Linear Vector Spaces

Transcription

1 CHAPTER 1 Linear Vector Spaces Definition A linear vector space over a field F is a triple (V, +, ), where V is a set, + : V V V and : F V V are maps with the properties : (i) ( x, y V ), x + y = y + x. (ii) ( x, y, z V ), x + (y + z) = (x + y) + z. (iii) 0 V V : ( x V ), x + 0 V = 0 V + x = x. (iv) x V, x V : x + ( x) = ( x) + x = 0 V. (v) ( λ F, x, y V ) λ (x + y) = λx + λy. (vi) ( λ, µ F, x V ) (λ + µ) x = λx + µx. (vii) ( λ, µ F, x V ) (λµ) x = λ (µx). (viii) ( x V ) 1 F x = x. Note that in the above and the following we will abbreviate λ x by λx. We sometimes talk of the map as the action of the field F on the vector space V. Sometimes we will refer to the set V as the vector space (where the + and is obvious from the context). Elements of the set V are called vectors, while those of F are called scalars. If the field F is either R or C (which are the only cases we will be interested in), we call V a real vector space or a complex vector space, respectively. Example (1) The set R n of n-tuples (x 1, x 2,..., x n ) of real numbers with + and defined by is a linear vector space over the field R. (x 1, x 2,..., x n ) + (y 1, y 2,..., y n ) = (x 1 + y 1, x 2 + y 2,..., x n + y n ) (2) The set C n of n-tuples of complex numbers (similar). λ (x 1, x 2,..., x n ) = (λx 1, λx 2,..., λx n ) (3) Let Mat m n (F ) be the set of all m n F -valued matrices. Then Mat m n (F ) is a vector space under usual addition of matrices and multiplication by scalars. (4) Let R n+1 [X] be the set of all polynomials up to degree n, i.e. all expressions of the form a 0 +a 1 X +a 2 X a n X n, where a i R, is a vector space over R with the usual definitions of addition of polynomials and multiplication of polynomials by numbers. (Note that we use the suffix n + 1 and not n - because the number of parameters you need to specify a polynomial is n some authors refer to this set as R n [X]. (5) R is a vector space over R! Similarly C is one over C. Note that C is also a vector space over R - though a different one from the previous example! Also note that R is not a vector space over C. Theorem If V is a vector space over F, then (1) ( λ F ) λ0 V = 0 V. (2) ( x V ) 0 F x = 0 V. (3) If λx = 0 V then either λ = 0 F or x = 0 V. (4) ( x V, λ F ) ( λ) x = (λx) = λ ( x). Proof. Exercise! Definition Let (V, +, ) be a vector space over a field F. If a nonempty subset W V is a vector space over the same field F with the operations + and being restricted to W W and F W, respectively, then (W, +, ) is called a subspace of (V, +, ). We usually say that W is a subspace of V. 1

2 1. LINEAR VECTOR SPACES 2 Theorem A nonempty W V is a subspace of V if ( x, y W, λ, µ F ) : λx + µy W Example (1) Every vector space is (trivially) a subspace of itself. (2) Every subspace of V has the element 0 V. Thus the smallest subspace of V is the singleton set {0 V }. (3) In R 3 the subspaces are (i) the origin, (ii) lines passing through the origin, (iii) planes passing through the origin, and (iv) the whole space. Theorem The intersection of any collection of subspaces of V is a subspace of V. Proof. Let C be a collection of subspaces of V and let T = W C W be their intersection. Since 0 V W, W C, 0 V T, so that T. Let x, y T. Then W C, x, y W and thus λx + µy W λ, µ F. So λx + µy T, and thus T is a subspace. Note that this is not true for a union. Given a subset S V, we define S as the intersection of all subspaces of V that contain S. This is a subspace of V and is the smallest subspace containing S. If (x, y) R 2 is not the origin and S = {(x, y)}, then S is the line joining (x, y) to the origin. If S =, then S is {0 V }. Definition Let V be a vector space over F and let S be a nonempty subset V. Then we say that v V is a linear combination of elements of S if x 1, x 2,..., x r S and λ 1, λ 2,..., λ r F such that This helps us to define v = λ 1 x 1 + λ 2 x λ r x r Definition Given a nonempty subset S V, the span of S, denoted Span S is defined to be the set of all linear combinations of elements of S. Since linear combinations of linear combinations are linear combinations, it is obvious that Span S is a subspace. If S =, we define Span S to be the singleton vector space {0 V }. Theorem S = Span S Proof. The result follows trivially for S =. Let us now consider S. x S, since 1 F x = x Span S, we have S Span S. Thus Span S is a subspace containing S. Since S is the smallest subspace containing S, we have S Span S. Let v Span S. Then, x 1,..., x r S, λ 1,..., λ r F such that v = r λ ix i. If W is a subspace of V containing S, then x 1,..., x r W and thus v W. Since S is a subspace containing S, it follows that v S and hence Span S S. Definition A subset S of a vector space V is called a spanning set if Span S = V. Example In R 3, the vectors (1, 0, 0), (0, 1, 0) and (0, 0, 1) form a spanning set, since any vector (x, y, z) R 3 can be written as a linear combination of these three. Of course, this is not the only choice - for example, the set {(1, 1, 1), (1, 2, 1), (0, 1, 3), (0, 0, 1), ( 1, 5, 7)} is also a spanning set. Definition A nonempty subset S of a vector space V is said to be linearly independent if the only way of expressing 0 V as a linear combination of elements of S is the trivial way, i.e. if x 1, x 2,..., x n S and λ 1, λ 2,..., λ n F then λ 1 x 1 + λ 2 x λ n x n = 0 = λ 1 = λ 2 =... = λ n = 0. A subset is called linearly dependent if it is not independent. Example In R 3, the vectors (1, 0, 0), (0, 1, 0) and (0, 0, 1) form a linearly independent set. Theorem No linearly independent subset of a vector space V can contain the vector 0 V. We can restate the theorem above as every subset of a vector space containing the null vector is linearly dependent.

3 1. LINEAR VECTOR SPACES 3 Theorem A superset of a linearly dependent set is linearly dependent. Every subset of a linearly independent set is linearly independent. Proof. Let S be linearly dependent and let T S. Then, x 1,..., x n S such that there is a nontrivial linear combination of these elements that yields 0 V. Since x 1,..., x n T as well, T is linearly dependent. The proof of the next part of the statement follows trivially. Theorem Let V be a vector space over a field F. If S V has at least two elements then it is linearly dependent iff one of its elements can be written as a linear combination of the others. Proof. (only if) Let S be linearly dependent. Then, x 1,..., x n S and λ 1,..., λ n F such that λ 1 x λ n x n = 0, with not all the λ i = 0. Since λ i 0, we can relabel if necessary so that λ 1 0. Then, we can write ( x 1 = λ ) ( 2 x λ ) n x n. λ 1 λ 1 (if) If it is possible to write one of the elements of S, say x 1, as a linear combination of the rest Then we can write x 1 = µ 2 x µ n x n which is a non-trivial linear combination. ( 1) x 1 + µ 2 x µ n x n = 0 V Definition A linearly independent spanning set of a vector space is called its basis. Example In R n the n vectors (1, 0, 0,..., 0), (0, 1, 0,..., 0),...,(0, 0,..., 0, 1) form a basis. This is called the natural basis of R n. Theorem A nonempty subset S of a vector space V is a basis of V iff every element of V can be expressed in a unique way as a linear combination of elements of S. Proof. (only if) Let S be a basis. Then Span S = V and thus x V we can expand x as a linear combination i λ ix i of elements of S. To see that this is unique, note that i λ ix i = i µ ix i implies i (λ i µ i ) x i = 0 V and since the x i s are independent, we have λ i µ i = 0 for all i, so that λ i = µ i. (if) Suppose that every element of V can be expressed in a unique way as a linear combination of elements of S. Then Span S = V. Again, since i 0 F x i = 0 V and the expansion is unique for all x V, the only linear combination that gives the null vector is the trivial one - so that S is also linearly independent. Example Every element in R n+1 [X] can be expanded as a unique linear combination a 0 + a 1 X + a 2 X a n X n of elements of the subset { 1, X, X 2,..., X n}, and hence this is a basis. Example For i = 1,..., n let a i = {a i1, a i2,..., a in }. Then {a 1, a 2,..., a n } is a basis of R n iff the matrix A = [a ij ] n n is invertible. To see this, let x = (x 1, x 2,..., x n ) and consider the expansion x = λ 1 a λ n a n. This corresponds to the system of equations A t λ 1. λ n = x 1. x n. So, an unique linear combination is possible if and only if A t is invertible, i.e. A is invertible. Theorem Let V be a vector space that is spanned by a finite set {v 1,..., v n }. If I = {w 1,..., w m } is a linearly independent subset of V then necessarily m n.

4 1. LINEAR VECTOR SPACES 4 Proof. Consider w 1 I. Since {v 1,..., v n } is a spanning set, λ 1,..., λ n F such that w 1 = λ 1 v λ n v n. Since w 1 0 V (because I is linearly independent), not all λ i = 0 F. By a change of indices if necessary, we can take λ 1 0 F. Then v 1 = λ 1 1 w 1 λ 1 1 λ 2v 2... λ 1 1 λ nv n, which shows that V = Span {v 1, v 2,..., v n } Span {w 1, v 2,..., v n } = V. Again, w 2 = µ 1 w 1 + µ 2 v µ n v n and not all of µ 2,..., µ n = 0 F (otherwise we would have w 2 = µ 1 w 1 - contradicing the independence of I). Again, we can relabel if necessary to get µ 2 0 F, and show that v 2 = µ 1 2 w 2 µ 1 2 µ 1w 1 µ 1 2 µ 3v 3... µ 1 2 µ nv n. This helps us to show that Span {w 1, w 2, v 3,..., v n } = V. If n < m, we could proceed in the same way and finally get Span {w 1, w 2,..., w n } = V. Then we will have a contradiction, because we would be able to express the rest of the w s - w n+1,..., w m as linear combinations of w 1,..., w n. Thus m n. In other words, you can not have more linearly independent vectors than there are in a spanning set. Corollary If V has a finite basis B then every basis of V is finite and has the same number of elements as B. Proof. Let B be another basis for V. If B is an infinite set then, since any subset of a linearly independent set is linearly independent, we can get finite linearly independent subsets of B with arbitrarily many elements, thus contradicting the last theorem. So, B must be finite. Since B is a spanning set and B is linearly independent, we have B B. By reversing the two bases, we have B B. Thus B = B. Definition A finite dimensional vector space V is one which has a finite basis. The number of elements in any basis of V is called the dimension of V, and is denoted by dim V. Example A basis for the vector space Mat m n (F ) is the set {e pq : 1 p m, 1 q n} where e pq is a m n matrix with the elements (e pq ) ij = δ ip δ jq. Thus dim Mat m n (F ) = mn. Theorem Let V be a finite dimensional vector space. If G is a finite spanning set of V and if I G is linearly independent, then there is a basis B of V such that I B G. Proof. If Span I = V then B = I and we have nothing left to prove. Suppose Span I V. Then I G and G \ Span I (If G Span I, we would have Span I Span G = V ). Thus, g 1 G such that g 1 / Span I. Then I = I {g 1 } G is linearly independent. If Span I = V,we are done. If not, g 2 G such that g 2 / Span I and we get a linearly independent set I = I {g 1, g 2 }. If Span I = V we are done. If not, we continue. Since G is a finite set, this process will come to an end. Finally we will get a linearly independent set I {g 1, g 2,..., g r } G which spans V. This is the basis B and obviously I B G. Corollary Every linearly independent subset I of a finite dimensional vector space V can be extended to form a basis. Proof. Since V is finite dimensional, it has a finite basis B. The subset I B I is a finite set that spans V. Now we can use the theorem above. Corollary Every finite spanning set of a finite dimensional vector space contains a basis. Proof. Let G be a finite spanning set of V. Since G is linearly independent, according to the theorem we have a basis B of V such that B G. Corollary If V is of dimension n then every linearly independent set consisting of nelements is a basis of V. Proof. If I is a linearly independent set containing n elements, then B, a basis of V containing I. B = n and thus B = I. But

5 1. LINEAR VECTOR SPACES 5 Corollary If S is a subset of a finite dimensional vector space V, then the following statements are equivalent. (1) S is a basis. (2) S is a maximal independent set ( i.e. if I is an independent set with S I, then S = I). (3) S is a minimal spanning set ( i.e. if G S spans V then G = S). Proof. (1) = (2) : If I is independent with S I, there is a basis B with I B. Then I B - but I = B. Thus I = B. (2) = (1): By the first Corollary above, there exists a basis B with S B. But B is independent and thus, by hypothesis B = S. (1) = (3) : If Span G = V then B G where B is a basis of V. Thus B G S. But since both B and S are bases, we have B = S, and thus B = S. (3) = (1) : Since S spans V, there is a basis B with B S. But B then also spans V and according to the hypothesis, B = S. Corollary If V is of dimension n then every subset containing more than n elements is linearly dependent. Again, no subset with less than n elements can span V. Theorem Let V be afinite dimensional vector space. If W is a subspace of V, then W is also finite dimensional, and dim W dim V. Moreover, we have dim W = dim V iff W = V. Proof. Let dim V = n. If I is a linearly independent subset of W then it is also a linearly independent subset of V and thus I n. Thus, a maximal such subset B exists and is, then, a basis of W. Hence W is also of finite dimension and dim W dim V. Again, if dim W = dim V, we have B = n. Thus B is a linearly independent subset of V with nelements. So, Bis also a basis of V and thus W = SpanB = V.

6 CHAPTER 2 Linear mappings Definition If V and W are vector spaces over the same field F, then a map f : V W is called a linear map if (1) ( x, y V ) f (x + y) = f (x) + f (y), and (2) ( x V ) ( λ F ) f (λx) = λf (x). Example The map f : R 2 R 3 defined by is linear. Example The n maps pr i : R n R defined by are linear. pr i is called the i-th projection map. (x, y) (x + y, x y, y) (x 1,..., x i,..., x n ) x i Example The map D : R n+1 [X] R n+1 [X] defined by is linear. a 0 + a 1 X + a 2 X a n X n a 1 + 2a 2 X na n X n 1 Theorem If the map f : V W is linear, then (1) f (0 V ) = 0 W, and (2) f ( x) = f (x) Proof. (1) f (0 V ) = f (0 F x) = 0 F f (x) = 0 W. (2) 0 W = f (0 V ) = f (x + ( x)) = f (x) + f ( x) - which gives the result if we add f (x) to both sides. Definition Given a linear map f : V W and λ F we define the map λf : V W by ( v V ) (λf) (v) = λf (v) Definition Given two linear maps f, g : V W we define the sum map f + g : V W by ( v V ) (f + g) (v) = f (v) + g (v) It is easy to check that the maps λf and f + g defined above are linear. If we denote the set of all linear maps from V to W by Lin (V, W ), this shows that Lin (V, W ) is a vector space over F under the addition and multiplication by scalar operation defined above. Definition If f : V W is linear we define f : 2 V 2 W by and f : 2 W 2 V by ( X V ) f (X) = {f (x) : x X} ( Y W ) f (Y ) = {x V : f (x) Y }. We call f (X) the direct image of X under f, while f (Y ) is called the inverse image of Y under f. Note that most authors denote f (X) and f (Y ) by f (X) and f 1 (Y ), respectively. It is easy to see that the maps f and f are inclusion preserving, i.e. if X 1 X 2 V then f (X 1 ) f (X 1 ) and if Y 1 Y 2 W then f (Y 1 ) f (Y 2 ). 6

7 2. LINEAR MAPPINGS 7 Theorem Both f and f carry subspaces into subspace, i.e. if X is a subspace of V, then f (X) is a subspace of W and if Y is a subspace of W then f (Y ) is a subspace of V. Proof. Let X be a subspace of V. Let w 1, w 2 f (X) and λ 1, λ 2 F. Then x 1, x 2 X such that f (x 1 ) = w 1 and f (x 2 ) = w 2. Then Thus, f (X) is a subspace. λ 1 w 1 + λ 2 w 2 = λ 1 f (x 1 ) + λ 2 f (x 2 ) = f (λ 1 x 1 + λ 2 x 2 ) f (X) Again, let Y be a subspace of W. Let v 1, v 2 f (Y ) and λ 1, λ 2 F. Then y 1, y 2 Y such that f (v 1 ) = y 1 and f (v 2 ) = y 2. Then f (λ 1 v 1 + λ 2 v 2 ) = λ 1 f (v 1 ) + λ 2 f (v 2 ) = λ 1 y 1 + λ 2 y 2 Y and thus λ 1 v 1 + λ 2 v 2 f (Y ), which proves that f (Y ) is a subspace. Definition The subspace f (V ) of W is called the image of f and is denoted by Im f. Definition The supspace f {0 W } is called the kernel of f and is denoted by Ker f. Example For the map in example , the image is and the Kernel is Im f = {(x + y, x y, y) : x, y R} = Span {(1, 1, 0), (1, 1, 1)}. Ker f = {(0, 0)}. Example For the i-th projection map pr i (example ), we have Im pr i = R, Ker pr i = {x R n : x i = 0}. Example For the differentiation map D on R n+1 [X] (example ) we have Im D = R n [X], Ker D = R. Theorem If f : V W is linear then f is injective iff Ker f = {0 V }. Proof. (only if) Let f be injective. Since f (0 V ) = 0 W for all linear maps, 0 V Ker f. On the other hand, if x Ker f, then f (x) = 0 W = f (0 V ). Since f is injective, we have x = 0 V. (if) Let Ker f = {0 V }. Then if f (x) = f (y) we have f (x y) = f (x) f (y) = 0 W so that x y Ker f = x y = 0 V = x = y. Thus f is injective. Theorem If f : V W is linear and injective and if {v 1,..., v n } is a linearly independent subset of V, then {f (v 1 ),..., f (v n )} is linearly independent. Proof. Let λ 1 f (v 1 ) λ n f (v n ) = 0 W. Then f (λ 1 v λ n v n ) = 0 W and hence λ 1 v λ n v n Ker f. Since f is injective, Ker f = {0 V } so that λ 1 v λ n v n = 0 V and since {v 1, v 2,..., v n } is linearly independent, we have λ 1 =... = λ n = 0. Thus {f (v 1 ),..., f (v n )} is linearly independent. Theorem Let V and W be finite dimensional vector spaces over the same field F. If f : V W is linear we have dim V = dim Im f + dim Ker f Proof. Let {v 1,..., v n } be a basis of Ker f V and {w 1,..., w m } be a basis of Im f W. Since w i Im f, v i V such that f (v i ) = w i. Consider the set J = {v 1,..., v m, v 1,..., v n } V. Let x V. Since f (x) Im f, we can write ( m m m ) f (x) = λ i w i = λ i f (vi ) = f λ i vi so that f (x m λ ivi ) = 0 W. Thus x m λ ivi Ker f so that m m x λ i vi = µ j v j = x = λ i vi + Thus J is a spanning set. µ j v j.

8 2. LINEAR MAPPINGS 8 Again, if m λ ivi + ( n µ m jv j = 0 V, we have f λ ivi + ) n µ jv j = 0 W so that m λ iw i = 0 W. Since {w 1,..., w m } is linearly independent we have λ i = 0. But then n µ jv j = 0 so that µ j = 0. Thus J is linearly independent and is hence a basis. Thus dim V = J = m + n = dim Im f + dim Ker f Definition If f is a linear map dim Im f is called the rank of f and dim Ker f is called the nullity of f. Thus theorem can be written as rank + nullity = dimension of departure space. Theorem Let V and W be vector spaces of dimension n over the same field F. If f : V W is linear then the following statements are equivalent (1) f is injective. (2) f is surjective. (3) f is bijective. (4) f carries bases to bases, i.e. if {v 1,..., v n } is a basis for V then {f (v 1 ),..., f (v n )} is a basis of W. Proof. (1) = (2) = (3) : Let f : V W be linear and injective. Then Ker f = {0 V } so that dim Ker f = 0. According to theorem , we have dim Im f = n and thus Im f = W, so that f is surjective. Thus f is both injective and surjective, so that it is bijective. (2) = (1) = (3) : Let f : V W be linear and surjective. Then Im f = W so that dim Im f = dim W = n and thus dim Ker f = 0 from theorem Thus Ker f = {0 V } and so f is injective and hence bijective. That (3) = (1) and (3) = (2) is obvious. (1) = (4) : Let f : V W be linear and injective and let {v 1,..., v n } be a basis for V. Then {f (v 1 ),..., f (v n )} are n distinct (because f is injective) liearly independent vectors (from theorem ) in W. Thus this provides a basis of W (corollary ). (4) = (1) : Let f : V W be linear and carry bases of V to bases of W, i.e. if {v 1,..., v n } is a basis for V then {f (v 1 ),..., f (v n )}is a basis of W. Let x Ker f and x = n λ iv i. Then 0 V = f (x) = n λ 1if (v i ) and thus λ i = 0 for i = 1, 2,..., n so that x = 0 V. So Ker f = {0 V } and so, according to theorem , f is injective. Definition Let V and W be linear vector spaces over the same field F. A linear map f : V W is called a linear isomorphism if it is a bijection. Two vector spaces V and W are called isomorphic, denoted by V = W if there is a isomorphism from one to the other. Remark The relation is isomorphic to between vector spaces is an equivalence relation. Theorem Let V be a vector space of dimension n 1 over a field F. Then V = F n. Proof. Let {v 1, v 2,... v n } be a basis of V. Consider the map f : V F n defined by ( ) f λ i v i = (λ 1, λ 2,..., λ n ). i Since {v 1, v 2,... v n } is a basis, every vector x V has a unique expansion of the form i λ ix i. Thus the map f is an isomorphism. It is also obviously linear. Theorem If V and W are vector spaces of dimension n over the same field F, then V = W.

9 2. LINEAR MAPPINGS 9 Proof. By the previous theorem, we have V = F n and W = F n. Then linear isomorphisms f : V F n and g : W F n. Then, the map g 1 f : V W is a linear bijection, and hence an isomorphism. For a general map f between two spaces V and W, we have to specify what f does to each element of V. We now show that for a linear map, we have to only worry about what the map does to each element of a basis of V. Theorem Let V and W be finite dimensional vector spaces over the same field F. If {v 1, v 2,..., v n } be a basis of V and let w 1, w 2,..., w n be vectors (not necessarily distinct) in W then there is a unique linear map f : V W such that f (v i ) = w i for i = 1, 2,..., n. Proof. Since every element of V can be uniquely expressed in the form i λ iv i, we can define a map f : V W by ( n ) f λ i v i = λ i w i It is easy to check that this map is linear. Moreover, for each i, we have f (v i ) = f δ ij v j = δ ij w j = w i. To show that this map f is unque, let g : V W be a linear map with the property that g (v i ) = w i. Then for v V, v = i λ iv i, we have ( n ) g (v) = g λ i v i = λ i g (v i ) = λ i w i = f (v). Corollary A linear map is completely and uniquely determined by its action on the basis of its departure space. Corollary Two linear mappings f, g : V W are equal iff they agree on a basis of V.

10 CHAPTER 3 Linear maps and matrices Definition Let V be a finite dimensional vector space. An ordered basis of V is a finite sequence (v i ) 1 i n of elements of V such that the set {v 1, v 2,..., v n } is a basis of V. We have seen that a linear map f between two spaces V and W is uniquely determined by its action on a basis of V. Given ordered bases (v i ) m and (w j ) n on V and W, respectively we define the matrix form of f with respect to the bases (v i ) m and (w j ) n, denoted by Mat f, as a matrix with the elements a ij given by f (v i ) = a ji w j, i = 1, 2,..., m Note the (slightly counterintuitive) order of the indices on a. Also note that if V and W are m and n dimensional, respectively, then for a linear map f : V W we have Matf Mat n m (F ). For the special map id V : V V which maps every v V to itself, it is easy to see that Matid V = I n if V is a n dimensional vector space and any given ordered basis (v i ) n of V is used. Theorem Let V and W be linear vector spaces over the same field F. Then for all linear maps f, g : V W and λ F, and with respect to fixed ordered bases Mat (f + g) = Matf + Matg Mat (λf) = λmatf Proof. Let Matf = [x ij ] n m and Matg = [y ij ] n m relative to fixed ordered bases (v i ) m of V and (w j ) n of W. Then for i = 1, 2,..., m f (v i ) = x ji w j, g (v i ) = y ji w j so that (f + g) (v i ) = f (v i ) + g (v i ) = and thus Mat (f + g) = [x ij + y ij ] n m = Matf + Matg. Similarly so that Mat (λf) = [λx ij ] n m = λmatf. (λf) (v i ) = λf (v i ) = λ (x ji + y ji ) w j, x ji w j, This theorem implies that the map ϑ : Lin (V, W ) Mat n m (F ) defined by f Matf is a linear map. It is also clearly a bijection (why?). Thus we have Theorem If V, W are vector spaces of dimension m, n over F then Lin (V, W ) = Mat n m (F ). Theorem Consider linear maps f : U V and g : V W with fixed ordered bases (u i ) m, (v j ) n and (w k ) p of U, V and W respectively. We have Mat (g f) = Matg Matf 10

11 3. LINEAR MAPS AND MATRICES 11 Proof. Let Matf = [x ij ] n m and Matg = [y ij ] p n. Then g f (u i ) = g (f (u i )) = g x ji v j = = Thus Mat (g f) = [z ij ] p m where x ji g (v j ) = ( p x ji k=1 p y kj x ji w k k=1 z ki = y kj x ji. y kj w k ) Corollary A square matrix is invertible iff it represents an isomorphism. Proof. Let A be an invertible n n square matrix. Thus, there exists a n n square matrix B such that BA = I n. Let V be a n dimensional vector space and let (v i ) n be an ordered basis of V. Let f, g : V V be linear maps represented by the matrices A, B with respect to the ordered basis (v i ) n. Then g f is represented by the matrix BA = I n and thus g f = id V and so f is invertible and is thus an isomorphism. Conversely, let the matrix A represent the linear isomorphism f : V V. Since f is a bijection, it has an inverse f 1 such that f 1 f = id V. Then the matrix B representing f 1 obeys BA = Matf 1 Matf = Mat ( f 1 f ) = Matid V = I n. Thus B = A 1 and the matrix A is invertible. We next consider the question of change of basis and its effect on the matrix representative of a linear map. Definition Given two ordered bases (v i ) n and (v i ) n of a given vector space V the transition matrix from (v i ) n to (v i ) n is defined as the matrix representative of the identity map id V with respect to the bases (v i ) n and (v i ) n. Remark If V = R n and (v i ) n is the natural ordered basis on V then it is easier to determine the transition matrix from (v i ) n to (v i) n, than the other way around. From corollary it is obvious that Theorem Transition matrices are invertible. Theorem If a linear mapping f : V W is represented relative to ordered bases (v i ) m, (w i ) n by the n m matrix A then relative to new ordered bases (v i ) m, (w i ) n the matrix representing f is the n m matrix Q 1 AP where Q is the transition matrix from (w i ) n to (w i) n and P is the transition matrix from (v i ) m to (v i) m. Proof. From the following diagram V ; (v i ) m f; A W ; (w i ) n id V ; P id W ; Q V ; (v i ) m f; B W ; (w i ) n using we get so that B = Q 1 AP. f id V = id W f = f AP = QB

12 3. LINEAR MAPS AND MATRICES 12 The converse of this theorem is also true. Theorem Let (v i ) m, (w i ) n be ordered bases of V, W, respectively. Suppose that A, B are n m matrices such that there are invertible matrices P, Q such that B = Q 1 AP. Then there are ordered bases (v i ) m, (w i ) n of V, W and a linear map f : V W such that A is the matrix of f relative to (v i ) m, (w i ) n and B is the matrix of f relative to (v i ) m, (w i ) n. Proof. If P = [p ij ] m m and Q = [q ij ] n n, define m (i = 1,..., m) v i = p ji v j ; (i = 1,..., n) w i = q ji w j. Since P is invertible, there is (corollary ) an isomorphism f P : V V that is represented by P relative to the ordered basis (v i ) m. Since, by definition v i = f P (v i ), it follows that (v i ) m is an ordered basis of V (by theorem ). P is the transition matrix from (v i ) m to (v i) m. Similarly (w i ) n is an ordered basis of and Q is the transition matrix from (w i ) n to (w i) n. Now let f : V W be the linear map whose matrix, relative to the bases (v i ) m and (w i ) n, is A. Then by theorem the matrix of f relative to the bases (v i ) m and (w i ) n is Q 1 AP = B. Definition If A, B are n n matrices then Bis called similar to A if there is an invertible matrix P such that B = P 1 AP. The next two theorems are very simple to prove. Theorem For n n matrices, the relation is similar to is an equivalence relation. Theorem Two n n matrices A, B are similar iff they represent the same linear map relative to possibly different bases.