1 Linear transformations; the basics

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1 Linear Algebra Fall 2013 Linear Transformations 1 Linear transformations; the basics Definition 1 Let V, W be vector spaces over the same field F. A linear transformation (also known as linear map, or linear mapping, linear operator) is a map T : V W such that 1. T (x + y) = T X + T y for all x, y V (For linear operators it is customary to write tx for the value of T on x, rather than T (x).) 2. T (cx) = ct x for all c F, x V. Exercise 1 Let T : V W be a mapping (function) from V to W. 1. Prove that T is linear if and only if T (ax + by) = at x + bt y for all a, b F, x, y V. 2. Assume T is linear. Prove T 0 = 0 (the first 0 is the zero of V, the second of W ). A few simple examples. 1. Let V, W be vector spaces over a field F. The zero or null mapping, defined by x 0 for all x V, is linear. 2. Let V be a vector space over a field F. The identity map I : V V defined by Ix = x for all x V is linear. Exercise 2 Let T : V W be a linear transformation. 1. The kernel of T is the set ker(t ) = {x V : T x = 0}. Prove: the kernel of T is a subspace of V. 2. Prove that the range of T, the set R(T ) = {T x : x V } is a subspace of W. Theorem 1 Let V, W be vector spaces over the field F ; assume dim V = n (finite). Let{e 1,..., e n } be a basis of V and let w 1,..., w n be vectors in W. There exists a unique linear transformation T : V W such that te j = w j for j = 1,..., n. Theorem 2 Let V, W be vector spaces over the field F ; assume dim V = n (finite), and let T : V W be linear. The following are equivalent: 1. T is one-to-one. 2. ker(t ) = {0}. 3. There exists a basis {e 1,..., e n } of V such that the vectors T e 1,..., T e n are linearly independent in W. 4. For all bases {e 1,..., e n } of V, the vectors T e 1,..., T e n are linearly independent in W. Theorem 3 Let V, W be vector spaces over the field F ; assume dim V = n (finite), and let T : V W be linear. The following are equivalent: 1. T is onto (R(T ) = W ). 2. There exists a basis {e 1,..., e n } of V such that {T e 1,..., T e n } spans W. 3. For all bases {e 1,..., e n } of V, the set {T e 1,..., T e n } spans W.

2 2 THE ALGEBRA OF LINEAR TRANSFORMATIONS 2 Theorem 4 Let V, W be vector spaces over the field F ; assume dim V = n (finite), and let T : V W be linear. The following are equivalent. 1. T is one-to-one and onto. 2. There exists a basis {e 1,..., e n } of V such that {T e 1,..., T e n } is a basis of W. 3. For all bases {e 1,..., e n } of V, the set {T e 1,..., T e n } is a basis of W. 4. T is one-to-one and dim W = n. 5. T is onto and dim W = n. In particular, if V = W, then a linear map T : V V is one-to-one if and only if it is onto. Theorem 5 Let V, W be vector spaces over the field F, assume at least one of them is finite dimensional, and let T : V W be linear. If T is one-to-one and onto, then dimv = dim W. Moreover, T 1 : W V is linear. Definition 2 A one-to-one onto mapping from a vector space V onto a vector space W is called an isomorphism, we say V, W are isomorphic. 2 The algebra of linear transformations Definition 3 If V, W are vector spaces over a field F, we denote by L(V, W ) (very occasionally by L F (V, W ) if we need to emphasize F ) the set of all linear transformations from V to W. We write L(V ) if W = V ; that is L(V ) = L(V, V ). The set L(V, W ) becomes a vector space over F defining operations by if S, T L(V, W ), c F, x V. (S + T )x = Sx + T x, (ct )x = c(t x), Notation. If n N, write J n = {1,..., n} = {i N, 1 i n}. Exercise 3 Assume V, W are finite dimensional vector spaces over the field F, let {e 1,..., e n } be a basis of V and {f 1,..., f m } a basis of W. By Theorem 1 for every pair (i, j) J m J n there exists a unique T ij L(V, W ) such that { fi, if k = j, T ij e k = 0, otherwise. Prove {T ij : (i, j) J m J n } is a basis of L(V, W ). hence dim L(V, W ) = nm. Definition 4 Let X, Y, Z be vector spaces over a field F. If T L(X, Y ) and S L(Y, Z), we define the product ST by ST = S T. That is, for every x X, ST x Z is given by ST x = S(T x). The following properties of the product are immediately verified. 1. If T L(X, Y ) and S L(Y, Z), then ST L(X, Z). 2. If X, Y, Z, W are vector spaces, T L(X, Y ), S L(Y, Z), R L(Z, W ), then (RS)T = R(ST ). 3. If S, T L(X, Y ) and R L(Y, Z), then R(S + T ) = RS + RT. 4. If T L(X, Y ) and R, S L(Y, Z), then (R + S)T = RT + ST. An algebra over a field F is a vector space over F in which a product has been defined that is associative and distributes with respect to addition. The algebra is commutative if the product is commutative; it is an algebra with identity if there exists a multiplicative identity. It is then clear that if V is a vector space over the field F, if S, T L(V ), then the definition given above defines ST L(V ). With this definitions, L(V ) is an algebra with identity. It is usually NOT commutative.

3 2 THE ALGEBRA OF LINEAR TRANSFORMATIONS 3 Exercise 4 Assume A is an algebra with identity e over some field F. An element x A is invertible iff there exists y A such that xy = e = yx. In this case one writes y = x 1. Obviously, if y = x 1, then x = y 1. An element x has a left inverse in A iff there exists y A such that yx = e; it has a right inverse in A iff there exists y A such that xy = e; 1. Prove: Let x, y, z A; assume xy = e = zx. Then y = z = x 1. (If an element has a left inverse and a right inverse, then it is invertible.) 2. If x, y are invertible, so is xy and (xy) 1 = y 1 x 1. We will concentrate now a bit on the algebra L(V ). We will also consider the algebra Halmos calls P, but which I prefer to denote by the more common notation F [X]; it consists of all polynomials with coefficients in F. As a vector space over F, F [X] is infinite dimensional. One has to be careful to not confuse a polynomial with coefficients in F with a polynomial function from F to F. Assume, for example F = Z 2 and consider the polynomials p(x) = X and q(x) = X 2. Then p q. However, for every element of a Z 2 (there are only two such elements) one clearly has p(a) = q(a). Let us begin with a simple observation, but call it a theorem because it is important. Theorem 6 Let V be a finite dimensional vector space over a field F. Let T L(V ). The following are equivalent: 1. T is invertible. 2. T has a left inverse in L(V ). 3. T has a right inverse in L(V ). Exercise 5 The previous theorem is false if the dimension is not finite. Among the many vector spaces that play a role in functional analysis, sequence spaces play a prominent role. One of these is l. The space l consists of all sequences {a n } of real numbers such that sup n N a n <. If {a n }, {b n } are sequences in l, c R, one defines {a n } + {b n } = {a n + b n }, c{a n } = {ca n }. 1. Prove that with the operations as defined, l is a vector space over R. 2. Define transformations S, T by S{a 1, a 2, a 3,...} = {0, a 1, a 2, a 3,...} (add a 0 at the front) T {a 1, a 2, a 3,...} = {a 2, a 3,...} (cut of the first term) Show that S, T are linear, ST = I, thus S has a right inverse, T a left inverse, but neither S nor T is invertible. Definition 5 Let V be a vector space over a field F and let T L(V ). Then T 2, T 3,... are defined in the obvious way; namely T 2 = T T, T 3 = T T 2 ; etc. One completes these definitions by setting T 0 = I (the identity map). If now p is a polynomial in F [X], say p(x) = a n X n + a n 1 X n a 1 X + a 0, one defines p(t ) = a n T n + + a 1 T + a 0 I L(V ). Exercise 6 Let T L(V ). Prove that the map p p(t ) is an algebra homomorphism from F [X] to L(V ). In other words, if c F, p, q F [X], then cp maps to cp(t ), p + q maps to p(t ) + q(t ) and pq maps to p(t )q(t ). All the exercises so far were quite straightforward. Here is a more challenging one. Exercise 7 Assume V is a finite dimensional vector space over F, dim V = n. A linear transformation A L(V ) is said to be nilpotent if there is k N such that A k = 0 (is the zero transformation). It is possible to have A k = 0 without having A = 0. Here is an example in R 3. Define A(x 1, x 2, x 3 ) = (0, x 1, x 2 ). Then A 2 (x 1, x 2, x 3 ) = A(0, x 1, x 2 ) = (0, 0, x 1 ) and A 3 (x 1, x 2, x 3 ) = (0, 0, 0) = 0, That is A 3 = Assume A L(V ) is nilpotent. Prove that I + A is invertible. (I is the identity transformation.) 2. In the definition of being nilpotent no assumption was made about how large the power k could be. Prove that if there is k such that A k = 0 for A L(V ), then the smallest such kis n. That is prove: Ais nilpotent if and only if A n = 0.

4 3 A VISIT TO MATRIXLAND 4 3 A visit to matrixland For what will amount to almost the most general possible example of a linear transformation, we need to introduce some notation. An m n matrix of elements of F is precisely what its name indicates. Rigorously, it is an assignment of a field element usually denoted by a ij (or similar notation) to each (i, j) {1,..., m} {1,..., n}. To be able to work efficiently with these objects it turns out to be convenient to write them out in the form of a rectangle of numbers. Thus the 2 3 matrix of real entries defined by (1, 1) 1, (1, 2) π, (1, 3) 7, (2, 1) 0, (2, 2) 3, (2, 3) 5 is usually written in the form ( ) 1 π But you know all this! I ll denote by M m,n (F ) the set of all m n matrices with entries in the field F. If m = n (square matrices), I ll just write M n (F ) rather than M n,n (F ). Theorem 7 Let V, W be finite dimensional vector spaces, of dimensions n, m, respectively. Fix a basis B = {e 1,..., e n } of V and a basis C = {f 1,..., f m } of W. Having fixed these bases, we can assign to every A = (a ij ) 1 i m,1 j n M m,n (F ) a map T A : V W as follows. If x V, write x = n ξ je j. Define n T A x = a ij ξ j f i. The map T A is linear. The mapping A T A : M m,n (F ) L(V, W ) is one-to-one and onto. If T L(V, W ), then T = T A, where A is the matrix whose columns are the components of T e 1,..., T e n with respect to the basis C. Proof. Proving that if A M m,n (F ), then the map T A is linear is totally straightforward, requires no special tricks or too much pondering. It is left as an exercise. All that remains to prove is that the map A T A is one-to-one and onto. It is one-to-one: This computation is also useful for the onto part. Taking x = e k = n δ kje j we get n T A e j = a ij δ kj f i = a ik f i. It follows that if we have matrices A, B M m,n (F ) such that T A = T B, meaning T A x = T B x for all x V, then for k = 1, 2,..., n, a ik f i = T A e k = T B e k = b ik f i, hence a ik = b ik for 1 = 1,..., m, since {f 1,..., f m } is linearly independent. Thus A = B. It is onto: Let T L(V, W ). The computation we did above shows what the matrix A has to be. For j = 1, 2,..., n, we have that T e j W, hence it is a linear combination of elements of B. In other words, there exist elements a ij F, 1 i m, 1 j n such that By linearity, if x = n ξ je j V, then T x = n ξ j T e j = T e j = a ij f i, j = 1, 2,..., n. ( n m ) n ξ j a ij f i = a ij ξ j f i = T A x.

5 3 A VISIT TO MATRIXLAND 5 Well, this is not the greatest way to explain the correspondence operator matrix. For a better way, let us assume that we have three spaces V, W, X, we have fixed bases in all three, in V, W as in the Theorem 7 (same notation), in X the basis D = {g 1,..., g r }. Let A = (a ik ) 1 i m,1 k n M m,n and consider the operator T A L(V, W ). Let B = (b kj ) 1 k n,1 j r M n,r and consider the operator T B L(X, V ). Let x = r ξ jg j X. Then ( n r n r r n ) T A T B x = T A b kj ξ j e k = a ik b kj ξ j f i = a ik b kj ξ j f i. Introducing the matrix C = (c ij ) 1 i m,1 j r M m,r (F ), where we define c ij = n a ikb kj, we see that T A T B = T C. Definition 6 Let A M m,n (F ), B M n,r (F ). In this case, the case in which the number of columns of A equals the number of rows of B, we define the product AB M m,r (F ) by AB = C, where C = (c ij ) 1 i m,1 j r is defined as above, by n c ij = a ik b kj, 1 i m, 1 j r. We also define addition and scalar multiplication of matrices. You know all this (I hope), but let s write it down for the record. Definition 7 Let m, n N and let A = (a ij ) 1 i m,1 j n, B = (b ij ) 1 i m,1 j n M m,n (F ). We define A + B = (a ij + b ij ) 1 i m,1 j n M m,n (F ). If A = (a ij ) 1 i m,1 j n M m,n (F ) and c F, we define ca = (ca ij ) 1 i m,1 j n M m,n (F ). These definitions have been made in part at least so that the following theorems are true. Theorem 8 1. Let m, n N. With the operations of addition and scalar multiplication just defined, M n,m (F ) is an mn dimensional vector space over F. 2. Let n N. With the operations as defined, the matrix operations satisfy: (a) If A M m,n (F ), B M n,r (F ), C M r,p (F ), then (AB)C = A(BC). (b) If A, B M m,n (F ), C M n,r (F ), then (A + B)C = AC + BC. (c) If A M m,n (F ), B, C M n,r (F ), then A(B + C) = AB + AC. In particular, if m = n, M n (F ) is an algebra with identity over F, the identity matrix being I = (δ ij ) 1 i,j n, where δ ij are the Kronecker deltas. In this algebra a square matrix A is invertible if and only if it has a left inverse, if and only if it has a right inverse. Theorem 9 Let V, W be finite dimensional vector spaces, of dimensions n, m, respectively. Fix a basis B = {e 1,..., e n } of V and a basis C = {f 1,..., f m } of W. The map A T A is a vector space isomorphism M m,n (F ) L(V, W ); if m = n it is also an algebra isomorphism. We prove both theorems simultaneously. Assume first V, W are finite dimensional vector spaces, of dimensions n, m, respectively and B = {e 1,..., e n } is a basis of V, C = {f 1,..., f m } of W. If A M m,n (F ), we define T A as above. We already saw that the map A T A is one-to-one from M m,n (F ) onto L(V, W ). It is quite immediate to see that, with A + B, CA as defined for matrices, one has T A+B = T A + T B, T ca = ct A

6 3 A VISIT TO MATRIXLAND 6 for matrices A, B M m,n (F ), c F. This completes the proof that A T A : M m,n (F ) L(V, W ) is a vector space isomorphism. For fun, let us bring in two more vector spaces, X of dimension R in which we fix a basis D = {g 1,..., g r } and Y of dimension p, where we fix the basis E = {h 1,..., h p }. Fixing these bases defines the correspondence between matrices and operators. I ll use the same notation, T A, regardless of where the matrix A is. For example, if A M r,p (F ), then T A L(Y, X). By the computations we did before we see that the matrix product has been defined precisely so that if A M m,n, B M n,r, then T A T B = T AB. Since operator multiplication, being just composition of functions, is associative, we have if A M m,n, B M n,r, C M r,p : T (AB)C = T AB T C = (T A T B )T C = T A (T B T C ) = T A T BC = T A(BC), since the map A T A is one-to-one, we see that (AB)C = A(BC). If A, B M m,n (F ), C M n,r (F ), then we have T (A+B)C = T A+B T C = (T A + T B )T C = T A T C + T B T C = T AC + T BC = T AC+BC, hence (A + B)C = AC + BC. Similarly one proves that A M m,n (F ), B, C M n,r (F ), implies A(B + C) = AB + AC. If we now restrict to the case m = n = r = p, V = W = X = Y, B = C = D = E, we proved that M n (F ) is an algebra over F. It is immediate that if I is the identity map from V to V then (with somewhat overloaded notation) I = T I. In fact, the matrix corresponding to I is the matrix whose j-th column contains the components of Ie j with respect to the basis; since Ie j = e j, that column has 0 s everywhere except in j-th place where the entry is 1. Assume A M n (F ) and A has a right inverse; that is assume there is B M n (F ) and AB = I. Then T A T B = T I = I, thus T A has a right inverse hence, by Theorem 6, it is invertible. It follows that T B = T 1 A, hence also I = T B T A = T BA, hence BA = I. Similarly one sees that if BA = I then AB = I, hence B = A 1. At this point either all assertions of Theorems 8 and 9 have been proved, or they are immediately provable by any graduate student who is willing to make the effort of proving them. Therefore I declare both theorems proved. Exercise 8 Verify that B = {(1, 1, 0), (1, 0, 1), (0, 1, 1)}, C = {(1, 1, 1), (1, 1, 0), (1, 0, 0)} are bases of R 3. Determine the matrix A of the identity operator with respect to these bases; that is A M 3 (R) such that I = T A. Exercise 9 Let F be a field, assume A M m,n (F ), B M n,m (F ), and AB = I (so I is the m m identity matrix). 1. Prove: m n. 2. Assume F is a field with an infinite number of elements (you may assume, if you wish, F = Q or F = R, or F = C) and m < n. If A M m,n (F ) has a right inverse; i.e., there exists B M n,m (F ) such that AB = I, then A has an infinity of right inverses. A challenging (perhaps) digression, related to the last exercise. A ring is a set R with two operations, and addition (a, b) a + b and a product (a, b) ab such that (R, +) is an abelian (commutative) group, the product is associative and distributes with respect to addition. If there exists e R such that ae = ea = a for all a R, then R is said to be a ring with identity. In this case an element a R is said to have a left inverse if there is b R{ such that} ba = e; it has a right { inverse } if there is b R such that ab = e. And, of course, an element b such ba = e left that is said to be a of a. As mentioned above for matrices, if a has both a left and right ab = e right inverse, these two have to coincide and be the unique inverse. Here is a nice exercise due to Kaplansky: Exercise 10 (Optional) Let R be a ring with identity. Prove that if an element of R has two right inverses, it has an infinity of right inverses. Let s return to the matter at hand. Having defined matrix products, we can give a much nicer definition of the relation between matrices and linear transformations. Let V, W be finite dimensional vector spaces, of dimensions n, m, respectively. Fix a basis B = {e 1,..., e n } of V and a basis C = {f 1,..., f m } of W. If x V we will write x = ξ 1 ξ 2.. ξ n

7 3 A VISIT TO MATRIXLAND 7 iff x = n ξ je j. We use a similar notation for vectors in W ; y W and y = means y = m η if i. If A M m,n, then the correspondence A T A can be described simply by η 1 η 2. η m T A x = Ax. That is, to compute T A x, write x as the n 1 column vector of its components with respect to B, multiply be the matrix A on the left; the result is an m 1 column vector; the components of T A x with respect to the basis C. This interpretation is particularly simple if V = F n, W = F m and we use the canonical bases in those spaces. Then there is really no real distinction between A and T A. One of the most important situations is when V = W. In this case one usually takes the basis B equal to the basis C. To emphasize this I write it as a definition. Definition 8 Let V be a finite dimensional vector space over the field F and let B = {e 1,..., e n } be a basis of V. If T L(V ), the matrix of T with respect to the basis B is the n n matrix A = (a ij ) 1 i,j n such that the j-th column of A consists of the components of T e j with respect to the basis B. Or, in more detail, T e j = n a ije i for j = 1,..., n. Exercise 11 Here is something I mentioned in class. A theorem know as the Cayley-Hamilton theorem asserts that if V is a finite dimensional vector space over a field F and if T L(V ), then there is a very particular p F [X] such that p(t ) = 0. We ll get to this. But now, using ONLY material covered in class and in these notes prove simply that given T L(V ) there has to exist some p F X such that p(t ) = 0. Strong hints: If p(t )x = 0 and q(t )y = 0, then pq(t )y = pq(t )x = 0. A linear operator vanishing on a well selected finite set of vectors will vanish everywhere. If dim V = n, any set of n + 1 vectors is linearly dependent. We now return to Halmos.