofs are similar to those of Proposition 1.3, but need care in distinguishing multiplication of scalars and scalar multiplication of vectors. ¤ Here are some important examples of vector spaces which will be met throughout the course. Example 1.7. Let F be any field of scalars. For n = 1, 2, 3, . . ., let F n denote the set of all n-tuples t = (t1 , . . . , tn ) of elements of F . For s ∈ F and u, v ∈ F n , we define (u1 , . . . , un ) + (v1 , . . . , vn ) =(u1 + v1 , . . . , un + vn ), s · (v1 , . . . , vn ) =s(v1 , . . . , vn ) = (sv1 , . . . , svn ). The zero vector is 0 = (0, . . . , 0). The cases F = R and F = C will be the most important in these notes. Example 1.8. Let F be a field. The set Mm×n (F ) of all m × n matrices with entries in F forms a vector space over F with addition and multiplication by scalars defined in the usual way. We identify the set Mn×1 (F ) with F n , to obtain the vector space of Example 1.7. We also set Mn (F ) = Mn×n (F ). In all cases, the zero vector of Mm×n (F ) is the zero matrix Om×n . Example 1.9. The complex numbers C can be viewed as a vector space over R with the usual addition and scalar multiplication being the usual multiplication. More generally, if V is a complex vector space then we can view it as a real vector space by only allowing real numbers as scalars. For example, Mm×n (C) becomes a real vector space in this way as does Cn . We then refer to the real vector space V as the underlying real vector space of the complex vector space V . Example 1.10. Let F be a field. Consider the set F [X] consisting of all polynomials in X with coefficients in F . We define addition and multiplication by scalars in F by (a0 + a1 X + · · · + ar X r ) + (b0 + b1 X + · · · + br X r ) =(a0 + b0 ) + (a1 + b1 )X + · · · + (ar + br )X r , s · (a0 + a1 X + · · · + ar X r ) =(sa0 ) + (sa1 )X + · · · + (sar )X r . The zero vector is the zero polynomial 0 = 0 + 0X + · · · + 0X r = 0.

1.2. VECTOR SPACES AND SUBSPACES

5

Example 1.11. Take R to be the field of scalars and consider the set D(R) consisting of all infinitely differentiable functions R −→ R, i.e., functions f : R −→ R for which all possible derivatives f (n) (x) exist for n > 1 and x ∈ R. We define addition and multiplication by scalars as follows. Let t ∈ R and f, g ∈ D(R), then f + g and s · f are the functions given by the following rules: for x ∈ R, (f + g)(x) = f (x) + g(x), (s · f )(x) = sf (x). Of course we really ought check that all the derivatives of f + g and s · f exist so that these function are in D(R), but this is an exercise in Calculus. The zero vector here is the constant function which sends every real number to 0, and this is usually written 0 in Calculus, although this might seem confusing in our context! More generally, for a given real number a, it is also standard to write a for the constant function which sends every real number to a. Note that Example 1.10 can also be viewed as a vector space consisting of functions since every polynomial over a field can be thought of as the function obtained by evaluation of the variable at values in F . When F = R, R[X] ⊆ D(R) and we will see that this is an example of a vector subspace which will be defined soon after some motivating examples. Example 1.12. Let V be a vector space over a field of scalars F . Suppose that W ⊆ V contains 0 and is closed under addition and multiplication by scalars, i.e., for s ∈ F and u, v ∈ W , we have u + v ∈ W, su ∈ W. Then W is also a vector space over F . Example 1.13. Consider the case where F = R and V = R2 . Then the line L with equation 2x − y = 0 consists of all vectors of the form (t, 2t) with t ∈ R, i.e., L = {(t, 2t) ∈ R2 : t ∈ R}. Clearly this passes through the origin so it contains 0. It is also closed under addition and scalar multiplication: if (t, 2t), (t1 , 2t1 ), (t2 , 2t2 ) ∈ L and s ∈ R then (t1 , 2t1 ) + (t2 , 2t2 ) = (t1 + t2 , 2t1 + 2t2 ) = (t1 + t2 , 2(t1 + t2 )) ∈ L, s(t, 2t) = (st, 2(st)) ∈ L. More generally, any line with equation of form ax + by = 0 with (a, b) 6= 0 = (0, 0) is similarly closed under addition and multiplication by scalars. Example 1.14. Consider the case where F = R and V = R3 . If (a, b, c) ∈ R3 is non-zero, then consider the plane P with equation ax + by + cz = 0, thus as a set P is given by P = {(x, y, z) ∈ R3 : ax + by + cz = 0}.

6

1. VECTOR SPACES AND SUBSPACES

Then P contains 0 and is closed under addition and scalar multiplication since whenever (x, y, z), (x1 , y1 , z1 ), (x2 , y2 , z2 ) ∈ P and s ∈ R, the sum (x1 , y1 , z1 ) + (x2 , y2 , z2 ) = (x1 + x2 , y1 + y2 , z1 + z2 ) satisfies a(x1 + x2 ) + b(y1 + y2 ) + c(z1 + z2 ) = (ax1 + by1 + cz1 ) + (ax2 + by2 + cz2 ) = 0, while the scalar multiple s(x, y, z) = (sx, sy, sz) satisfies a(sx) + b(sy) + c(sz) = s(ax + by + cz) = 0, showing that (x1 , y1 , z1 ) + (x2 , y2 , z2 ), s(x, y, z) ∈ P. R3

A line in through the origin consists of all vectors of the form tu for t ∈ R, where u is some non-zero direction vector. This is also closed under addition and scalar multiplication. Definition 1.15. Let V be a vector space over a field of scalars F . Suppose that the subset W ⊆ V is non-empty and is closed under addition and multiplication by scalars, thus it forms a vector space over F . Then W is called a (vector ) subspace of V . Remark 1.16. • In order to verify that W is non-empty it is usually easiest to show that it contains the zero vector 0. • The conditions for W to be closed under addition and scalar multiplication are equivalent to the single condition that for all s, t ∈ F and u, v ∈ W , su + tv ∈ W. • If W ⊆ V is a subspace, then for any w ∈ W , all of its scalar multiples (including 0w = 0 and (−1)w = −w) are also in W . Example 1.17. Let V be a vector space over a field F . • The sets {0} and V are both subspaces of V , usually thought of as its uninteresting subspaces. • If v ∈ V is a non-zero vector, then {v} is never a subspace of V since (for example) 0v = 0 is not an element of {v}. • The smallest subspace of V containing a given non-zero vector v is the line spanned by v which is the set of all scalar multiplies of v, {tv : t ∈ F }. Before giving more examples, here are some non-examples. Example 1.18. Consider the real vector space R2 and the following subsets with the same addition and scalar multiplication. (a) V1 = {(x, y) ∈ R2 : x > 0} is not a real vector space: for example, given any vector (x, y) ∈ V1 with x > 0, there is no additive inverse −(x, y) ∈ V1 since this would have to be (−x, −y) which has −x < 0. So V1 is not closed under addition. (b) V2 = {(x, y) ∈ R2 : y = x2 } is not a real vector space: for example, (1, 1) ∈ V2 but 2(1, 1) = (2, 2) does not satisfy y = x2 since x2 = 4 and y = 2. So V2 is not closed under scalar multiplication. (c) V3 = {(x, y) ∈ R2 : x = 2} is not a real vector space: for example, (2, 0), (2, 1) ∈ V3 but (2, 0) + (2, 1) = (4, 1) ∈ / V3 , so V3 is not closed under addition.

1.2. VECTOR SPACES AND SUBSPACES

7

Here are some further examples of subspaces. Example 1.19. Suppose that F is a equations a11 x1 + .. (1.1) . a x + m1 1

field and that we have a system of homogenous linear ··· .. .

+

a1n xn .. .

···

+ amn xn

= 0 .. . = 0

x1 . n . for scalars aij ∈ F . Then the set W of solutions (x1 , . . . , xn ) = . in F of the system (1.1) xn is a subspace of F n . This is most easily checked by recalling that the system is equivalent to a single matrix equation

(1.2)

Ax = 0,

where A = [aij ] and then verifying that given solutions x, y ∈ W and s ∈ F , A(x + y) = Ax + Ay = 0 + 0 = 0, and A(sx) = s(Ax) = s(A0) = s0 = 0. This example indicates an extremely important relationship between vector space ideas and linear equations, and this is one of the main themes in the subject of Linear Algebra. Example 1.20. Recall the vector space F [X] of Example 1.10. Consider the following subsets of F [X]: W0 = F [X], W1 = {f (X) ∈ F [X] : f (0) = 0}, W2 = {f (X) ∈ F [X] : f (0) = f 0 (0) = 0}, and in general, for n > 1, Wn = {f (X) ∈ F [X] : for k = 0, . . . , n − 1, f (k) (0) = 0}. Show that each Wn is a subspace of F [X] and that Wn is a subspace of Wn−1 . Solution. Let n > 1. Suppose that s, t ∈ F and f (X), g(X) ∈ Wn . Then for each k = 0, . . . , n − 1 we have f (k) (0) = 0 = g (k) (0), hence

dk (sf (X) + tg(X)) = sf (k) (X) + tg (k) (X), dX k and evaluating at X = 0 gives dk (sf + tg)(0) = sf (k) (0) + tg (k) (0) = 0. dX k This shows that Wn is closed under addition and scalar multiplication, therefore it is a subspace of F [X]. Clearly Wn ⊆ Wn−1 and as it is closed under addition and scalar multiplication, it is a subspace of Wn−1 . ¤

8

1. VECTOR SPACES AND SUBSPACES

We can combine subspaces to form new subspaces. Given subsets Y1 , . . . , Yn ⊆ X of a set X, their intersection is the subset Y1 ∩ Y2 ∩ · · · ∩ Yn = {x ∈ X : for k = 1, . . . , n, x ∈ Yk } ⊆ X, while their union is the subset Y1 ∪ Y2 ∪ · · · ∪ Yn = {x ∈ X : there is a k = 1, . . . , n such that x ∈ Yk } ⊆ X, Proposition 1.21. Let V be a vector space. If W1 , . . . , Wn ⊆ V are subspaces of V , then W1 ∩ · · · ∩ Wn is a subspace of V . Proof. Let s, t ∈ F and u, v ∈ W1 ∩ · · · ∩ Wn . Then for each k = 1, . . . , n we have u, v ∈ Wk , and since Wk is a subspace, su + tv ∈ Wk . But this means that su + tv ∈ W1 ∩ · · · ∩ Wn , i.e., W1 ∩ · · · ∩ Wn is a subspace of V .

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On the other hand, the union of subspace does not always behave so well. Example 1.22. Consider the real vector space R2 and the subspaces W 0 = {(s, 0) : s ∈ R},

W 00 = {(0, t) : t ∈ R}.

Show that W 0 ∪ W 00 is not a subspace of R2 and determine W 0 ∩ W 00 . Solution. The union of these subspaces is W 0 ∪ W 00 = {v ∈ R2 : v = (s, 0) or v = (0, s) for some s ∈ R}. Then (1, 0) ∈ W 0 and (0, 1) ∈ W 00 , but (1, 1) = (1, 0) + (0, 1) is visibly not in W 0 ∪ W 00 , so this is not closed under addition. (However, it is closed under scalar multiplication, so care is required when checking this sort of example.) If x ∈ W 0 ∩ W 00 then x = (s, 0) and x = (0, t) for some s, t ∈ R, which is only possible if s = t = 0 and so x = (0, 0). Therefore W 0 ∩ W 00 = {0}. ¤ Here is another way to think about Example 1.19. Example 1.23. Suppose that F is a equations a11 x1 + .. . a x + m1 1

field and that we have a system of homogenous linear ··· .. .

+

a1n xn .. .

···

+ amn xn

= 0 .. . = 0

with the aij ∈ F . For each r = 1, . . . , m, let Wr = {(x1 , . . . , xn ) : ar1 x1 + · · · + arn xn = 0} ⊆ F n , which is a subspace of F n .

1.2. VECTOR SPACES AND SUBSPACES

9

Then the set W of solutions (x1 , . . . , xn ) of the above linear system is given by the intersection W = W1 ∩ · · · ∩ Wm , therefore W is a subspace of F n . Example 1.24. In the real vector space R4 , determine the intersection U ∩V of the subspaces U = {(w, x, y, z) ∈ R4 : x + 2y = 0 = w + 2x − z},

V = {(w, x, y, z) ∈ R4 : w + x + 2y = 0}.

Solution. Before doing this we first use elementary row operations to find the general solutions of the systems of equations used to define U and V . For U we consider the system ) ( x + 2y = 0 w + 2x − z = 0 with augmented matrix # " # " # " 1 2 0 −1 0 1 0 −4 −1 0 0 1 2 0 0 ∼ ∼ , 2 0 0 1 2 0 −1 0 R1 ↔R2 0 1 2 0 0 R1 →R1 −2R2 0 1 where the last matrix is a reduced echelon matrix. So the general solution vector of this system is (4s + t, −2s, s, t) where s, t ∈ R. Thus we have U = {(4s + t, −2s, s, t) ∈ R4 : s, t ∈ R}. Similarly, the general solution vector of the equation w + x + 2y = 0 is (−r − 2s, r, s, t) where r, s, t ∈ R and we have V = {(−r − 2s, r, s, t) ∈ R4 : r, s, t ∈ R}. Now to find U ∩ V we have to solve the system x + 2y = 0 w + 2x − z = 0 w + x + 2y = 0 with augmented matrix 1 0 1 2 0 0 1 2 0 −1 0 ∼ ∼ 0 1 2 0 0 1 2 0 −1 0 0 R3 →R3 −R1 R1 ↔R2 0 1 1 2 0 0 1 1 2 0 0 1 0 −4 −1 0 1 ∼ ∼ 1 2 0 0 0 0 R1 →R1 −2R2 R3 →(1/4)R3 R3 →R3 +R1 0 0 4 1 0 0

2 0 −1 0 1 2 0 0 −1 2 1 0 0 −4 1 2 0 1

−1 0 0 0 1/4 0

1 0 0 0 0 ∼ 0 1 0 −1/2 0 , R1 →R1 +4R3 R2 →R2 −2R3 0 0 1 1/4 0

where the last matrix is reduced echelon. The general solution vector is (0, t/2, −t/4, t) where t ∈ R, thus we have U ∩ V = {(0, t/2, −t/4, t) : t ∈ R} = {(0, 2s, −s, 4s) : s ∈ R}.

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CHAPTER 2

Spanning sequences, linear independence and bases When working with R2 or R3 we often use the fact that every vector v can be expressed in terms of the standard basis vectors e1 , e2 or e1 , e2 , e3 i.e., v = xe1 + ye2

or v = xe1 + ye2 + ze3 ,

where v = (x, y) or v = (x, y, z). Something similar is true for every vector space and involves the concepts of spanning sequence and basis. 2.1. Linear combinations and spanning sequences We will always assume that V is a vector space over a field F . The first notion we will require is that of a linear combination. Definition 2.1. If t1 , . . . , tr ∈ F and v1 , . . . , vr is a sequence of vectors in V (some of which may be 0 or may be repeated), then the vector t1 v1 + · · · + tr vr ∈ V is said to be a linear combination of the sequence v1 , . . . , vr (or of the vectors vi ). Sometimes it is useful to allow r = 0 and view the 0 as a linear combination of the elements of the empty sequence ∅. A sequence is often denoted (v1 , . . . , vr ), but we will not use that notation to avoid confusion with the notation for elements of F n . We will sometimes refer to a sequence v1 , . . . , vr of vectors in V as a sequence in V . There are various operations which can be performing on sequences such as deleting terms, reordering terms, multiplying terms by non-zero scalars, combining two sequences into a new one by concatinating them in either order. Notice that 0 can be expressed as a linear combination in other ways. For example, when F = R and V = R2 , 3(1, −1) + (−3)(2, 0) + 3(1, 1) = 0,

1(0, 0) = 0,

1(5, 0) + 0(0, 0) + (−5)(1, 0) = 0,

so 0 is a linear combination of each of the sequences (1, −1), (2, 0), (1, 1),

0,

(5, 0), 0, (1, 0).

For any sequence v1 , . . . , vr we can always write 0 = 0v1 + · · · + 0vr , so 0 is a linear combination of any sequence (even the empty sequence of length 0). Sometimes it is useful to allow infinite sequences as in the next example. 11

12

2. SPANNING SEQUENCES, LINEAR INDEPENDENCE AND BASES

Example 2.2. Let F be a field and consider the vector space of polynomials F [X]. Then every f (X) ∈ F [X] has a unique expression of the form f (X) = a0 + a1 X + · · · + ar X r + · · · , where ai ∈ F and for large i, ai = 0. Thus every polynomial in F [X] can be uniquely written as a linear combination of the sequence 1, X, X 2 , . . . , X r , . . .. Such a special set of vectors is called a basis and soon we will develop the theory of bases. Bases are extremely important and essential for working in vector spaces. In applications such as to Fourier series, there are bases consisting of trigonometric functions, while in other context orthogonal polynomials play this rˆole. Definition 2.3. Let S : w1 , . . . be a sequence (possibly infinite) of elements of V . Then S is a spanning sequence for V (or is said to span V ) if for every element v ∈ V there is an expression of the form v = s1 w1 + · · · + sk wk , where s1 , . . . , sk ∈ F , i.e., every vector in V is a linear combination of the sequence S. Notice that in this definition we do not require any sort of uniqueness of the expansion. Definition 2.4. Let S : w1 , . . . be a sequence of elements of V . Then the linear span of S, Span(S) ⊆ V , is the subset consisting of all linear combinations of S, i.e., Span(S) = {t1 w1 + · · · + tr wr : r = 0, 1, 2, . . . , ti ∈ F }. Remark 2.5. • By definition, S is a spanning sequence for Span(S). • We always have 0 ∈ Span(S). • If S = ∅ is the empty sequence, then we make the definition Span(S) = {0}. Lemma 2.6. Let S : w1 , . . . be a sequence of elements of V . Then the following hold. (a) Span(S) ⊆ V is a subspace and S is a spanning sequence of Span(S). (b) Every subspace of V containing the elements of S contains Span(S) as a subspace, therefore Span(S) is the smallest such subspace. (c) If S 0 is a subsequence of S, then Span(S 0 ) ⊆ Span(S). In (c), we use the idea that S 0 is a subsequence of S if it is obtained from S by removing some terms and renumbering. For example, if S : u, v, w is a sequence then the following are subsequences: S 0 : u, v,

S 00 : u, w,

S 000 : v.

Proof. (a) It is easy to see that any sum of linear combinations of elements for S is also such a linear combination, similarly for scalar multiples. (b) If a subspace W ⊆ V contains all the elements of S, then all linear combinations of S are in W , therefore Span(S) ⊆ W . (c) This follows from (b) since the elements of S 0 are all in Span(S). ¤ Definition 2.7. The vector space V is called finite dimensional if it has a spanning sequence S : w1 , . . . , w` of finite length `.

2.2. LINEAR INDEPENDENCE AND BASES

13

Much of the theory we will develop will be about finite dimensional vector spaces although many important examples are infinite dimensional. Here are some further useful properties of sequences and the subspaces they span. Proposition 2.8. Let S : w1 , . . . be a sequence in V . Then the following hold. (a) If S 0 : w10 , . . . is a sequence obtained by permuting (i.e., reordering) the terms of S, then Span(S 0 ) = Span(S). Hence the span of a sequence is unchanged by permuting its terms. (b) If r > 1 and wr ∈ Span(w1 , . . . , wr−1 ), then for the sequence S 00 : w1 , . . . , wr−1 , wr+1 , . . . , we have Span(S 00 ) = Span(S). In particular, we can remove 0’s and any repetitions of vectors without changing the span of a sequence. (c) If s > 1 and S 000 is obtained from S by replacing ws by a vector of the form ws000 = (x1 w1 + · · · + xs−1 ws−1 ) + ws , where x1 , . . . , xs−1 ∈ F , then Span(S 000 ) = Span(S). (d) If t1 , . . . is sequence of non-zero scalars, then for the sequence T : t1 w1 , t2 w2 , . . . we have Span(T ) = Span(S). Proof. (a),(b),(d) are simple consequences of the commutativity, associativity and distributivity of addition and scalar multiplication of vectors. (c) follows from the obvious fact that Span(w1 , . . . , ws−1 , ws000 ) = Span(w1 , . . . , ws−1 , ws ). ¤ 2.2. Linear independence and bases Now we introduce a notion which is related to the uniqueness issue. Definition 2.9. Let S : w1 , . . . be a sequence in V . • S is linearly dependent if for some r > 1 there are scalars s1 , . . . , sr ∈ F not all zero and for which s1 w1 + · · · + sr wr = 0. • S is linearly independent if it is not linearly dependent. This means that whenever s1 , . . . , sr ∈ F satisfy s1 w1 + · · · + sr wr = 0, then s1 = · · · = sr = 0. Example 2.10. • Any sequence S containing 0 is linearly dependent since 1 · 0 = 0. • If a vector v occurs twice in S then 1v + (−1)v = 0, so again S is linearly dependent. • If a vector in S can be expressed as a linear combination of the list obtained by deleting it, then S is linearly dependent. • In the real vector space R2 , the sequence (1, 2), (3, −1), (0, 7) is linearly dependent since 3(1, 2) + (−1)(3, −1) + (−1)(0, 7) = (0, 0), while its subsequence (1, 2), (3, −1) is linearly independent.

14

2. SPANNING SEQUENCES, LINEAR INDEPENDENCE AND BASES

Lemma 2.11. Suppose that S : w1 , . . . is a linearly dependent sequence in V . Then there is some k > 1 and scalars t1 , . . . , tk−1 ∈ F for which wk = t1 w1 + · · · + tk−1 wk−1 . Hence if S 0 is the sequence obtained by deleting wk then Span(S 0 ) = Span(S). Proof. Since S is linearly dependence, there must be an r > 1 and scalars s1 , . . . , sr ∈ F where not all of the si are zero and they satisfy s1 w1 + · · · + sr wr = 0. Let k be the largest value of i for which si 6= 0. Then s1 w1 + · · · + sk wk = 0. with sk 6= 0. Multiplying by s−1 k gives −1 s−1 k (s1 w1 + · · · + sk wk ) = sk 0 = 0,

whence −1 −1 −1 −1 −1 s−1 k s1 w1 + sk s2 w2 + · · · + sk sk wk = sk s1 w1 + sk s2 w2 + · · · + sk sk−1 wk−1 + wk = 0.

Thus we have −1 wk = (−s−1 k s1 )w1 + · · · + (−sk sk−1 )wk−1 .

The equality of the two spans is clear.

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The next result is analogous to Proposition 2.8 for spanning sequences. Proposition 2.12. Let S : w1 , . . . be a sequence in V . Then the following hold. (a) If S 0 : w10 , . . . is a sequence obtained by permuting the terms of S, then S 0 is linearly dependent (respectively linearly independent) if S is. (b) If s > 1 and S 00 is obtained from S by replacing ws by a vector of the form ws00 = (x1 w1 + · · · + xs−1 ws−1 ) + ws , where x1 , . . . , xs−1 ∈ F , then S 00 is linearly dependent (respectively linearly independent) if S is. (c) If t1 , . . . is sequence of non-zero scalars, then for the sequence T : t1 w1 , t2 w2 , . . . , we have T linearly dependent (resp. linearly independent) if S is. Proof. Again these are simple consequences of properties of addition and scalar multiplication. ¤ Proposition 2.13. Suppose that S : w1 , . . . be a linearly independent sequence in V . If for some k > 1, the scalars s1 , . . . , sk , t1 , . . . , tk ∈ F satisfy s1 w1 + · · · + sk wk = t1 w1 + · · · + tk wk , then s1 = t1 , . . . , sk = tk . Hence every vector has at most one expression as a linear combination of S.

2.2. LINEAR INDEPENDENCE AND BASES

15

Proof. The equation can be rewritten as (s1 − t1 )w1 + · · · + (sk − tk )wk = 0. Now linear independence shows that each coefficient satisfies (si − ti ) = 0.

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Proposition 2.14. Let S : w1 , . . . be a spanning sequence for V . Then there is a subsequence T of S which is a spanning sequence for V and is also linearly independent. Proof. We indicate a proof for the case when S is finite, the general case is slightly more involved. Notice that Span(S) = V and for some n > 1, the sequence is S : w1 , . . . , wn . If S is already linearly independent then we can take T = S. Otherwise, by Lemma 2.11, one of the wi can be expressed as a linear combination of the others; let k1 be the largest such i. Then wk1 can be expressed as a linear combination of the list S1 : w1 , . . . wk1 −1 , wk1 +1 , . . . , wn and we have Span(S1 ) = Span(S) = V , so S1 is a spanning sequence for V . The length of S1 is n − 1 < n, so this sequence is shorter than S. Clearly we can repeat this obtaining new spanning sequences S1 , S2 , . . . for V , where the lengths of Sr is n − r. Eventually we must get to a spanning sequence Sk which cannot be reduced further, but this must be linearly independent. ¤ Example 2.15. Taking F = R and V = R2 , show that the sequence S : (1, 2), (1, 1), (−1, 0), (1, 3) spans R2 and find a linearly independent spanning subsequence of S. Solution. Following the idea in the proof of Proposition 2.14, first observe that (1, 3) = 3(1, 1) + 2(−1, 0), so we can replace S by the sequence S 0 : (1, 2), (1, 1), (−1, 0) for which Span(S 0 ) = Span(S). Next notice that (−1, 0) = (1, 2) + (−2)(1, 1), so we can replace S 0 by the sequence S 00 : (1, 2), (1, 1) for which Span(S 00 ) = Span(S 0 ) = Span(S). For every (a, b) ∈ R2 , the equation x(1, 2) + y(1, 1) = (a, b) corresponds to the matrix equation

" #" # " # 1 1 x a = , 2 1 y b

" # 1 1 and since det = −1 6= 0, this has the unique solution 2 1 " # " #−1 " # " #" # x 1 1 a −1 1 a = = . y 2 1 b 2 −1 b Thus S 00 spans R2 and is linearly independent. The next definition is very important.

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16

2. SPANNING SEQUENCES, LINEAR INDEPENDENCE AND BASES

Definition 2.16. A linearly independent spanning sequence of a vector space V is called a basis of V . The next result shows why bases are useful and leads to the notion of coordinates with respect to a basis which will be introduced in Section 2.3. Proposition 2.17. Let V be a vector space over F . A sequence S in V is a basis of V if and only if every vector of V has a unique expression as a linear combination of S. Proof. This follows from Proposition 2.13 together with the fact that a basis is a spanning sequence. ¤ Theorem 2.18. If S is a spanning sequence of V , then there is a subsequence of S which is a basis. In particular, if V is finite dimensional then it has a finite basis. Proof. The first part follows from Proposition 2.14. For the second part, start with a finite spanning sequence then find a subsequence which is a basis, and note that this sequence is finite. ¤ Example 2.19. Consider the vector space V = R3 over R. Show that the sequence S : (1, 0, 1), (1, 1, 1), (1, 2, 1), (1, −1, 2) is a spanning sequence of V and find a subsequence which is a basis. Solution. The following method for the first part uses the general approach to solving systems of linear equations. Given any numbers h1 , h2 , h3 ∈ R, consider the equation x1 (1, 0, 1) + x2 (1, 1, 1) + x3 (1, 2, 1) + x4 (1, −1, 2) = (h1 , h2 , h3 ) = h. This is equivalent to the system x + x + x + x = h 1 2 3 4 1 x2 + 2x3 − x4 = h2 x1 + x2 + x3 + 2x4 = h3 which has augmented matrix

Performing elementary row 1 1 [A | h] ∼ 0 1 R3 →R3 −R1 0 0

1 1 1 1 h1 [A | h] = 0 1 2 −1 h2 . 1 1 1 2 h3

operations we find that 1 1 h1 1 ∼ 2 −1 h2 0 R1 →R1 −R2 0 1 h3 − h1 0 1 ∼ 0 R1 →R1 −2R3 R2 →R2 +R3 0

0 −1 2 h1 − h2 1 2 −1 h2 0 0 1 h3 − h1

0 −1 0 3h1 − h2 − 2h3 1 2 0 −h1 + h2 + h3 , 0 0 1 −h1 + h3

where the last matrix is reduced echelon. So the system is consistent with general solution x1 = s + 3h1 − h2 − 2h3 ,

x2 = −2s − h1 + h2 + h3 ,

x3 = s,

x4 = −h1 + h3

So every element of R3 can indeed be expressed as a linear combination of S.

(s ∈ R).

2.2. LINEAR INDEPENDENCE AND BASES

17

Notice that when h = 0 we have the solution x1 = s,

x2 = −2s,

x3 = s,

x4 = 0

(s ∈ R).

Thus shows that S is linearly dependent. Taking for example s = 1, we see that (1, 0, 1) − 2(1, 1, 1) + (1, 2, 1) + 0(1, −1, 2) = 0, hence (1, 2, 1) = (−1)(1, 0, 1) + 2(1, 1, 1) + 0(1, −1, 2), so the sequence T : (1, 0, 1), (1, 1, 1), (1, −1, 2) also spans R3 . Notice that there is exactly one solution of the above system of equations for h = 0 with s = 0, so T is linearly independent and therefore it is a basis for R3 . ¤ The approach of the last example leads to a general method. We state it over any field F since the method works in general. For this we need to recall the idea of a reduced echelon matrix and the way this is used to determine the general solution of a system of linear equations. Theorem 2.20. Suppose that a1 , . . . , an is a sequence of vectors in F m . Let A = [a1 · · · an ] be the m × n matrix with aj as its j-th column and let A0 be its reduced echelon matrix. (a) a1 , . . . , an is a spanning sequence of F m if and only if the number of non-zero rows of A0 is m. (b) If the j-th column of A0 does not contain a pivot, then the corresponding vector aj is expressible as a linear combination of those associated with columns which do contain a pivot. Such a linear combination can found by setting the free variable for the j-th column equal to 1 and all the other free variables equal to 0. (c) The vectors ai for those i where the i-th column of A0 contains a pivot, form a basis for Span(a1 , . . . , an ). Here is another kind of situation. Example 2.21. Consider the vector space R3 over R. Find a basis for the plane W with equation x − 2y + 3z = 0. Solution. This h i time we need to start with a system of one equation and its associated matrix 1 −2 3 which is reduced echelon, hence we have the general solution x = 2s − 3t,

y = s,

z=t

(s, t ∈ R).

So every vector (x, y, z) in W is a linear combination (x, y, z) = s(2, 1, 0) + t(−3, 0, 1) of the sequence (2, 1, 0), (−3, 0, 1) obtained by setting s = 1, t = 0 and s = 0, t = 1 respectively. In fact this expression is unique since these two vectors are linearly independent. ¤ Example 2.22. Consider the vector space D(R) over R discussed in Example 1.11. Let W ⊆ D(R) be the set of all solutions of the differential equation dy + 3y = 0. dx Show that W is a subspace of D(R) and find a basis for it.

18

2. SPANNING SEQUENCES, LINEAR INDEPENDENCE AND BASES

Solution. It is easy to see that W is a subspace. Notice that if y1 , y2 are two solutions then d(y1 + y2 ) dy1 dy2 + 3(y1 + y2 ) = + + 3y1 + 3y2 dx dx ¶ µ µdx ¶ dy1 dy2 = + 3y1 + + 3y2 = 0, dx dx hence y1 + y2 is a solution. Also, if y is a solution and s ∈ R, then d(sy) dy + 3(sy) = s + s(3y) dx dx µ ¶ dy =s + 3y = 0, dx so sy is a solution. Clearly the constant function 0 is a solution. The general solution is y = ae−3x

(a ∈ R),

so the function e−3x spans W . Furthermore, if for some t ∈ R, se−3x = 0, where the right hand side means the constant function, this has to hold for every x ∈ R. Taking x = 0 we find that t = te0 = 0, so t = 0. Hence the sequence e−3x is a basis of W .

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A good exercise is to try to generalise this, for example by considering the solutions of the differential equation d2 y − 6y = 0. dx2 There is a brief review of the solution of ordinary differential equations in Appendix A. Here is another example. Example 2.23. Let W ⊆ D(R) be the set of all solutions of the differential equation d2 y + 4y = 0. dx2 Show that W is a subspace of D(R) and find a basis for it. Solution. Checking W is a subspace is similar to Example 2.22 The general solution has the form y = a cos 2x + b sin 2x

(a, b ∈ R),

so the functions cos 2x, b sin 2x span W . Suppose that for some s, t ∈ R, s cos 2x + t sin 2x = 0, where the right hand side means the constant function. This has to hold whatever the value of x ∈ R, so try taking values x = 0, π/4. These give the equations s cos 0 + t sin 0 = 0,

s cos π/2 + t sin π/2 = 0,

i.e., s = 0, t = 0. So cos 2x, sin 2x are linearly independent and hence cos 2x, sin 2x is a basis. Notice that this is a finite dimensional vector space. ¤

2.2. LINEAR INDEPENDENCE AND BASES

19

Example 2.24. Let R∞ denote the set consisting of all sequences (an ) of real numbers which satisfy an = 0 for all large enough n, i.e., (an ) = (a1 , a2 , . . . , ak , 0, 0, 0, . . .). This can be viewed as a vector space over R with addition and scalar multiplication given by (an ) + (bn ) = (an + bn ), t · (cn ) = (tcn ). Then the sequences er = (0, . . . , 0, 1, 0, . . .) with a single 1 in the r-th place, form a basis for R∞ . Solution. First note that (xn ) = (x1 , x2 , . . . , xk , 0, 0, 0, . . .) can be expressed as (xn ) = x1 e1 + · · · + xk ek , so the ei span R∞ . If t1 e1 + · · · + tk ek = (0), then (tn ) = (t1 , . . . , tk , 0, 0, 0, . . .) = (0, . . . , 0, . . .), and so tk = 0 for all k. Thus the ei are also linearly independent.

¤

Here is another important method for constructing bases. Theorem 2.25. Let V be a finite dimensional vector space and let S be a linearly independent sequence in V . Then there is a basis S 0 of V in which S is a subsequence. In particular, S and S 0 are both finite. Proof. We will indicate a proof for the case where S is a finite sequence. Choose a finite spanning sequence, say T : w1 , . . . , wk . By Proposition 2.14 we might as well assume that T is linearly independent. If S does not span V then certainly S, T (the sequence obtained by joining S and T together) spans (since T does). Now choose the largest value of ` = 1, . . . , k for which the sequence S` = S, w1 , . . . , w` is linearly independent. Then S, T is linearly dependent and if ` < k, we can write wk as a linear combination of S, w1 , . . . , wk−1 , so this is still a spanning set for V . Similarly we can keep discarding the vectors wk−j for j = 0, . . . , (k − ` − 1) and at each stage we have a spanning sequence S, w1 , . . . , wk−j−1 . Eventually we are left with just S` which is still a spanning set for V as well as being linearly independent, and this is the required basis. ¤ Theorem 2.26. Let V be a finite dimensional vector space and suppose that it has a basis with n elements. Then any other basis is also finite and has n elements. This result allows us to make the following important definition.

20

2. SPANNING SEQUENCES, LINEAR INDEPENDENCE AND BASES

Definition 2.27. Let V be a finite dimensional vector space. Then the number of elements in any basis of V is finite and is called the dimension of the vector space V and is denoted dimF V or just dim V if the field is clear. Example 2.28. Let F be any field. Then the vector space F n over F has dimension n, i.e., dimF F n = n. Proof. For each k = 1, 2, . . . , n let ek = (0, . . . , 0, 1, 0, . . . , 0) (with the 1 in the k-th place). Then the sequence e1 , . . . , en spans F n since for any vector (x1 , . . . , xn ) ∈ F n , (x1 , . . . , xn ) = x1 e1 + · · · + xn en . Also, if the right hand side is 0, then (x1 , . . . , xn ) = (0, . . . , 0) and so x1 = · · · = xn = 0, showing that this sequence is linearly independent. Therefore, e1 , . . . , en is a basis for F n and dimF F n = n. The basis e1 , . . . , en of F n is often called the standard basis of F n . ¤ Proof of Theorem 2.26. Let B : b1 , . . . , bn be a basis and suppose that S : w1 , . . . is another basis for V . We will assume that S is finite, say S : w1 , . . . , wm , the more general case is similar. The sequence w1 , b1 , . . . , bn must be linearly dependent since B spans V . As B is linearly independent, there is a number k1 = 1, . . . , n for which bk1 = s0 w1 + s1 b1 + · · · + sk1 −1 bk1 −1 for some scalars s0 , s1 . . . , sk1 −1 . By reordering B, we can assume that k1 = n, so bn = s0 w1 + s1 b1 + · · · + sn−1 bn−1 , where s0 6= 0 since B is linearly independent. Thus the sequence B1 : w1 , b1 , . . . , bn−1 spans V . If it were linearly dependent then there would be an equation of the form t0 w1 + t1 b1 + · · · + tn−1 bn−1 = 0 in which not all the ti are zero; in fact t0 6= 0 since B is linearly independent. From this we obtain a non-trivial linear combination of the form t01 b1 + · · · + t0n bn = 0, which is impossible. Thus B1 is a basis. Now we repeat this argument with w2 , w3 , and so on, at each stage forming (possibly after reordering and renumbering the bi ’s) a sequence Bk : w1 , . . . , wk , b1 , . . . , bn−k which is linearly independent and a spanning sequence for V , i.e., it is a basis. If the length of S were finite (say equal to `) and less than the length of B (i.e., ` < n), then the sequence B` : w1 , . . . , w` , b1 , . . . , bn−` would be a basis. But then B` = S and so we could not have any bi at the right hand end since S is a spanning sequence, so B` would be linearly dependent. Thus we must be able to find at least n terms of S, and so eventually we obtain the sequence Bn : w1 , . . . , wn which is a basis. This shows that S must have finite length which equals the length of B, i.e., any two bases have finite and equal length . ¤ The next result is obtained from our earlier results.

2.2. LINEAR INDEPENDENCE AND BASES

21

Proposition 2.29. Let V be a finite dimensional vector space of dimension n. Suppose that R : v1 , . . . , vr is a linearly independent sequence in V and that S : w1 , . . . , ws is a spanning sequence in V . Then the following inequalities are satisfied: r 6 n 6 s. Furthermore, R is a basis if and only if r = n, while S is a basis if and only if s = n. Theorem 2.30. Let V be a finite dimensional vector space and suppose that W ⊆ V is a vector subspace. Then W is finite dimensional and dim W 6 dim V, with equality precisely when W = V . Proof. Choose a basis S for W . Then by Theorem 2.25, this extends to a basis S 0 for and these are both finite sets with dim W and dim V elements respectively. As S is subsequence of S 0 , we have dim W 6 dim V, with equality exactly when S = S 0 , which can only happen when W = V .

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A similar inequality holds when the vector space V is infinite dimensional. Example 2.31. Recall from Example 2.23 the infinite dimensional real vector space D(R) and the subspace W ⊆ D(R). Then dim W = 2 since the vectors cos 2x, sin 2x were shown to form a basis. The next result gives are some other characterisations of bases in a vector space V . Let X and Y be sets; then X is a proper subset of Y (indicated by X ( Y ) if X ⊆ Y and X = 6 Y. Proposition 2.32. Let S be a sequence in V . • Let S be a spanning sequence of V . Then S is a basis if and only if it is a minimal spanning sequence, i.e., no proper subsequence of S is a spanning sequence. • Let S be a linearly independent sequence. Then is a basis if and only if it is a maximal linearly independent set, i.e., no sequence in V containing S as a proper subsequence is linearly independent. Proof. Again this follows from things we already know.

¤

Example 2.33. Recall Example 1.8: given a field of scalars F , for each pair m, n > 1 we have the vector space over F consisting of all m × n matrices with entries in F , namely Mm×n (F ). For r = 1, . . . , m and s = 1, . . . , n let E rs be the m × n matrix all of whose entries is 0 except for the (r, s) entry which is 1. Now any matrix [aij ] ∈ Mm×n (F ) can be expressed in the form n m X X

ars E rs ,

r=1 s=1

E rs

so the matrices span Mm×n (F ). But the only way that the expression on the right hand side can be the zero matrix is if [aij ] = Om×n , i.e., if aij = 0 for every entry aij . Thus the E rs form a basis and so dimF Mm×n (F ) = mn.

22

2. SPANNING SEQUENCES, LINEAR INDEPENDENCE AND BASES

In particular, dimF Mn (F ) = n2 and dimF Mn×1 (F ) = n = dimF M1×n (F ). The next result provides another useful way to find a basis for a subspace of F n . Proposition 2.34. Suppose that S : w1 , . . . , wm is a sequence of vectors in F n and that W = Span(S). Arrange the coordinates of each wr as the r-th row of an m × n matrix A. Let A0 be the reduced echelon form of A, and use the coordinates of the r-th row of A0 to form the vector wr0 . If A0 has k non-zero rows, then S 0 : w10 , . . . , wk0 is a basis for W . Proof. To see this, notice that we can recover A from A0 by reversing the elementary row operations required to obtain A0 from A, hence the rows of A0 are all in Span(A) and vice versa. This shows that Span(A0 ) = Span(A) = W . Furthermore, it is easy to see that the non-zero rows of A0 are linearly independent. ¤ Example 2.35. Consider the real vector space R4 and let W be the subspace spanned by the vectors (1, 2, 3, 0), (2, 0, 1, −2), (0, 1, 0, 4), (2, 1, 1, 2). Find a basis for W . Solution. Let

1 2 A= 0 2

2 0 1 1

3 0 1 −2 . 0 4 1 2

Then 1 2 3 0 1 2 3 0 0 −4 −5 −2 0 1 0 4 A ∼ ∼ R2 →R2 −2R1 0 1 0 4 R2 ↔R3 0 −4 −5 −2 R4 →R4 −2R1 0 −3 −5 2 0 −3 −5 2 1 0 3 −8 1 0 0 1 0 4 0 1 ∼ ∼ R1 →R1 −2R2 0 0 −5 14 R4 →R4 −R3 0 0 R3 →R3 +4R2 R4 →R4 +3R2 0 0 −5 14 0 0 1 1 0 3 −8 0 1 0 4 0 ∼ ∼ R3 →(−1/5)R4 0 0 1 −14/5 R1 →R1 −3R3 0 0 0 0 0 0

3 0 −5 0 0 1 0 0

−8 4 14 0

0 2/5 0 0 . 1 −14/5 0 0

Since the last matrix is reduced echelon, the sequence (1, 0, 0, 2/5), (0, 1, 0, 0), (0, 0, 1, −14/5) is a basis of W . An alternative basis is (5, 0, 0, 2), (0, 1, 0, 0), (0, 0, 5, −14). ¤ 2.3. Coordinates with respect to bases Let V be a finite dimensional vector space over a field F and let n = dim V . Suppose that v1 , . . . , vn form a basis of V . Then by Proposition 2.17, every vector x ∈ V has a unique expression of the form x = x1 v1 + · · · + xn vn ,

2.4. SUMS OF SUBSPACES

23

where x1 , . . . , xn ∈ F are the coordinates of x with respect to this basis. Of given the course, x1 . vectors v1 , . . . , vn , each vector x is determined by the coordinate vector .. with respect xn to this basis. If v10 , . . . , vn0 is another basis and if the coordinates of x with respect it are x01 , . . . , x0n ∈ F , then x = x01 v10 + · · · + x0n vn0 , If we write vk0 = a1k v1 + · · · + ank vn for ars ∈ F , we obtain x = x01

n X

ar1 vr + · · · + x0n

n X

r=1 n X

r=1 n X

s=1

s=1

a1s x0s )v1 + · · · + (

=(

arn vr ,

ans x0s )vn .

Thus we obtain the equation n n X X x1 v1 + · · · + xn vn = ( a1s x0s )v1 + · · · + ( ans x0s )vn . s=1

s=1

and so for each r = 1, . . . , n we have (2.1a)

xr =

n X

ars x0s .

s=1

This is more efficiently expressed in the matrix form x1 a11 · · · a1n x01 . . .. .. .. .. = .. (2.1b) . . . . xn an1 · · · ann x0n Clearly we could also express the coordinates x0r in terms of the xs . Thus implies that the matrix [aij ] is invertible and in fact we have −1 x01 a11 · · · a1n x1 . . . . . .. = .. .. .. .. . (2.1c) x0n an1 · · · ann xn 2.4. Sums of subspaces Let V be a vector space over a field F . In this section we will study another way to combine subspaces of a vector space by taking a kind of sum. Definition 2.36. Let V 0 , V 00 be two vector subspaces of V . Then their sum is the subset V 0 + V 00 = {v0 + v00 : v0 ∈ V 0 , v00 ∈ V 00 }. More generally, for subspaces V1 , . . . , Vk of V , V1 + · · · + Vk = {v1 + · · · + vk : for j = 1, . . . , k, vj ∈ Vj }. Notice that V1 + · · · + Vk contains each Vj as a subset.

24

2. SPANNING SEQUENCES, LINEAR INDEPENDENCE AND BASES

Proposition 2.37. For subspaces V1 , . . . , Vk of V , the sum V1 + · · · + Vk is also a subspace of V which is the smallest subspace of V containing each of the Vj as a subset. Proof. Suppose that x, y ∈ V1 +· · ·+Vk . Then for j = 1, . . . , k there are vectors xj , yj ∈ Vj for which x = x1 + · · · + xk , y = y1 + · · · + yk . We have x + y = (x1 + · · · + xk ) + (y1 + · · · + yk ) = (x1 + y1 ) + · · · + (xk + yk ). Since each Vj is a subspace it is closed under addition, hence xj + yj ∈ Vj , therefore x + y ∈ V1 + · · · + Vk . Also, for s ∈ F , sx = s(x1 + · · · + xk ) = sx1 + · · · + sxk , and each Vj is closed under multiplication by scalars, hence sx ∈ V1 + · · · + Vk . Therefore V1 + · · · + Vk is a subspace of V . Any subspace W ⊆ V which contains each of the Vj as a subset has to also contain every element of the form v1 + · · · + vk (vj ∈ Vj ) and hence V1 + · · · + Vk ⊆ W . Therefore V1 + · · · + Vk is the smallest such subspace.

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Definition 2.38. Let V 0 , V 00 be two vector subspaces of V . Then V 0 + V 00 is a direct sum if V 0 ∩ V 00 = {0}. More generally, for subspaces V1 , . . . , Vk of V , V1 +· · ·+Vk is a direct sum if for each r = 1, . . . , k, Vr ∩ (V1 + · · · + Vr−1 + Vr+1 + · · · + Vk ) = {0}. Such direct sums are sometimes denoted V 0 ⊕ V 00 and V1 ⊕ · · · ⊕ Vk . Theorem 2.39. Let V 0 , V 00 be subspaces of V for which V 0 + V 00 is a direct sum V 0 ⊕ V 00 . Then every vector v ∈ V 0 ⊕ V 00 has a unique expression v = v0 + v00

(v0 ∈ V 0 , v00 ∈ V 00 ).

More generally, if V1 , . . . , Vk are subspaces of V for which V1 +· · ·+Vk is a direct sum V1 ⊕· · ·⊕Vk , then every vector v ∈ V1 ⊕ · · · ⊕ Vk has a unique expression v = v1 + · · · + vk

(vj ∈ Vj , j = 1, . . . , k).

Proof. We will give the proof for the first case, the general case is more involved but similar. Let v ∈ V 0 ⊕ V 00 . Then certainly there are vectors v0 ∈ V 0 , v00 ∈ V 00 for which v = v0 + v00 . Suppose there is a second pair of vectors w0 ∈ V 0 , w00 ∈ V 00 for which v = w0 + w00 . Then (w0 − v0 ) + (w00 − v00 ) = 0, and w0 − v0 ∈ V 0 , w00 − v00 ∈ V 00 . But this means that w0 − v0 = v00 − w00 where the left hand side is in V 0 and the right hand side is in V 00 . But the equality here means that each of these terms is in V 0 ∩ V 00 = {0}, hence w0 = v0 and w00 = v00 . Thus such an expression for v must be unique. ¤

2.4. SUMS OF SUBSPACES

25

Proposition 2.40. Let V 0 , V 00 be subspaces of V . If these are finite dimensional, then so is V 0 + V 00 and dim(V 0 + V 00 ) = dim V 0 + dim V 00 − dim(V 0 ∩ V 00 ). Proof. Notice that V 0 ∩ V 00 is a subspace of each of V 0 and V 00 and of V 0 + V 00 . Choose a basis for V 0 ∩ V 00 , say v1 , . . . , va . This extends to bases for V 0 and V 00 , say v1 , . . . , va , v10 , . . . , vb0 and v1 , . . . , va , v100 , . . . , vc00 . Every element of V 0 + V 00 is a linear combination of the vectors v1 , . . . , va , v10 , . . . , vb0 , v100 , . . . , vc00 , we will show that they are linearly independent. Suppose that for scalars s1 , . . . , sa , s01 , . . . , s0b , s001 , . . . , s00c , the equation s1 v1 + · · · + sa va + s01 v10 + · · · + s0b vb0 + s001 v100 + · · · + s00c vc00 = 0 holds. Then we have s001 v100 + · · · + s00c vc00 = −(s1 v1 + · · · + sa va + s01 v10 + · · · + s0b vb0 ) ∈ V 0 , but the left hand side is also in V 00 , hence both sides are in V 0 ∩ V 00 . By our choice of basis for V 0 ∩ V 00 , we find that s001 v100 + · · · + s00c vc00 is a linear combination of the vectors v1 , . . . , va and since v1 , . . . , va , v100 , . . . , vc00 is linearly independent the only way that can happen is if all the coefficients are zero, hence s001 = · · · = s00c = 0. A similar argument also shows that s01 = · · · = s0b = 0. We are left with the equation s1 v1 + · · · + sa va = 0 which also implies that s1 = · · · = sa = 0 since the vi form a basis for V 0 ∩ V 00 . Now we have dim(V 0 ∩ V 00 ) = a,

dim V 0 = a + b,

dim V 00 = a + c,

and dim(V 0 + V 00 ) = a + b + c = dim V 0 + dim V 00 − dim(V 0 ∩ V 00 ).

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Definition 2.41. Let V be a vector space and W ⊆ V be a subspace. A subspace W 0 ⊆ V for which W + W 0 = V and W ∩ W 0 = {0} is called a linear complement of W in V . For such a linear complement we have V = W ⊕ W 0 . We remark that although linear complements always exist they are by no means unique! The next result summarises the situation. Proposition 2.42. Let W ⊆ V be subspace. (a) There is a linear complement W 0 ⊆ V of W in V . (b) If V is finite dimensional, then any subspace W 00 ⊆ V satisfying the conditions • W ∩ W 00 = {0}, • dim W 00 = dim V − dim W is a linear complement of W in V . (c) Any subspace W 000 ⊆ V satisfying the conditions • W + W 000 = V , • dim W 000 = dim V − dim W is a linear complement of W in V .

26

2. SPANNING SEQUENCES, LINEAR INDEPENDENCE AND BASES

Proof. (a) We show how to find a linear complement in the case where V is finite dimensional with n = dim V . By Theorem 2.30, we know that dim W 6 dim V . So let W have a basis w1 , . . . , wr , where r = dim W . Now by Theorem 2.25 there is an extension of this to a basis of V , say 0 w1 , . . . , wr , w10 , . . . , wn−r . 0 Now let W 0 = Span(w10 , . . . , wn−r ). Then by Lemma 2.6, W 0 is a subspace of V and it has 0 w10 , . . . , wn−r as a basis, so dim W 0 = n − r. Now it is clear that W + W 0 = V . Suppose that v ∈ W ∩ W 0 . Then there are expressions

v = s1 w1 + · · · + sr wr , 0 v = s01 w10 + · · · + s0n−r wn−r ,

hence there is an equation 0 s1 w1 + · · · + sr wr + (−s01 )w10 + · · · + (−s0n−r )wn−r = 0.

But the vectors wi , wj0 form a basis so are linearly independent. Therefore s1 = · · · = sr = s01 = · · · = s0n−r = 0. This means that v = 0, hence W ∩ W 0 = {0} and the sum W + W 0 is direct. So we have shown that W 0 is a linear complement of W in V . (b) and (c) can be verified in similar fashion. ¤ Example 2.43. Consider the vector space R2 over R. If V 0 ⊆ R2 and V 00 ⊆ R2 are two distinct lines through the origin, show that V 0 + V 00 is a direct sum and that V 0 ⊕ V 00 = R2 . Solution. As these lines are distinct and V 0 ∩ V 00 is subspace of each of V 0 and V 00 , we have dim(V 0 ∩ V 00 ) < dim V 0 = dim V 00 = 1, hence dim(V 0 ∩ V 00 ) = 0. This means that V 0 ∩ V 00 = {0} which is geometrically obvious. Now notice that dim(V 0 + V 00 ) = dim V 0 + dim V 00 − 0 = 1 + 1 = 2, hence dim(V 0 + V 00 ) = dim R2 and so V 0 + V 00 = R2 .

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Example 2.44. Consider the vector space R3 over R. If V 0 ⊆ R3 is a line through the origin and V 00 ⊆ R2 is a plane through the origin which does not contain V 0 , show that V 0 + V 00 is a direct sum and that V 0 ⊕ V 00 = R3 . Solution. V 0 ∩ V 00 is subspace of each of V 0 and V 00 , so we have dim(V 0 ∩ V 00 ) 6 dim V 0 = 1 < 2 = dim V 00 . In fact the inequality must be strict since otherwise V 0 ⊆ V 00 , hence dim(V 0 ∩ V 00 ) = 0. This means that V 0 ∩ V 00 = {0} which is also geometrically obvious. Now notice that dim(V 0 + V 00 ) = dim V 0 + dim V 00 − 0 = 1 + 2 = 3, therefore dim(V 0 + V 00 ) = dim R3 and so V 0 + V 00 = R3 .

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2.4. SUMS OF SUBSPACES

27

Example 2.45. Consider the vector space R4 over R and the subspaces V 0 = {(s + t, s, −s, −t) : s, t ∈ R},

V 00 = {(u, 2v, −v, −u) : u, v ∈ R}.

Show that dim(V 0 + V 00 ) = 3. Solution. V 0 is spanned by the vectors (1, 1, −1, 0), (1, 0, 0, −1), which are linearly independent and hence form a basis. V 00 is spanned by (1, 0, 0, −1), (0, 2, −1, 0) which are linearly independent and hence form a basis. Thus we find that dim V 0 = 2 = dim V 00 . Now suppose that (w, x, y, z) ∈ V 0 ∩ V 00 . Then there are s, t, u, v ∈ R for which (w, x, y, z) = (s + t, s, −s, −t) = (u, 2v, −v, −u). By comparing coefficients, this leads to the system of four linear equations s + t − u = 0 s − 2v = 0 (2.2) − s + v = 0 − t + u = 0 which has associated augmented matrix 1 1 −1 0 0 1 1 0 0 −2 0 0 ∼ −1 0 0 1 0 R2 →R2 −R1 0 R3 →R3 +R1 0 −1 1 0 0 0 1 0 0 0 1 −1 ∼ R1 →R1 −R2 0 0 0 R3 →R3 +R2 R4 →R4 +R2 0 0 0

1 −1 0 0 1 1 −1 0 0 0 −1 1 −2 0 1 −1 1 0 ∼ 1 −1 1 0 R2 ↔R3 0 −1 1 −2 0 −1 1 0 0 0 −1 1 0 0 −1 0 1 0 0 −1 0 1 0 1 0 0 1 −1 ∼ R ↔R −1 0 3 4 0 0 0 1 0 1 0 0 0 0 −1 0 1 0 0 0 0 0 1 −1 0 0 ∼ , R1 →R1 +R3 0 0 0 1 0 R2 →R2 −R3 R4 →R4 +R3 0 0 0 0 0

where the last matrix is reduced echelon. The general solution of the system (2.2) is s = 0,

t = u,

u ∈ R,

v = 0.

hence we have V 0 ∩ V 00 = {(u, 0, 0, −u) : u ∈ R}, and this has basis (1, 0, 0, −1), hence dim(V 0 ∩ V 00 ) = 1. This means that these two planes intersect in a line in R4 . Now we see that dim(V 0 + V 00 ) = dim V 0 + dim V 00 − dim(V 0 ∩ V 00 ) = 2 + 2 − 1 = 3.

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CHAPTER 3

Linear transformations 3.1. Functions Before introducing linear transformations, we will review some basic ideas about functions, including properties of composition and inverse functions which will be required later. Let X, Y and Z be sets. Definition 3.1. A function (or mapping) f : X −→ Y from X to Y involves a rule which associates to each x ∈ X a unique element f (x) ∈ Y . X is called the domain of f , denoted dom f , while Y is called the codomain of f , denoted codom f . Sometimes the rule is indicated by writing x 7→ f (x). Given functions f : X −→ Y and g : Y −→ Z, we can form their composition g ◦ f : X −→ Z which has the rule g ◦ f (x) = g(f (x)). Note the order of composition here! g◦f

X

f

/Y

g

Ã /Z

We often write gf for g ◦ f when no confusion will arise. If X, Y , Z are different it may not be possible to define f ◦ g since dom f = X may not be the same as codom g = Z. Example 3.2. For any set X, the identity function IdX −→ X has rule x 7→ IdX (x) = x. Example 3.3. Let X = Y = R. Then the following rules define functions R −→ R: x 7→ x + 1,

x 7→ x2 ,

x 7→

x2

1 , +1

3

x 7→ sin x, x 7→ e5x .

Example 3.4. Let X = Y = R+ , the set of all positive real numbers. Then the following rules define two functions R+ −→ R: √ √ x 7→ x, x → 7 − x. Example 3.5. Let X and Y be sets and suppose that w ∈ Y is an element. The constant function cw : X −→ Y taking value w has the rule x 7→ cw (x) = w. For example, if X = Y = R then c0 is the function which returns the value c0 (x) = 0 for every real number x ∈ R. Definition 3.6. A function f : X −→ Y is • injective (or an injection or one-to-one) if for x1 , x2 ∈ X, 29

30

3. LINEAR TRANSFORMATIONS

f (x1 ) = f (x2 ) implies x1 = x2 , • surjective (or a surjection or onto) if for every y ∈ Y there is an x ∈ X for which y = f (x), • bijective (or injective or a one-to-one correspondence) if it is both injective and surjective. We will use the following basic fact. Proposition 3.7. The function f : X −→ Y is a bijection if and only if there is an inverse function Y −→ X which is the unique function h : Y −→ X satisfying h ◦ f = IdX ,

f ◦ h = IdY .

If such an inverse exists it is usual to denote it by f −1 : Y −→ X, and then we have f −1 ◦ f = IdX ,

f ◦ f −1 = IdY .

Later we will see examples of all these notions in the context of vector spaces. 3.2. Linear transformations In this section, let F be a field of scalars. Definition 3.8. Let V and W be two vector spaces over F and let f : V −→ W be a function. Then f is called a linear transformation or linear mapping if it satisfies the two conditions • for all vectors v1 , v2 ∈ V , f (v1 + v2 ) = f (v1 ) + f (v2 ), • for all vectors v ∈ V and scalars t ∈ F , f (tv) = tf (v). These two conditions together are equivalent to the single condition • for all vectors v1 , v2 ∈ V , and scalars t1 , t2 ∈ F , f (t1 v1 + t2 v1 ) = t1 f (v1 ) + t2 f (v1 ). Remark 3.9. For a linear transformation f : V −→ W , we always have f (0) = 0, where the left hand 0 means the zero vector in V and the right hand 0 means the zero vector in W . The next example introduces an important kind of linear transformation associated with a matrix. Example 3.10. Let A ∈ Mm×n be an m × n matrix. Define fA : F n −→ F m by the rule fA (x) = Ax. Then for s, t ∈ F and u, v ∈ F n , fA (su + tv) = A(su + tv) = sAu + tAv = sfA (u) + tfA (v). So fA is a linear transformation. Observe that the standard basis vectors e1 , . . . , en (viewed as column vectors) satisfy fA (ej ) = j-th column of A,

3.2. LINEAR TRANSFORMATIONS

so

h A = fA (e1 ) · · ·

i h fA (en ) = Ae1 · · ·

31

i Aen .

Now it easily follows that fA (e1 ), . . . , fA (en ) spans Im fA since every element of the latter is a linear combination of this sequence. Theorem 3.11. Let f : V −→ W and g : U −→ V be two linear transformations between vector spaces over F . Then the composition f ◦ g : U −→ W is also a linear transformation. Proof. For s1 , s2 ∈ F and u1 , u2 ∈ V , we have to show that f ◦ g(s1 u1 + s2 u2 ) = s1 f ◦ g(u1 ) + s2 f ◦ g(u2 ). Using the fact that both f and g are linear transformations we have f ◦ g(s1 v1 + s2 v2 ) = f (s1 g(u1 ) + s2 g(u2 )) = s1 f (g(u1 )) + s2 f (g(u2 )) = s1 f ◦ g(u1 ) + s2 f ◦ g(u2 ), as required.

¤

Theorem 3.12. Let V and W be vector spaces over F . Suppose that f, g : V −→ W are linear transformations and s, t ∈ F . Then the function sf + tg : V −→ W ;

(sf + tg)(v) = sf (v) + tg(v)

is a linear transformation. Definition 3.13. Let f : V −→ W be a linear transformation. • The kernel (or nullspace) of f is the following subset of V : Ker f = {v ∈ V : f (v) = 0} ⊆ V. • The image (or range) of f is the following subset of W : Im f = {w ∈ W : there is an v ∈ V s.t. w = f (v)} ⊆ W. Theorem 3.14. Let f : V −→ W be a linear transformation. Then (a) Ker f is a subspace of V , (b) Im f is a subspace of W . Proof. (a) Let s1 , s2 ∈ F and v1 , v2 ∈ Ker f . We must show that s1 v1 + s2 v2 ∈ Ker f , i.e., that f (s1 v1 + s2 v2 ) = 0. Since f is a linear transformation and f (v1 ) = 0 = f (v2 ), we have f (s1 v1 + s2 v2 ) = s1 f (v1 ) + s2 f (v2 ) = s1 0 + s2 0 = 0, hence f (s1 v1 + s2 v2 ) ∈ Ker f . Therefore Ker f is a subspace of V . Now suppose that t1 , t2 ∈ F and w1 , w2 ∈ Im f . We must show that t1 w1 + t2 w2 ∈ Im f , i.e., that t1 w1 + t2 w2 = f (v) for some v ∈ V . Since w1 , w2 ∈ Im f , there are vectors v1 , v2 ∈ V for which f (v1 ) = w1 ,

f (v2 ) = w2 .

32

3. LINEAR TRANSFORMATIONS

As f is a linear transformation, t1 w1 + t2 w2 = t1 f (v1 ) + t2 f (v2 ) = f (t1 v1 + t2 v2 ), so we can take v = t1 v1 + t2 v2 to get t1 w1 + t2 w2 = f (v). Hence Im f is a subspace of W .

¤

For linear transformations the following result holds, where we make use of notions introduced in Section 3.1. Theorem 3.15. Let f : V −→ W be a linear transformation. (a) f is injective if and only if Ker f = {0}. (b) f is surjective if and only if Im f = W . (c) f is bijective if and only if Ker f = {0} and Im f = W . Proof. (a) Notice that f (0) = 0. Hence if f is injective then f (v) = 0 implies v = 0, i.e., Ker f = {0}. We need to show the converse, i.e., if Ker f = {0} then f is injective. So suppose that Ker f = {0}, v1 , v2 ∈ V , and f (v1 ) = f (v2 ). Since f is a linear transformation, f (v1 − v2 ) = f (v1 + (−1)v2 ) = f (v1 ) + (−1)f (v2 ) = f (v1 ) − f (v2 ) = 0, hence f (v1 − v2 ) = 0 and therefore v1 − v2 = 0, i.e., v1 = v2 . This shows that f is injective. (b) This is immediate from the definition of surjectivity. (c) This comes by combining (a) and (b). ¤ Theorem 3.16. Let f : V −→ W be a bijective linear transformation. Then the inverse function f −1 : W −→ V is a linear transformation. Proof. Let t1 , t2 ∈ F and w1 , w2 ∈ W . We must show that f −1 (t1 w1 + t2 w2 ) = t1 f −1 (w1 ) + t2 f −1 (w2 ).

3.2. LINEAR TRANSFORMATIONS

33

The following calculation does this: ¡ ¢ f f −1 (t1 w1 + t2 w2 ) − t1 f −1 (w1 ) + t2 f −1 (w2 ) =f ◦ f −1 (t1 w1 + t2 w2 ) − f (t1 f −1 (w1 ) + t2 f −1 (w2 )) = IdW (t1 w1 + t2 w2 ) − f (t1 f −1 (w1 ) + t2 f −1 (w2 )) =(t1 w1 + t2 w2 ) − (t1 f ◦ f −1 (w1 ) + t2 f ◦ f −1 (w2 )) =(t1 w1 + t2 w2 ) − (t1 IdW (w1 ) + t2 IdW (w2 )) =(t1 w1 + t2 w2 ) − (t1 w1 + t2 w2 ) = 0. Since f is injective this means that f −1 (t1 w1 + t2 w2 ) − (t1 f −1 (w1 ) + t2 f −1 (w2 )) = 0, giving the desired formula.

¤

Definition 3.17. A linear transformation f : V −→ W which is a bijective function is called an isomorphism from V to W , and these are said to be isomorphic vector spaces. Notice that for an isomorphism f : V −→ W , the inverse f −1 : W −→ V is also an isomorphism. Definition 3.18. Let f : V −→ W be a linear transformation. Then the rank of f is rank f = dim Im f, and the nullity of f is null f = dim Ker f, whenever these are finite. Theorem 3.19 (The rank-nullity theorem). Let f : V −→ W be a linear transformation with V finite dimensional. Then dim V = rank f + null f. Proof. Start by choosing a basis for Ker f , say u1 , . . . , uk . Notice that k = null f . By Theorem 2.25, this can be extended to a basis of V , say u1 , . . . , uk , uk+1 , . . . , un , where n = dim V . Now for s1 , . . . , sn ∈ F , f (s1 u1 + · · · + sn un ) = s1 f (u1 ) + · · · + sn f (un ) = s1 0 + · · · + sk 0 + sk+1 f (uk+1 ) + · · · + sn f (un ) = sk+1 f (uk+1 ) + · · · + sn f (un ). This shows that Im f is spanned by the vectors f (uk+1 ), . . . , f (un ). In fact, this last expression is only zero when s1 u1 + · · · + sn un lies in Ker f which can only happen if sk+1 = · · · = sn = 0. So the vectors f (uk+1 ), . . . , f (un ) are linearly independent and therefore form a basis for Im f . Thus rank f = n − k and the result follows. ¤ We can also reformulate the results of Theorem 3.15.

34

3. LINEAR TRANSFORMATIONS

Theorem 3.20. Let f : V −→ W be a linear transformation. (a) f is injective if and only if null f = 0. (b) f is surjective if and only if rank f = dim W . (c) f is bijective if and only if f = 0 and rank f = dim W . Corollary 3.21. Let f : V −→ W be a linear transformation with V and W finite dimensional vector spaces. The following hold: (a) if f is an injection, then dimF V 6 dimF W ; (b) if f is a surjection, then dimF V > dimF W ; (c) if f is a bijection, then dimF V = dimF W . Proof. This makes use of the rank-nullity Theorem 3.19 and Theorem 3.20. (a) Since Im f is a subspace of W , we have dim Im f 6 dim W , hence if f is injective, dim V = rank f 6 dim W. (b) This time, if f is surjective we have dim V > rank f = dim W. Combining (a) and (b) gives (c).

¤

Example 3.22. Let D(R) be the vector space of differentiable real functions considered in Example 1.11. (a) For n > 1, define the function Dn : D(R) −→ D(R);

Dn (f ) =

dn f . dxn

Show that this is a linear transformation and that Dn = D ◦ Dn−1 = Dn−1 ◦ D and so Dn is the n-fold composition D ◦ · · · ◦ D. Identify the kernel and image of Dn . (b) More generally, for real numbers a0 , a1 , . . . , an , define the function a0 + a1 D + · · · + an Dn : D(R) −→ D(R); (a0 + a1 D + · · · + an Dn )(f ) = a0 f + a1 D(f ) + · · · + an Dn (f ), and show that this is a linear transformation and identify its kernel and image. Solution. (a) When n = 1, D has the effect of differentiation on functions. For scalars s, t and functions f, g in D(R), d(sf + tg) df dg D(sf + tg) = =s +t = sD(f ) + tD(g). dx dx dx Hence D is a linear transformation. The formulae for Dn are clear and compositions of linear transformations are also linear transformations by Theorem 3.11. To understand the kernel of Dn , notice that f ∈ Ker Dn if and only if Dn (f ) = 0, so we are looking for the general solution of the differential equation dn f =0 dxn which has general solution f (x) = c0 + c1 x + · · · cn−1 xn−1

(c0 , . . . , cn−1 ∈ R).

3.2. LINEAR TRANSFORMATIONS

35

To find the image of Dn , first notice that for any real function g we have µZ x ¶ Z x d g(t) dt = g(x). D g(t) dt = dx 0 0 This shows that Im D = D(R), i.e., D is surjective. More generally, we find that Im Dn = D(R). (b) This uses Theorem 3.12. The point is that a0 + a1 D + · · · + an Dn = a0 IdD(R) +a1 D + · · · an Dn . We have f ∈ Ker(a0 + a1 D + · · · + an Dn ) if and only if df dn f + · · · + an n = 0, dx dx so the kernel is the same as the set of solutions of the latter differential equation. Again we find that this linear transformation is surjective and we leave this as an exercise. ¤ a0 f + a1

Example 3.23. Let F be a field of scalars. For n > 1, consider the vector space Mn (F ). Define the trace function tr : Mn (F ) −→ F by tr([aij ]) =

n X

arr = a11 + · · · + ann .

r=1

Show that tr is linear transformation and find bases for its kernel and image. Solution. For [aij ], [bij ] ∈ Mn (F ) and s, t ∈ F we have tr(s[aij ] + t[bij ]) = tr([saij + tbij ]) =

n X (sarr + tbrr ) r=1

=

n X

sarr +

r=1 n X

=s

n X

tbrr

r=1 n X

arr + t

r=1

brr

r=1

= s tr([aij ]) + t tr([bij ]), hence tr is a linear transformation. Notice that dimF Mn (F ) = n2 and dimF F = 1. Also, for t ∈ F , t 0 ··· 0 . . . . . . . ... = t. tr . . 0 0 ··· 0 This shows that Im tr = F , i.e., tr is surjective and Im tr = F has basis 1. By the rank-nullity Theorem 3.19, null tr = dimF Ker tr = n2 − 1. Now the matrices E rs (r, s = 1, . . . , n − 1, r 6= s) and E rr − E (r+1)(r+1) (r = 1, . . . , n − 1) lie in Ker tr and are linearly independent. There are n(n − 1) + (n − 1) = (n + 1)(n − 1) = n2 − 1 of these, so they form a basis of Ker tr.

¤

36

3. LINEAR TRANSFORMATIONS

Example 3.24. Let F be a field and let m, n > 1. Consider the transpose function (−)T : Mm×n (F ) −→ Mn×m (F ) defined by [aij ]T = [aji ], i.e., the (i, j)-th entry of [aij ]T is the (j, i)-th entry of [aij ]. Show that (−)T is a linear transformation which is an isomorphism. What is its inverse? Solution. For [aij ], [bij ] ∈ Mm×n (F ), ([aij ] + [bij ])T = [aij + bij ]T = [aji + bji ] = [aji ] + [bji ] = [aij ]T + [bij ]T , and if s ∈ F , (s[aij ])T = [saij ]T = [saji ] = s[aji ] = s[aji ]T . Thus (−)T is a linear transformation. If [aij ] ∈ Ker(−)T then [aji ] = On×m (the n × m zero matrix), hence aij = 0 for i = 1, . . . , m and j = 1, . . . , n and so [aji ] = Om×n . Thus null(−)T = 0 and by the rank-nullity theorem, rank(−)T = dim Mm×n (F ) = mn = dim Mn×m (F ). This shows that Im(−)T = Mn×m (F ). Hence (−)T is an isomorphism. Its inverse is (−)T : Mn×m (F ) −→ Mm×n (F ).

¤

Similar arguments apply to the next example. Example 3.25. Let m, n > 1 and consider the Hermitian conjugate function (−)∗ : Mm×n (C) −→ Mn×m (C) defined by [aij ]∗ = [aji ], i.e., the (i, j)-th entry [aij ]∗ is the complex conjugate of the (j, i)-th entry of [aij ]. If Mm×n (C) and Mn×m (C) are viewed as real vector spaces, then (−)∗ is a linear transformation which is an isomorphism. There are some useful rules for dealing with transpose and Hermitian conjugates of products of matrices. Proposition 3.26 (Reversal rules). (a) If F is any field, the transpose operation satisfies (AB)T = B T AT for all matrices A ∈ M`×m (F ) and B ∈ Mm×n (F ). (b) For all complex matrices P ∈ M`×m (C) and Q ∈ Mm×n (C), (P Q)∗ = Q∗ P ∗ .

3.3. WORKING WITH BASES AND COORDINATES

37

Proof. (a) Writing A = [aij ] and B = [bij ], we obtain "m #T " m # X X (AB)T = ([aij ][bij ])T = air brj = ajr bri r=1

r=1

=

"m X

# bri ajr

r=1

= [bji ][aji ] = [bij ]T [aij ]T = B T AT . (b) is proved in a similar way but with the appearance of complex conjugates in the formulae. ¤ Definition 3.27. For an arbitrary field F , an matrix A ∈ Mn (F ) is called • symmetric if AT = A, • skew-symmetric if AT = −A. A complex matrix B ∈ Mn (C) is called • Hermitian if B ∗ = A, • skew-Hermitian if B ∗ = −A. 3.3. Working with bases and coordinates When performing calculations with a linear transformation it is often useful to use coordinates with respect to bases of the domain and codomain. In fact, this reduces every linear transformation to one associated to a matrix as described in Example 3.10. Lemma 3.28. Let f, g : V −→ W be linear transformations and let S : v1 , . . . , vn be a basis for V . Then f = g if and only if the values of f and g agree on the elements of S. Hence a linear transformation is completely determined by its values on the elements of a basis for its domain. Proof. Every element v ∈ V can be uniquely expressed as a linear combination v = s1 v1 + · · · + sn vn Now f (v) = f (s1 v1 + · · · + sn vn ) = s1 f (v1 ) + · · · + sn f (vn ) and similarly g(v) = g(s1 v1 + · · · + sn vn ) = s1 g(v1 ) + · · · + sn g(vn ). So the functions f and g agree on v if and only if for each i, f (vi ) = g(vi ). We conclude that f and g agree on V if and only if they agree on S. ¤ Here is another useful result involving bases. Lemma 3.29. Let V and W be finite dimensional vector spaces over F , and let S : v1 , . . . , vn be a basis for V . Suppose that f : V −→ W is a linear transformation. Then (a) if f (v1 ), . . . , f (vn ) is linearly independent in W , then f is an injection; (b) if f (v1 ), . . . , f (vn ) spans W , then f is surjective; (b) if f (v1 ), . . . , f (vn ) forms a basis for W , then f is bijective.

38

3. LINEAR TRANSFORMATIONS

Proof. (a) The equation f (s1 v1 + · · · + sn vn ) = 0 is equivalent to s1 f (v1 ) + · · · + sn f (vn ) = 0 and so if the vectors f (v1 ), . . . , f (vn ) are linearly independent, then this equation has unique solution s1 = · · · = sn = 0. This shows that Ker f = {0} and hence by Theorem 3.15 we see that f is injective. (b) Im f ⊆ W contains every vector of the form f (t1 v1 + · · · + tn vn ) = t1 f (v1 ) + · · · + tn f (vn ) (t1 , . . . , tn ∈ F ). If the vectors f (v1 ), . . . , f (vn ) span W , this shows that W ⊆ Im f and so f is surjective. (c) This follows by combining (a) and (b).

¤

Theorem 3.30. Let S : v1 , . . . , vn be a basis for the vector space V . Let W be a second vector space and let T : w1 , . . . , wn be any sequence of vectors in W . Then there is a unique linear transformation f : V −→ W satisfying f (vi ) = wk

(i = 1, . . . , n).

Proof. For every vector v ∈ V , there is a unique expression v = s1 v1 + · · · + sk vk , where si ∈ F . Then we can set f (v) = s1 w1 + · · · + sk wk since the right hand side is clearly well defined in terms of v. In particular, if v = vi , then f (vi ) = wi and that f is a linear transformation f : V −→ W . Uniqueness follows from Lemma 3.28. ¤ A useful consequence of this result is the following. Theorem 3.31. Suppose that the vector space V is a direct sum V = V1 ⊕V2 of two subspaces and let f1 : V1 −→ W and f2 : V2 −→ W be two linear transformations. Then there is a unique linear transformation f : V −→ W extending f1 , f2 , i.e., satisfying f (u) if u ∈ V , 1 1 f (u) = f2 (u) if u ∈ V2 . Proof. Use bases of V1 and V2 to form a basis of V , then use Theorem 3.30.

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Here is another useful result. Theorem 3.32. Let V and W be finite dimensional vector spaces over F with n = dimF V and m = dimF W . Suppose that S : v1 , . . . , vn and T : w1 , . . . , wm are bases for V and W respectively. Then there is a unique transformation f : V −→ W with the following properties: (a) if n 6 m then f (vj ) = wj for j = 1, . . . , n; (b) if n > m then

w j f (vj ) = 0

when j = 1, . . . , m, when j > m.

3.3. WORKING WITH BASES AND COORDINATES

39

Furthermore, f also satisfies: (c) if n 6 m then f is injective; (d) if n > m then f is surjective; (e) if m = n then f is an isomorphism. Proof. (a) Starting with the sequence w1 , . . . , wn , we can apply Theorem 3.30 to show that there is a unique linear transformation f : V −→ W with the stated properties. (b) Starting with the sequence w1 , . . . , wm , 0, . . . , 0 of length n, we can apply Theorem 3.30 to show that there is a unique linear transformation f : V −→ W with the stated properties. (c), (d) and (e) now follow from Lemma 3.29. ¤ The next result provides a convenient way to decide if two vector spaces are isomorphic: simply show that they have the same dimension. Theorem 3.33. Let V and W be finite dimensional vector spaces over the field F . Then there is an isomorphism V −→ W if and only if dimF V = dimF W . Proof. If there is an isomorphism f : V −→ W , then dimF V = dimF W by Corollary 3.21(c). For the converse, if dimF V = dimF W then by Theorem 3.32(e) there is an isomorphism V −→ W . ¤ Now we will see how to make use of bases for the domain and codomain to work with a linear transformation by using matrices. Let V and W be two finite dimensional vector spaces. Let S : v1 , . . . , vn and T : w1 , . . . , wm be bases for V and W respectively. Suppose that f : V −→ W is a linear transformation. Then there are unique scalars tij ∈ F (i = 1, . . . , m, j = 1, . . . , n) for which (3.1)

f (vj ) = t1j w1 + · · · + tmj wm .

We can form the m × n matrix

(3.2)

t11 · · · . .. . T [f ]S = [tij ] = .. tm1 · · ·

t1n .. . . tmn

Remark 3.34. (a) Notice that this matrix depends on the bases S and T as well as f . (b) The coefficients are easily found: the j-th column is made up from the coefficients that occur when expressing f (vj ) in terms of the wi . 0 In the next result we will also assume that there are bases S 0 : v10 , . . . , vn0 and T 0 : w10 , . . . , wm for V and W respectively, for which

vj0 = a1j v1 + · · · + anj vn =

n X

aij vi

i=1 n X

wk0 = b1k v1 + · · · + bmk wm =

i=1

bik wi

(j = 1, . . . , n), (k = 1, . . . , m).

40

3. LINEAR TRANSFORMATIONS

Working with these bases we obtain the matrix

t011 · · · .. .. = [t0ij ] = . . t0m1 · · ·

T 0 [f ]S 0

We will set

a11 · · · . .. A = [aij ] = .. . an1 · · ·

a1n .. . , ann

t01n .. . . t0mn

b11 · · · . .. B = [bij ] = .. . bm1 · · ·

b1m .. . . bmm

Theorem 3.35. (a) Let x ∈ V have coordinates x1 , . . . , xn with respect to the basis S : v1 , . . . , vn of V and let f (x) ∈ W have coordinates y1 , . . . , ym with respect to the basis T : w1 , . . . , wm of W . Then x1 t11 · · · t1n x1 y1 . . . . . . . . .. .. . . (3.3a) . . . = T [f ]S . = . xn tm1 · · · tmn xn ym (b) The matrices T [f ]S and

T 0 [f ]S 0

are related by the formula

(3.3b)

T 0 [f ]S 0

= B −1 T [f ]S A.

Proof. These follow from calculations with the above formulae.

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The formulae of (3.3), allow us to reduce calculations to ones with coordinates and matrices. Example 3.36. Consider the linear transformation f : R3 −→ R2 ;

f (x, y, z) = (2x − 3y + 7z, x − z)

between the real vector spaces R3 and R2 . (a) Write down the matrix of f with respect to the standard bases of R3 and R2 . (b) Determine the matrix T [f ]S , where S and T are the bases S : (1, 0, 1), (0, 1, 1), (0, 0, 1),

T : (3, 1), (2, 1).

Solution. (a) In terms of the standard basis vectors (1, 0, 0), (0, 1, 0), (0, 0, 1) and (1, 0), (0, 1) we have f (1, 0, 0) = (2, 1) = 2(1, 0) + 1(0, 1), f (0, 1, 0) = (−3, 0) = −3(1, 0) + 0(0, 1), f (0, 0, 1) = (7, −1) = 7(1, 0) + (−1)(0, 1), " # 2 −3 7 hence the matrix is . 1 0 −1 (b) We have (1, 0, 1) = 1(1, 0, 0) + 0(0, 1, 0) + 1(0, 0, 1), (0, 1, 1) = 0(1, 0, 0) + 1(0, 1, 0) + 1(0, 0, 1), (0, 0, 1) = 0(1, 0, 0) + 0(0, 1, 0) + 1(0, 0, 1),

3.4. APPLICATION TO MATRICES AND SYSTEMS OF LINEAR EQUATIONS

so we take

41

1 0 0 A = 0 1 0 . 1 1 1

We also have (3, 1) = 3(1, 0) + 1(0, 1), and so we take

(2, 1) = 2(1, 0) + 1(0, 1),

" # 3 2 B= . 1 1

The inverse of B is B −1

# # " " 1 −2 1 −2 1 , = = 3 − 2 −1 −1 3 3

so we have T [f ]S

=B

−1

" # 2 −3 7 A 1 0 −1

#" # 1 0 0 1 −2 2 −3 7 = 0 1 0 −1 3 1 0 −1 1 1 1 " # 9 6 9 = . −9 −7 −10 "

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Theorem 3.37. Suppose that f : V −→ W and g : U −→ V are linear transformations and that R, S, T are bases for U, V, W respectively. Then the matrices T [f ]S , S [g]R , T [f ◦ g]R are related by the matrix equation T [f ◦ g]R = T [f ]S S [g]R . Proof. This follows by a calculation using the formula (3.1). Notice that the right hand matrix product does make sense because T [f ]S is dim W × dim V , S [g]R and is dim V × dim U , while T [f ◦ g]R is dim W × dim U . ¤ 3.4. Application to matrices and systems of linear equations Let F be a field and consider an m × n matrix A ∈ Mm×n (F ). Recall that by performing a suitable sequence of elementary row operations we may obtain a reduced echelon matrix A0 which is row equivalent to A, i.e., A0 ∼ A. This can be used to solve a linear system of the form Ax = b. It is also possible to perform elementary column operations on A, or equivalently, to form T A then perform elementary row operations until we get a reduced echelon matrix A00 , then transpose back to get another column reduced echelon matrix (A00 )T . Definition 3.38. The number of non-zero rows in A0 is called the row rank of A. The number of non-zero columns in (A00 )T is called the column rank of A. Example 3.39. Taking F = R, find the row and column ranks of the matrix 1 2 A = 0 3 . 5 1

42

3. LINEAR TRANSFORMATIONS

Solution. For the row rank we have 1 2 1 2 1 2 1 0 A = 0 3 ∼ 0 3 ∼ 0 1 ∼ 0 1 = A0 , 5 1 0 −9 0 −9 0 0 so the row rank is 2. For the column rank,

" # " # " # 1 0 5 1 0 5 1 0 5 A = ∼ ∼ = A00 , 2 3 1 0 3 −9 0 1 −3 T

which gives

1 0 (A00 )T = 0 1 5 −3

and the column rank is 2.

¤

Theorem 3.40. Let A ∈ Mm×n (F ). Then row rank of A = column rank of A. Proof of Theorem 3.40. We will make use of ideas from Example 3.10. Recall the linear transformation fA : F n −→ F m determined by fA (x) = Ax (x ∈ F n ). First we identify the value of the row rank of A in other terms. Let r be the row rank of A. Then Ker fA = {x ∈ F n : Ax = 0} is the set of solution of the associated system of homogeneous linear equations. Recall that once the reduced echelon matrix A0 has been found, the general solution can be expressed in terms of n − r basic solutions each obtained by setting one of the parameters equal to 1 and the rest equal to 0. These basic solutions form a spanning sequence for Ker fA which is linearly independent and hence is a basis for Ker fA . So we have (3.4)

row rank of A = r = n − dim Ker fA .

On the other hand, if c is the column rank of A, we find that the columns of (A00 )T are a spanning sequence for the subspace of F m spanned by the columns of A, and in fact, they are linearly independent. As the columns of A span Im fA , (3.5)

column rank of A = c = dim Im fA .

But now we can use the rank-nullity Theorem 3.19 to deduce that n = dim Ker fA + dim Im fA = (n − r) + c = n + (c − r), giving c = r as claimed.

¤

Remark 3.41. In this proof we saw that the columns of A span Im fA and also how to find a basis for Ker fA by the method of solution of linear systems using elementary row operations.

3.5. GEOMETRIC LINEAR TRANSFORMATIONS

43

Definition 3.42. The rank of A is the number rank A = rank fA = dim Im fA = column rank of A = row rank of A, where rank fA was defined in Definition 3.18. It agrees with the number of non-zero rows in the reduced echelon matrix similar to A. The kernel of A is Ker A = Ker fA and the image of A is Im A = Im fA . 3.5. Geometric linear transformations When considering a linear transformation f : Rn −→ Rn on the real vector space Rn it is often important or useful to understand its geometric content. Example 3.43. For θ ∈ R, consider the linear transformation ρθ : R2 −→ R2 given by ρθ (x, y) = (x cos θ − y sin θ, x sin θ + y cos θ). Investigate the geometric effect of ρθ on the plane. Solution. The effect on the standard basis vectors is ρθ (1, 0) = (cos θ, sin θ),

ρθ (0, 1) = (− sin θ, cos θ).

Using the dot product of vectors it is possible to check that the angle between ρθ (x, y) and (x, y) is always θ if (x, y) 6= (0, 0), and that the lengths of ρθ (x, y) and (x, y) are equal. So ρθ has the effect of rotating the plane about the origin through the angle θ measured in the anti-clockwise direction. Notice that the matrix of ρθ with respect to the standard basis of R2 is the rotation matrix " # cos θ − sin θ . ¤ sin θ cos θ Example 3.44. For θ ∈ R, consider the linear transformation σθ : R2 −→ R2 given by σθ (x, y) = (x cos θ + y sin θ, x sin θ − y cos θ). Investigate the geometric effect of σθ on the plane. Solution. The effect on the standard basis vectors is σθ (1, 0) = (cos θ, sin θ),

σθ (0, 1) = (sin θ, − cos θ).

The angle between every pair of vectors is preserved by σθ , as is the length of a vector. We also find that σθ (cos(θ/2), sin(θ/2)) = (cos(θ/2), sin(θ/2)), σθ (− sin(θ/2), cos(θ/2)) = (sin(θ/2), − cos(θ/2)) = −(− sin(θ/2), cos(θ/2)), from which it follows that σθ has the effect of reflecting the plane in the line through the origin and parallel to (cos(θ/2), sin(θ/2)). Notice that the matrix of σθ with respect to the standard basis of R2 is the reflection matrix " # cos θ sin θ . ¤ sin θ − cos θ

44

3. LINEAR TRANSFORMATIONS

# " 1 t Example 3.45. Let S = and let fS : R2 −→ R2 be the real linear transformation 0 1 given by fS (x) = Sx. Then fS has the geometric effect of a shearing parallel to the x-axis. Solution. This can be seen by plotting the effect of fS on points. For example, points on the x-axis are fixed by fS , while points on the y-axis get moved parallel to the x-axis into the line of slope t. ¤ " # d 0 Example 3.46. Let D = where d > 0 and let fD : R2 −→ R2 be the real linear 0 1 transformation defined by fD (x) = Dx. Then fD has the geometric effect of a dilation parallel to the x-axis. Solution. When applying fD , vectors along the x-axis get dilated by a factor d, while those on the y-axis are unchanged. ¤ Example 3.47. Let V a finite dimensional vector space over a field F . Suppose that V = V 0 ⊕ V 00 is the direct sum of two subspaces V 0 , V 00 (see Definition 2.38). Then there is a linear transformation pV 0 ,V 00 : V −→ V given by pV 0 ,V 00 (v0 + v00 ) = v0

(v0 ∈ V 0 , v00 ∈ V 00 ).

Then Ker pV 0 ,V 00 = V 00 and Im pV 0 ,V 00 = V 0 . pV 0 ,V 00 is sometimes called the projection of V onto V 0 with kernel V 00 . The case where V 00 is a line (i.e., dim V 00 = 1) is particularly important. Example 3.48. Find a formula for the projection of R3 onto the plane P through the origin and perpendicular to (1, 0, 1) with kernel equal to the line L spanned by (1, 0, 1). Solution. Let this projection be p : R3 −→ R3 . Then every vector x ∈ R3 can be expressed as

³ √ √ √ √ ´ √ √ √ √ x = x − [(1/ 2, 0, 1/ 2) · x](1/ 2, 0, 1/ 2) + [(1/ 2, 0, 1/ 2) · x](1/ 2, 0, 1/ 2),

where and

√ √ √ √ x − [(1/ 2, 0, 1/ 2) · x](1/ 2, 0, 1/ 2) ∈ P √ √ √ √ [(1/ 2, 0, 1/ 2) · x](1/ 2, 0, 1/ 2) ∈ L.

This corresponds to the direct sum decomposition R3 = P ⊕ L. The projection p is given by √ √ √ √ p(x) = x − [(1/ 2, 0, 1/ 2) · x](1/ 2, 0, 1/ 2).

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CHAPTER 4

Determinants For 2 × 2 and 3 × 3 matrices determinants are defined by the formulae ¯ # ¯ " ¯ ¯a a11 a12 ¯ 11 a12 ¯ (4.1) det =¯ ¯ = a11 a22 − a12 a21 ¯a21 a22 ¯ a21 a22 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯a a a a11 a12 a13 11 12 13 ¯ ¯ ¯a ¯ ¯a ¯ ¯a ¯ ¯ ¯ 11 a12 ¯ ¯ 11 a12 ¯ ¯ 11 a12 ¯ ¯ (4.2) det a21 a22 a23 = ¯a21 a22 a23 ¯ = a11 ¯ ¯ − a12 ¯ ¯ + a13 ¯ ¯. ¯ ¯a21 a22 ¯ ¯a21 a22 ¯ ¯a21 a22 ¯ ¯ ¯ ¯ a31 a32 a33 a31 a32 a33 We will introduce determinants for all n × n matrices over any field of scalars and these will have properties analogous to that of (4.2) which allows the calculation of 3 × 3 determinants from 2 × 2 ones. Definition 4.1. For the n×n matrix A = [aij ] with entries in F , for each pair r, s = 1, . . . , n, the cofactor matrix Ars is the (n − 1) × (n − 1) matrix obtained by deleting the r-th row and s-th column of A. Warning. When working with the determinant of a matrix [aij ] we will often write |A| for det(A) and display it as an array with vertical lines rather than square brackets as in (4.1) and (4.2). But beware, the two expressions a11 a12 a13 a21 a22 a23 a31 a32 a33 and

¯ ¯ ¯a ¯ a a a a a 11 12 13 11 12 13 ¯ ¯ ¯ ¯ ¯a21 a22 a23 ¯ = det a21 a22 a23 ¯ ¯ ¯a31 a32 a33 ¯ a31 a32 a33

are very different things: the first is a matrix, whereas the second is a scalar. They should not be confused! 4.1. Definition and properties of determinants Let F be a field of scalars. Theorem 4.2. For each n > 1, and every n × n matrix A = [aij ] with entries in F , there is a uniquely determined scalar det A = det(A) = |A| ∈ F which satisfies the following properties. (A) For the n × n identity matrix In , det(In ) = 1. 45

46

4. DETERMINANTS

(B) If A, B are two n × n matrices then det(AB) = det(A) det(B). (C) Let A be an n × n matrix. Then we have Expansion along the r-th row: n X (−1)r+j arj det(Arj )

det(A) =

j=1

for any r = 1, . . . , n. Expansion along the s-th column: det(A) =

n X

(−1)s+i ais det(Ais )

i=1

for any s = 1, . . . , n. The signs in these ¯ ¯+ ¯ ¯ ¯− ¯ ¯+ ¯ ¯ .. ¯.

expansions can be obtained from the pattern ¯ − + − · · · ¯¯ ¯ + − + ··· ¯ ¯ − + − · · · ¯¯ .. ¯ .. . . ¯

(D) For an n × n matrix A, det(AT ) = det(A). Notice that the formula det(A) =

n X

(−1)j−1 a1j det(A1j )

j=1

from (C) is a direct generalisation of (4.2). Here are some important results on determinants of invertible matrices that follow from the above properties. Proposition 4.3. Let A be an invertible n × n matrix. (i) det(A) 6= 0 and 1 . det(A−1 ) = det(A)−1 = det(A) (ii) If B is any n × n matrix then det(ABA−1 ) = det(B). Proof. (i) By (B), det(In ) = det(AA−1 ) = det(A) det(A−1 ), hence using (A) we find that det(A) det(A−1 ) = 1, so det(A−1 ) 6= 0 and det(A−1 ) = det(A)−1 =

1 . det(A)

4.1. DEFINITION AND PROPERTIES OF DETERMINANTS

47

(ii) By (B) and (i), det(ABA−1 ) = det(AB) det(A−1 ) = det(A) det(B) det(A−1 ) = det(B) det(A) det(A−1 ) = det(B).

¤

Let us relate this to elementary row operations and elementary matrices, whose basic properties we will assume. Suppose that R is an elementary row operation and E(R) is the corresponding elementary matrix. Then the matrix A0 obtained from A by applying R satisfies A0 = E(R)A, hence (4.3)

det(A0 ) = det(E(R)A) = det(E(R)) det(A).

Proposition 4.4. If A0 = E(R)A is the to the n × n matrix A, then − det(A) det(A0 ) = λ det(A) det(A) In particular,

−1 det(E(R)) = λ 1

result of applying the elementary row operation R

if R = (Rr ↔ Rs ), if R = (Rr → λRr ), if R = (Rr → Rr + λRs ).

if R = (Rr ↔ Rs ), if R = (Rr → λRr ), if R = (Rr → Rr + λRs ).

Proof. If R = Rr ↔ Rs , then it is possible to prove by induction on the size n that det(E(Rr ↔ Rs )) = −1. The initial case is n = 2 where

¯ ¯ ¯0 1¯ ¯ ¯ det(E(R1 ↔ R2 )) = ¯ ¯ = −1. ¯1 0¯

If R = Rr → λRr for λ 6= 0, then expanding along the r-th row gives det(E(Rr → λRr )) = λ det(In ) = λ. Finally, if R = Rr → Rr + λRs with r 6= s, then expanding along the r-th row gives det(E(Rr → Rr + λRs )) = det(In−1 ) + λ det(Inrs ) = 1 + λ × 0 = 1.

¤

Thus to calculate a determinant, first find any sequence R1 , . . . Rk of elementary row operations so that the combined effect of applying these successively to A is an upper triangular matrix L. Then L = E(Rk ) · · · E(R1 )A and so det(L) = det(E(Rk )) · · · det(E(R1 )) det(A), giving det(A) = det(E(Rk ))−1 · · · det(E(R1 ))−1 det(L),

48

4. DETERMINANTS

To calculate det(L) is easy, since

¯ ¯`1 ∗ ∗ ¯ ¯0 ` ∗ ¯ 2 ¯ ¯ 0 0 ` 3 det(L) = ¯ ¯ .. .. .. ¯. . . ¯ ¯0 0 0

∗ ∗ ∗

∗ ∗ ∗ .. .

0 ···

¯ ¯ ∗ ¯¯ ¯` ¯ ¯ 2 ∗ ∗ ∗ ∗¯ ¯ ¯ ¯ 0 `3 ∗ ∗ ∗ ¯ = `1 ¯ . . .. ¯ ¯. . . .. ¯ ¯. . ¯ . ¯¯ ¯0 0 0 ··· 0 `n ¯ ∗ ∗ ∗

¯ ∗ ¯¯ ¯ ∗¯ .. ¯¯ .¯ ¯ 0 `n ¯ ∗ ∗

by definition and repeating this gives det(L) = `1 `2 · · · `n . Using this we can obtain a useful criterion for invertibility of a square matrix.

Proposition 4.5. Let A be an n × n matrix with entries in a field of scalars F . Then det(A) = 0 if and only if A is singular. Equivalently, det(A) 6= 0 if and only if A is invertible.

Proof. By Proposition 4.3 we know that if A is invertible, then det(A) 6= 0. On the other hand, if det(A) 6= 0 then suppose that the reduced echelon form of A is A0 . If there is a zero row in A0 then det(A0 ) = 0. But if A0 = E(Rk ) · · · E(R1 )A, then

0 = det(A0 ) = det(E(Rk )) · · · det(E(R1 )) det(A),

where the scalars det(E(Rk )), . . . , det(E(R1 )), det(A) are all non-zero. Hence this cannot happen and so A0 = In , showing that A is invertible. ¤

There is an explicit formula, Cramer’s Rule, for finding the inverse of a matrix using determinants. However, it is not much use for computations with large matrices and methods based on row and column operations are usually much more efficient. In practise, evaluation of determinants is often simplified by using elementary row operations to create zero’s or otherwise reduced calculations.

Example 4.6. Evaluate the following determinants:

¯ ¯4 ¯ ¯0 ¯ ¯ ¯3 ¯ ¯2

5 3 0 7

1 2 1 0

¯ 2¯¯ 1¯¯ ¯, 3¯ ¯ 3¯

¯ ¯ ¯ x y 2 1 ¯¯ ¯ ¯ ¯ 2 x¯ . ¯ y ¯ ¯ ¯x + y 0 x¯

4.1. DEFINITION AND PROPERTIES OF DETERMINANTS

Solution. ¯ ¯4 5 ¯ ¯0 3 ¯ ¯ ¯3 0 ¯ ¯2 7

1 2 1 0

¯ ¯ x y2 ¯ ¯ 2 ¯ y ¯ ¯x + y 0

49

¯ ¯ ¯ 2¯¯ ¯¯4 0 3 2¯¯ 1¯¯ ¯¯5 3 0 7¯¯ ¯=¯ ¯ [transposing] 3¯ ¯1 2 1 0¯ ¯ ¯ ¯ 3¯ ¯2 1 3 3¯ ¯ ¯ ¯ 4 0 3 2¯¯ ¯ ¯−1 0 −9 −2¯ ¯ ¯ =¯ ¯ [R2 → R2 − 3R4 , R3 → R3 − 2R4 ] ¯−3 0 −5 −6¯ ¯ ¯ ¯ 2 1 3 3¯ ¯ ¯ ¯ 4 3 2¯¯ ¯ ¯ ¯ = ¯−1 −9 −2¯ [expanding along column 2] ¯ ¯ ¯−3 −5 −6¯ ¯ ¯ ¯4 3 2¯ ¯ ¯ ¯ ¯ = ¯1 9 2¯ [multiplying rows 2 and 3 by (−1)] ¯ ¯ ¯3 5 6¯ ¯ ¯ ¯ 4 3 2¯¯ ¯ ¯ ¯ = ¯−3 6 0¯ [R2 → R2 − R1 , R3 → R3 − 3R1 ] ¯ ¯ ¯ 0 −22 0¯ ¯ ¯ ¯ 4 2¯ ¯ ¯ = 22 ¯ ¯ = 22 × 6 = 132, [expanding along row 2] ¯−3 0¯ ¯ ¯ ¯ ¯ ¯ 1 ¯¯ ¯y 2 1 ¯ ¯x y 2 ¯ ¯ ¯ ¯ ¯ ¯ x¯ = (x + y) ¯ ¯ + x¯ ¯ [expanding along row 3] ¯ ¯ 2 x¯ ¯y 2 ¯ ¯ x = (x + y)(xy 2 − 2) + x(2x − y 3 ) = x2 y 2 − 2x + xy 3 − 2y + 2x2 − xy 3 = x2 y 2 − 2x − 2y + 2x2 .

Example 4.7. Solve the equation ¯ ¯1 2 ¯ ¯ ¯x x + 1 ¯ ¯1 1

¤

¯ −1 ¯¯ ¯ x2 ¯ = 0. ¯ 5¯

Solution. Expanding the determinant gives ¯ ¯ ¯ ¯ ¯1 2 −1 ¯¯ 2 −1 ¯¯ ¯¯1 ¯ ¯ ¯ ¯ ¯ x2 + x¯ [R2 → R2 − xR1 , R3 → R3 − R1 ] x2 ¯ = ¯0 1 − x ¯x x + 1 ¯ ¯ ¯ ¯ ¯1 6 ¯ 1 5 ¯ ¯0 −1 ¯ ¯ ¯1 − x x2 + x¯¯ ¯ =¯ ¯ [expanding along column 1] ¯ −1 6 ¯ = x2 + x + 6 − 6x = x2 − 5x + 6 = (x − 2)(x − 3). So the solutions are x = 2, 3.

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50

4. DETERMINANTS

Example 1 x matrix 1 y 1 z

¯ ¯ ¯1 x x2 ¯ ¯ ¯ ¯ ¯ 4.8. Evaluate the Vandermonde determinant ¯1 y y 2 ¯. Determine when the ¯ ¯ ¯1 z z 2 ¯ x2 y 2 is singular. z2

Solution.

¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯1 x x2 ¯ ¯1 x x2 x x2 ¯¯ ¯¯1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯1 y y 2 ¯ = ¯0 y − x y 2 − x2 ¯ = ¯0 y − x (y − x)(y + x)¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯1 z z 2 ¯ ¯0 z − x z 2 − x2 ¯ ¯0 z − x (z − x)(z + x)¯ ¯ ¯ 2 ¯ ¯1 x x ¯ ¯ ¯ ¯ = (y − x)(z − x) ¯0 1 y + x¯ ¯ ¯ ¯0 1 z + x ¯ ¯ ¯ ¯1 y + x¯ ¯ ¯ = (y − x)(z − x) ¯ ¯ ¯0 z − y ¯ = (y − x)(z − x)(z − y).

The matrix is singular precisely when its determinant vanishes, and this happens when at least one of (y − x), (z − x), (z − y) is zero, i.e., when two of the scalars x, y, z are equal. ¤ Remark 4.9. For n > 1, ¯ ¯1 ¯ ¯ ¯1 ¯. ¯. ¯. ¯ ¯1

there is a general formula for an n × n Vandermonde determinant, ¯ ¯ x1 x21 · · · xn−1 ¯ 1 ¯ Y x2 x22 · · · xn−1 ¯ 2 (xs − xr ). .. ¯¯ = .. . . ¯ 16r<s6n ¯ ¯ xn x2n · · · xn−1 n

The proof is similar to that for the case n = 3. Again we can see when the matrix 1 x1 x21 · · · xn−1 1 1 x2 x22 · · · xn−1 2 . .. .. . . . . 1 xn x2n · · · xn−1 n is singular, namely when xr = xs for some pair of distinct indices r, s. 4.2. Determinants of linear transformations Let V be a finite dimensional vector space over the field of scalars F and suppose that f : V −→ V is a linear transformation. Write n = dim V . If we choose a basis S : v1 , . . . vn for V , then there is a matrix S [f ]S and we can find its determinant det(S [f ]S ). A second basis T : w1 , . . . wn leads to another determinant det(T [f ]T ). Appealing to Theorem 3.35(b) we find that for a suitable invertible matrix P , T [f ]T

= P −1 S [f ]S P,

hence by Proposition 4.3, det(T [f ]T ) = det(P −1 S [f ]S P ) = det(S [f ]S ).

4.3. CHARACTERISTIC POLYNOMIALS AND THE CAYLEY-HAMILTON THEOREM

51

This shows that det(S [f ]S ) does not depend on the choice of basis, only on f , hence it is an invariant of f . Definition 4.10. Let f : V −→ V is a linear transformation on a finite dimensional vector space. Then the determinant of f is det f = det(f ) = det(S [f ]S ), where S is any basis of V . This value only depends on f and not on the basis S. Remark 4.11. Suppose that V = Rn . Then there is a geometric interpretation of det f (at least for n = 2, 3, but it makes sense in the general case if we suitably interpret volume in Rn ). The standard basis vectors e1 , . . . , en form the edges of a unit cube based at the origin, with volume 1. The vectors f (e1 ), . . . , f (en ) form edges of a parallellipiped based at the origin. The volume of this parallellipiped is | det f | (note that volumes are non-negative!). This is used in change of variable formulae for multivariable integrals. The value of det f (or more accurately of | det f |) indicates how much volumes are changed by applying f . The sign of det f indicates whether or not ‘orientations’ change and this is also encountered in multivariable integration formulae. 4.3. Characteristic polynomials and the Cayley-Hamilton theorem For a field F , let A be an n × n matrix with entries in F . Definition 4.12. The characteristic polynomial of A is the polynomial χA (t) ∈ F [t] (often denoted charA (t)) defined by χA (t) = det(tIn − A). Proposition 4.13. The polynomial χA (t) has degree n and is monic, i.e., χA (t) = tn + cn−1 tn−1 + · · · + c1 t + c0 , where c0 , c1 , . . . , cn−1 ∈ F . Furthermore, cn−1 = − tr A,

c0 = (−1)n det(A).

Proof. To see the first part, write

¯ ¯t − a ¯ 11 · · · ¯ . .. ¯ det(tIn − A) = ¯ .. . ¯ ¯ −an1 · · ·

¯ ¯ ¯ ¯ ¯ ¯ ¯ t − ann ¯ −a1n .. .

and then apply property (C) repeatedly. The formula for cn−1 follows from the fact that the terms contributing to the tn−1 term are obtained by multiplying n − 1 of the diagonal terms and then multiplying by the constant term in the remaining one. The result is then cn−1 = −(a11 + · · · + ann ) = − tr(A). For the constant term c0 , taking t = 0 and using Proposition 4.4 we obtain c0 = det(−A) = (−1)n det(A). Proposition 4.14. Let B be an invertible n × n matrix, then χBAB −1 (t) = χA (t).

¤

52

4. DETERMINANTS

Proof. This follows from the fact that ¡ ¢ χBAB −1 (t) = det(tIn − BAB −1 ) = det B(tIn − A)B −1 = det(tIn − A) by Proposition 4.3(ii).

¤

This allows us to make the following definition. See Definition 4.10 for a similar idea. Let V be a finite dimensional vector space over a field F and let f : V −→ V be a linear transformation. Definition 4.15. The characteristic polynomial of f is χf (t) = χS [f ]S (t) = det(tIn − S [f ]S ), where S is any basis of V . We now come to an important result which is useful in connection with the notion of eigenvalues to be considered in Chapter 5. Let p(t) = ak tk + ak−1 tk−1 + · · · + a1 t + a0 ∈ F [t] be a polynomial (so a0 , a1 , . . . , ak ∈ F ). For an n × n matrix A we write p(A) = ak Ak + ak−1 Ak−1 + · · · + a1 A + a0 In , which is also an n×n matrix. If p(A) = On then we say that A satisfies this polynomial identity. Similarly, if f : V −→ V is a linear transformation, p(f ) = ak f k + ak−1 f k−1 + · · · + a1 f + a0 IdV . Theorem 4.16 (Cayley-Hamilton theorem). (i) Let A be an n × n matrix. Then χA (A) = On . (ii) Let V be a finite dimensional vector space of dimension n and let f : V −→ V be a linear transformation. Then χf (f ) = 0, where the right hand side is the constant function taking value 0. Informally, these results are sometimes summarised by saying that a square matrix A or a linear transformation f : V −→ V satisfies its own characteristic equation. 0 0 2 Example 4.17. Show that the matrix A = 1 0 0 satisfies a polynomial identity of 0 1 3 degree 3. Hence show that A is invertible and express A−1 as a polynomial in A. Solution. The characteristic polynomial of A is ¯ ¯ ¯ t 0 −2 ¯¯ ¯ ¯ ¯ χA (t) = ¯−1 t 0 ¯ = t3 − 3t2 − 2. ¯ ¯ ¯ 0 −1 t − 3¯ So det(A) = 2 6= 0 and A is invertible. Also, A3 − 3A2 − 2I3 = O3 so A(A2 − 3A) = (A2 − 3A)A = A3 − 3A2 = 2I3 ,

giving

4.3. CHARACTERISTIC POLYNOMIALS AND THE CAYLEY-HAMILTON THEOREM

53

1 A−1 = (A2 − 3A). 2

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CHAPTER 5

Eigenvalues and eigenvectors 5.1. Eigenvalues and eigenvectors for matrices Let F be a field and A ∈ Mn (F ) = Mn×n (F ) be a square matrix. We know that there is an associated linear transformation fA : F n −→ F n given by fA (x) = Ax. We can ask whether there are any vectors which are fixed by fA , i.e., which satisfy the equation fA (x) = x. Of course, 0 is always fixed so we should ask this question with the extra requirement that x be non-zero. Usually there will be no such vectors, but a slightly more general situation that might occur is that fA will dilate some non-zero vectors, i.e., fA (x) = λx for some λ ∈ F and non-zero vector x. This amounts to saying that fA sends all vectors in some line through the origin into that line. Of course, such a line is a 1-dimensional subspace of F n . " # 2 1 . Example 5.1. Let F = R and consider the 2 × 2 matrix A = 0 3 (i) Show that the linear transformation fA : R2 −→ R2 does not fix any non-zero vectors. (ii) Show thatfA dilates all vectors on the x-axis by a factor of 2. (iii) Find all the non-zero vectors which fA dilates by a factor of 3. Solution. (i) Suppose that a vector x ∈ R2 satisfies Ax = x, then it also satisfies the homogenous equation # " −1 −1 (I2 − A)x = 0, i.e., x = 0. 0 −2 But the only solution of this is x = 0. (ii) We have " #" # " # " # 2 1 x 2x x = =2 , 0 3 0 0 0 so vectors on the x-axis get dilated under fA by a factor of 2. (iii) Suppose that Ax = 3x, then " # 1 −1 (3I2 − A)x = 0, i.e., x = 0. 0 0 The set of all solution vectors of this is the line L = {(t, t) : t ∈ R}.

¤

Definition 5.2. Let F be a field and A ∈ Mn (F ). Then λ ∈ F is an eigenvalue for A if there is a non-zero vector v ∈ F n for which Av = λv; such a vector v is called an eigenvalue associated with λ. Definition 5.2 crucially depends on the field F as the following example shows. 55

56

5. EIGENVALUES AND EIGENVECTORS

# " 0 2 Example 5.3. Let A = . 1 0 (i) Show that A has no eigenvalues in Q. (ii) Show that A has two eigenvalues in R. Solution. Notice that A2 − 2I2 = O2 , so if λ is a real eigenvalue with associated eigenvector v, then (A2 − 2I2 )v = 0, which gives (λ2 − 2)v = 0. √ √ Thus we must have λ2 = 2, hence λ = ± 2. However, neither of ± 2 is a rational number although both are real. Working in R2 we have "√ # "√ # " √ # " √ # √ √ − 2 2 2 − 2 A = 2 , A =− 2 . ¤ 1 1 1 1 " # 0 −2 A similar problem occurs with the real matrix which has the two imaginary eigen1 0 √ values ±i 2. Because of this it is preferable to work in an algebraically closed field such as the complex numbers. So from now on we will work with complex matrices and consider eigenvalues in C and eigenvectors in Cn . When discussing real matrices we need to take care to check for real eigenvalues. This is covered by the next result which can be proved easily from the definition. Lemma 5.4. Let A be an n × n real matrix which we can also view as a complex matrix. Suppose that λ ∈ R is an eigenvalue for which there is an associated eigenvector in Cn . Then there is an associated eigenvector in Rn . Theorem 5.5. Let A be an n × n complex matrix and let λ ∈ C. Then λ is an eigenvalue of A if and only if λ is a root of the characteristic polynomial χA (t), i.e., χA (λ) = 0. Proof. From Definition 5.2, λ is an eigenvalue for A if and only if λIn − A is singular, i.e., is not invertible. By Proposition 4.5, this is equivalent to requiring that χA (λ) = det(λIn − A) = 0.

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5.2. Some useful facts about roots of polynomials Recall that every complex (which includes real) polynomial p(t) ∈ C[t] of positive degree n admits a factorisation into linear polynomials p(t) = c(t − α1 ) · · · (t − αn ), where c 6= 0 and the roots αk are unique apart from the order in which they occur. It is often useful to write c(t − β1 )r1 · · · (t − βm )rm ,

5.2. SOME USEFUL FACTS ABOUT ROOTS OF POLYNOMIALS

57

where β1 , . . . , βm are the distinct complex roots of p(t) and rk > 1. Then rk is the (algebraic) multiplicity of the root βk . Again, this factorisation is unique apart form the order in which the distinct roots are listed. When p(t) is a real polynomial, i.e., p(t) ∈ R[t], the complex roots fall into two types: real roots and non-real roots which come in complex conjugate pairs. Thus for such a real polynomial we have p(t) = c(t − α1 ) · · · (t − αr )(t − γ1 )(t − γ 1 ) · · · (t − γs )(t − γ s ), where α1 , . . . , αr are the real roots (possibly with repetitions) and γ1 , γ 1 . . . , γs , γ s are the nonreal roots which occur in complex conjugate pairs γk , γ j . For a non-real complex number γ, (t − γ)(t − γ) = t2 − (γ + γ)t + γγ = t2 − (2 Re γ)t + |γ|2 , where Re γ is the real part of γ and |γ| is the modulus of γ. Proposition 5.6. Let p(t) = tn + an−1 tn−1 + · · · + a1 t + a0 ∈ C[t] be a monic complex polynomial which factors completely as p(t) = (t − α1 )(t − α2 ) · · · (t − αn ), where α1 , . . . , αn are the complex roots, occurring with multiplicities. The following formulae apply for the sum and product of these roots: n X

αi = α1 + · · · + αn = −an−1 ,

i=1

n Y

αi = α1 · · · αn = (−1)n a0 .

i=1

Example 5.7. Let p(t) be a complex polynomial of degree 3. Suppose that p(t) = t3 − 6t2 + at − 6, where a ∈ C, and also that 1 is a root of p(t). Find the other roots of p(t). Solution. Suppose that the other roots are α and β. Then by Proposition 5.6, −(α + β + 1) = −6,

i.e.,

β = 5 − α;

also, (−1) × 1 × αβ = −6,

i.e.,

αβ = 6.

Hence we find that (5 − α)α = 6,

i.e.,

α2 − 5α + 6 = 0.

The roots of this are 2, 3, therefore we obtain these as the remaining roots of p(t).

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Of course we can use this when working with the characteristic polynomial of an n×n matrix A when the roots are the eigenvalues λ1 , . . . , λn of A occurring with suitable multiplicities. If χA (t) = tn + cn−1 tn−1 + · · · + c1 t + c0 , then by Proposition 4.13, (5.1) (5.2)

λ1 + · · · + λn = tr A = −cn−1 , λ1 · · · λn = det A = (−1)n c0 .

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5.3. Eigenspaces and multiplicity of eigenvalues In this section we will work over the field of complex numbers C. Definition 5.8. Let A be an n × n complex matrix and let λ ∈ C be an eigenvalue of A. Then the λ-eigenspace of A is EigA (λ) = {v ∈ Cn : Av = λv} = {v ∈ Cn : (λIn − A)v = 0}, the set of all eigenvectors associated with λ together with the zero vector 0. If A is a real matrix and if λ ∈ R is a real eigenvalue, then the real λ-eigenspace of A is n n EigR A (λ) = {v ∈ R : Av = λv} = EigA (λ) ∩ R .

Proposition 5.9. Let A be an n × n complex matrix and let λ ∈ C be an eigenvalue of A. Then EigA (λ) is a subspace of Cn . If A is a real matrix and λ is a real eigenvalue, then EigR A (λ) is a subspace of the real vector n space R . Proof. By definition, 0 ∈ EigA (λ). Now notice that if u, v ∈ EigA (λ) and z, w ∈ C, then A(zu + wv) = A(zu) + A(wv) = zAu + wAv = zλu + wλv = λ(zu + wv). Hence EigA (λ) is a subspace of Cn . n The proof in the real case easily follows from the fact that EigR A (λ) = EigA (λ) ∩ R .

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Definition 5.10. Let A be an n × n complex matrix and let λ ∈ C be an eigenvalue of A. The geometric multiplicity of λ is dimC EigA (λ). Note that dimC EigA (λ) > 1. In fact there are some other constraints on geometric multiplicity which we will not prove. Theorem 5.11. Let A be an n × n complex matrix and let λ ∈ C be an eigenvalue of A. (i) Let the algebraic multiplicity of λ in the characteristic polynomial χA (t) be rλ . Then geometric multiplicity of λ = dimC EigA (λ) 6 rλ . (ii) If A is a real matrix, then dimC EigA (λ) = dimR EigR A (λ). 2 1 0 Example 5.12. For the real matrix A = −1 0 1, determine its complex eigenvalues. 2 0 0 For each eigenvalue λ, find the corresponding eigenspace EigA (λ). For each real eigenvalue λ, find EigR A (λ). Solution. We have ¯ ¯ ¯ ¯ ¯ ¯ ¯t − 2 −1 0¯¯ ¯ t −1¯ ¯ 1 −1¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ χA (t) = ¯ 1 t −1¯ = (t − 2) ¯ ¯ ¯+¯ ¯ 0 ¯ ¯ t¯ t ¯ ¯−2 ¯ −2 0 t¯ = (t − 2)t2 + (t − 2) = (t − 2)(t2 + 1) = (t − 2)(t − i)(t + i).

5.3. EIGENSPACES AND MULTIPLICITY OF EIGENVALUES

59

So the complex roots of χA (t) are 2, i, −i and there is one real root 2. Now we can find the three eigenspaces. EigA (2): Working over C, we need to solve 2 − 2 −1 0 (2I3 − A)z = 1 2 −1 z = 0, −2 0 2 i.e.,

0 −1 0 2 −1 z = 0. 1 −2 0 2

This is equivalent to the equation

1 0 −1 0 z = 0. 0 1 0 0 0

in which the matrix is reduced echelon, and the general solution is z = (z, 0, z) (z ∈ C). Hence, EigA (2) = {(z, 0, z) ∈ C3 : z ∈ C} and so dimC EigA (2) = 1. Since 2 is real, we can form 3 EigR A (2) = {(t, 0, t) ∈ R : t ∈ R}

which also has dimR EigR A (2) = 1. EigA (i): Working over C, we need to solve i − 2 −1 0 (iI3 − A)z = 1 i −1 z = 0, −2 0 i which is equivalent to the reduced echelon equation 1 0 −i/2 (1 + 2i)/2 z = 0. 0 1 0 0 0 This has general solution (iz, −(1 + 2i)z, 2z)

(z ∈ C),

so EigA (i) = {(iz, −(1 + 2i)z, 2z) ∈ C3 : z ∈ C} and dimC EigA (i) = 1. EigA (−i): Similarly, we have EigA (−i) = {(−iz, −(1 − 2i)z, 2z) ∈ C3 : z ∈ C} and dimC EigA (−i) = 1.

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Notice that in this example, each eigenvalue has geometric multiplicity equal to 1 which is also its algebraic multiplicity. Before giving a general result on this, here is an important observation.

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5. EIGENVALUES AND EIGENVECTORS

Proposition 5.13. Let A be an n × n complex matrix. (i) Suppose that λ1 , . . . , λk are distinct eigenvalues for A with associated eigenvectors v1 , . . . , vk . Then v1 , . . . , vk is linearly independent. (ii) The sum of the eigenspaces EigA (λ1 ) + · · · + EigA (λk ) is a direct sum, i.e., EigA (λ1 ) + · · · + EigA (λk ) = EigA (λ1 ) ⊕ · · · ⊕ EigA (λk ). Proof. (i) Choose the largest r 6 k for which v1 , . . . , vr is linearly independent. Then if r < k, v1 , . . . , vr+1 is linearly dependent. Suppose that for some z1 , . . . , zr+1 ∈ C, not all zero, (5.3)

z1 v1 + · · · + zr+1 vr+1 = 0.

Multiplying by A we obtain z1 Av1 + · · · + zr+1 Avr+1 = 0 and hence (5.4)

z1 λ1 v1 + · · · + zr+1 λr+1 vr+1 = 0.

Now subtracting λr+1 × (5.3) from (5.4) we obtain z1 (λ1 − λr+1 )v1 + · · · + zr (λr − λr+1 )vr + 0vr+1 = 0. But since v1 , . . . , vr is linearly independent, this means that z1 (λ1 − λr+1 ) = · · · = zr (λr − λr+1 ) = 0, and so since the λj are distinct, z1 = · · · = zr = 0. hence we must have zr+1 vr+1 = 0 which implies that zr+1 = 0 since vr+1 6= 0. So we must have r = k and v1 , . . . , vk is linearly independent. (ii) This follows by a similar argument to (i). ¤ Theorem 5.14. Let A be an n × n complex matrix. Suppose that the distinct complex eigenvalues of A are λ1 , . . . , λ` and that for each j = 1, . . . , `, geometric multiplicity of λj = algebraic multiplicity of λj . Then Cn is the direct sum of the eigenspaces EigA (λj ), i.e., Cn = EigA (λ1 ) ⊕ · · · ⊕ EigA (λ` ). In particular, Cn has a basis consisting of eigenvectors of A. −1 2 2 Example 5.15. Let A = 2 2 2. Find the eigenvalues of A and show that C3 has −3 −6 −6 a basis consisting of eigenvectors of A. Show that the real vector space R3 also has a basis consisting of eigenvectors of A.

5.3. EIGENSPACES AND MULTIPLICITY OF EIGENVALUES

61

Solution. The characteristic polynomial of A is ¯ ¯ ¯t + 1 −2 ¯ −2 ¯ ¯ ¯ ¯ χA (t) = ¯ −2 t − 2 −2 ¯ ¯ ¯ ¯ 3 6 t + 6¯ ¯ ¯ ¯t + 1 −2 ¯ 0 ¯ ¯ ¯ ¯ = ¯ −2 t − 2 −t ¯ ¯ ¯ ¯ 3 6 t¯ ¯ ¯ ¯ ¯ ¯t − 2 −t¯ ¯−2 −t¯ ¯ ¯ ¯ ¯ = (t + 1) ¯ ¯ + 2¯ ¯ ¯ 6 ¯ 3 t¯ t¯ ¯ ¯ ¯ ¯! Ã ¯t − 2 −1¯ ¯−2 −1¯ ¯ ¯ ¯ ¯ = t (t + 1) ¯ ¯ + 2¯ ¯ ¯ 6 ¯ 3 1¯ 1¯ ¡ ¢ = t (t + 1)(t + 4) + 2 = t(t2 + 5t + 6) = t(t + 2)(t + 3). Thus the roots of χA (t) are 0, −2, −3 which are all real numbers. Notice that these each have algebraic multiplicity 1. EigA (0): We need to find the solutions of u 1 −2 −2 u 0 (0I3 − A) v = −2 −2 −2 v = 0 w 3 6 6 w 0 which is equivalent to

1 0 0 u 0 0 1 1 v = 0 0 0 0 w 0

and the general solution is (u, v, w) = (0, z, −z) ∈ C3 . Thus a basis for EigA (0) is (0, 1, −1) which is also a basis for the real vector space EigR A (0). So dimC EigA (0) = dimR EigR A (0) = 1. EigA (−2): We need to find the solutions of u −1 −2 −2 u 0 ((−2)I3 − A) v = −2 −4 −2 v = 0 w 3 6 4 w 0 which is equivalent to

1 2 0 u 0 0 0 1 v = 0 0 0 0 w 0

and the general solution is (u, v, w) = (2z, −z, 0) ∈ C3 . Thus a basis for EigA (−2) is (2, −1, 0) which is also a basis for the real vector space EigR A (−2). So dimC EigA (−2) = dimR EigR A (−2) = 1. EigA (−3): We need to find the solutions of u −2 −2 −2 u 0 ((−3)I3 − A) v = −2 −5 −2 v = 0 w 3 6 3 w 0

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5. EIGENVALUES AND EIGENVECTORS

which is equivalent to

1 0 1 u 0 = 0 1 0 v 0 0 0 0 w 0

and the general solution is (u, v, w) = (z, 0, −z) ∈ C3 . Thus a basis for EigA (−2) is (1, 0, −1) which is also a basis for the real vector space EigR A (−3). So dimC EigA (−3) = dimR EigR A (−3) = 1. Then (0, 1, −1), (2, −1, 0), (1, 0, −1) is a basis for the complex vector space C3 and for the real vector space R3 . ¤ 0 1 2 Example 5.16. Let B = −1 0 2. Find the eigenvalues of B and show that C3 has a −2 −2 0 basis consisting of eigenvectors of B. For each real eigenvalue λ determine EigR B (λ). Solution. The characteristic polynomial of B is ¯ ¯ ¯ t −1 −2¯ ¯ ¯ ¯ ¯ χB (t) = ¯1 t −2¯ = t3 + 9t = t(t2 + 9) = t(t − 3i)(t + 3i), ¯ ¯ ¯2 2 t¯ so the eigenvalues are 0, 3i, −3i. We find that EigB (0) = {(2z, −2z, z) : z ∈ C}, EigB (3i) = {((1 + 3i)z, (−1 + 3i)z, −4z) : z ∈ C}, EigB (−3i) = {((1 − 3i)z, (−1 − 3i)z, −4z)) : z ∈ C}. For the real eigenvalue 0, EigR B (0) = {(2t, −2t, t) : t ∈ R}. Note that vectors (2t, −2t, t) with t ∈ R and t 6= 0 are the only real eigenvectors of B. Then (2, −2, 1), (1 + 3i, −1 + 3i, −4), (1 − 3i, −1 − 3i, −4) is a basis for the complex vector space C3 . ¤ 0 0 −1 Example 5.17. Let A = 0 2 0. Find the eigenvalues of A and show that C3 has a 1 0 2 basis consisting of eigenvectors of A. For each real eigenvalue λ determine EigR A (λ). Show that 3 C has no basis consisting of eigenvectors for A. Solution. The characteristic polynomial of A is ¯ ¯ ¯ ¯ t 0 1 ¯ ¯ ¯ ¯ χA (t) = ¯ 0 t − 2 0 ¯ = (t − 1)2 (t − 2), ¯ ¯ ¯−1 0 t − 2¯ hence the eigenvalues are 1 with algebraic multiplicity 2, and 2 with algebraic multiplicity 1. EigA (1): We have to solve the equation 1 0 1 u 0 0 v = 0 , 0 −1 −1 0 −1 w 0

5.4. DIAGONALISABILITY OF SQUARE MATRICES

which is equivalent to

63

1 0 1 u 0 0 1 0 v = 0 . 0 0 0 w 0

This gives EigA (1) = {(z, 0, −z) : z ∈ C}. Notice that dim EigA (1) = 1 which is less than the algebraic multiplicity of this eigenvalue. EigA (2): We have to solve the equation 0 u 2 0 1 0 0 0 v = 0 , 0 −1 0 0 w which is equivalent to

1 0 0 u 0 0 0 1 v = 0 . 0 0 0 w 0

This gives EigA (2) = {(0, z, 0) : z ∈ C}. Thus we obtain dim(EigA (1) + EigA (2)) = dim EigA (1) + dim EigA (2) − dim EigA (1) ∩ EigA (2) = 1 + 1 − 0 = 2. So the eigenvectors cannot span C3 . Similarly, EigR A (1) = {(s, 0, −s) : s ∈ R},

EigR A (2) = {(0, t, 0) : t ∈ R},

and R R R R R dimR (EigR A (1) + EigA (2)) = dimR EigA (1) + dimR EigA (2) − dimR EigA (1) ∩ EigA (2)

= 1 + 1 − 0 = 2.

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This shows that eigenvalues cannot span C3 . These examples show that when the algebraic multiplicity of an eigenvalue is greater than 1, its geometric multiplicity need not be the same as its algebraic multiplicity. 5.4. Diagonalisability of square matrices Now we will consider the implications of the last section for diagonalisability of square matrices. Definition 5.18. Let A, B ∈ Mn (F ) be n × n matrices with entries in a field F . Then A is similar to B if there is an invertible n × n matrix P ∈ Mn (F ) for which B = P −1 AP.

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5. EIGENVALUES AND EIGENVECTORS

Remark 5.19. Notice that if B = P −1 AP then A = P BP −1 = (P −1 )−1 BP −1 , so B is also similar to A. If we take P = In = In−1 , then A = In−1 AIn , so A is similar to itself. Finally, if A is similar to B (say B = P −1 AP ) and B is similar to C (say C = Q−1 BQ), then C = Q−1 (P −1 AP )Q = (Q−1 P −1 )A(P Q) = (P Q)−1 A(P Q), hence A is similar to C. These three observations show that the notion of similarity defines an equivalence relation on Mn (F ). Definition 5.20. Let λ1 , . . . , λn ∈ F be scalars. the n × n matrix λ1 . . diag(λ1 , . . . , λn ) = . 0

The diagonal matrix diag(λ1 , . . . , λn ) is ··· .. . ···

0 .. . λn

with the scalars λ1 , . . . , λn down the main diagonal and 0 everywhere else. An n × n matrix A ∈ Mn (F ) is diagonalisable (over F ) if it is similar to a diagonal matrix, i.e., if there is an invertible matrix P for which P −1 AP is a diagonal matrix. Now we can ask the question: When is a square matrix similar to a diagonal matrix? This is answered by our next result. Theorem 5.21. A matrix A ∈ Mn (F ) is diagonalisable over F if and only F n has a basis consisting of eigenvectors of A. Proof. If F n has a basis of eigenvectors of A, say v1 , . . . , vn with associated eigenvalues λ1 , . . . , λn , let P be the invertible matrix with these vectors as its columns. We have h i AP = Av1 · · · Avn h i = λ1 v1 · · · λn vn h i = v1 · · · vn diag(λ1 , . . . , λn ) = P diag(λ1 , . . . , λn ), from which we obtain P −1 AP = diag(λ1 , . . . , λn ), hence A is similar to the diagonal matrix diag(λ1 , . . . , λn ). On the other hand, if A is similar to a diagonal matrix, say P −1 AP = diag(λ1 , . . . , λn ), then AP = P diag(λ1 , . . . , λn ), and the columns of P are eigenvectors associated with the eigenvalues λ1 , . . . , λn and they also form a basis of F n . ¤

5.4. DIAGONALISABILITY OF SQUARE MATRICES

65

The process of finding a diagonal matrix similar to a given matrix is called diagonalisation. It usually works best over C, but sometimes a real matrix can be diagonalised over R if all its eigenvalues are real. Example 5.22. Diagonalise the real matrix 2 −2 3 A = 1 1 1 . 1 3 −1 Solution. First we ¯ ¯t − 2 ¯ ¯ χA (t) = ¯ −1 ¯ ¯ −1 ¯ ¯t − 2 ¯ ¯ =¯ 0 ¯ ¯ −1

find the complex eigenvalues of A. The characteristic polynomial is ¯ 2 −3 ¯¯ ¯ t − 1 −1 ¯ ¯ −3 t + 1¯ ¯ 2 −3 ¯¯ ¯ t + 2 −t − 2¯ ¯ −3 t + 1¯ ¯ ¯ ¯t − 2 2 −3 ¯¯ ¯ ¯ ¯ = (t + 2) ¯ 0 1 −1 ¯ ¯ ¯ ¯ −1 −3 t + 1¯ ¯ ¯ ¯ ¯! Ã ¯ 1 ¯ ¯ 2 −3¯ −1 ¯ ¯ ¯ ¯ = (t + 2) (t − 2) ¯ ¯−¯ ¯ ¯−3 t + 1¯ ¯ 1 −1¯ = (t + 2) ((t − 2)(t + 1 − 3) − (−2 + 3)) ¡ ¢ = (t + 2) (t − 2)2 − 1 = (t + 2)(t2 − 4t + 4 − 1) = (t + 2)(t2 − 4t + 3) = (t + 2)(t − 3)(t − 1).

Hence the roots of χA (t) are the real numbers −2, 1, 3 and these are the eigenvalues of A. Now solving the equation ((−2)I3 − A)x = 0 we find that the general real solution is x = (11t, t, −14t) with t ∈ R. So an eigenvector for the eigenvalue −2 is (11, 1, −14). Solving (I3 − A)x = 0 we find that the general real solution is x = (t, −t, −t) with t ∈ R. So an eigenvector for the eigenvalue 1 is (1, −1, −1). Solving (3I3 − A)x = 0 we find that the general real solution is x = (t, t, t) with t ∈ R. So an eigenvector for the eigenvalue 3 is (1, 1, 1). 3 The vectors (11, 1, −14), (1, −1, −1), (1, 1, 1) form a basis for the real vector space R and 11 1 1 the matrix P = 1 −1 1 satisfies −14 −1 1 11 1 1 −22 1 3 11 1 1 A 1 −1 1 = −2 −1 3 = 1 −1 1 diag(−2, 1, 3), −14 −1 1 28 −1 3 −14 −1 1 hence AP = P diag(−2, 1, 3). Therefore diag(−2, 1, 3) = P −1 AP , showing that A is diagonalisable.

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APPENDIX A

Complex solutions of linear ordinary differential equations It often makes sense to look for complex valued solutions of differential equations of the form dn f dn−1 f df (A.1) + a + · · · + a1 + a0 f = 0, n−1 n n−1 dx dx dx where the ar are real or complex numbers which do not depend on x. The solutions are functions defined on R and taking values in C. The number n is called the degree of the differential equation. When n = 1, for any complex number a, the differential equation df + af = 0 dx has as its solutions all the functions of the form te−ax , for t ∈ C. Here we recall that for a complex number z = x + yi with x, y ∈ R, ez = ex+yi = ex eyi = ex (cos y + i sin y) = ex cos y + iex sin y. When n = 2, for any complex numbers a, b, we consider the differential equation d2 f df + bf = 0, +a 2 dx dx where we assume that the polynomial z 2 + az + b has the complex roots α, β, which might be equal. (A.2)

• If α 6= β, then (A.2) has as its solutions all the functions of the form seαx + tseβx , for s, t ∈ C. • If α = β, then (A.2) has as its solutions all the functions of the form seαx + txeαx , for s, t ∈ C. When a, b are real, then either both roots are real or there are two complex conjugate roots α, α; suppose that α = u + vi where u, v ∈ R, hence the complex conjugate of α is α = u − vi. In this case we can write the solutions in the form • peux cos vx + qeux sin vx for p, q ∈ C if v 6= 0, • peux + qxeux for p, q ∈ C if v = 0. In either case, to get real solutions we have to take p, q ∈ R.

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Bibliography [BLA] [FLA] [LADR]

T. S. Blyth & E. F. Robertson, Basic Linear Algebra, Springer-Verlag (1998), ISBN 3540761225. T. S. Blyth & E. F. Robertson, Further Linear Algebra, Springer-Verlag (2002), ISBN 1852334258. S. Axler, Linear Algebra Done Right, 2nd edition, corrected 7th printing, Springer-Verlag (2004), ISBN 0387982582. [Schaum] S. Lipschutz & M. Lipson, Schaum’s Outline of Linear algebra, McGraw-Hill (2000), ISBN 0071362002. [Strang] G. Strang, Linear Algebra and Its Applications, 4th edition, BrooksCole, ISBN 0534422004.

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