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1 Chapter 2 Defn 1. (p. 65) Let V and W be vector spaces (over F ). We call a function T : V W a linear transformation form V to W if, for all x, y V and c F, we have (a) T (x + y) = T (x) + T (y) and (b) T (cx) = ct (x). Fact 1. (p. 65) 1. If T is linear, then T (0) = 0. Note that the first 0 is in V and the second one is in W. 2. T is linear if and only if T (cx + y) = ct (x) + T (y) for all x, y V and c F. 3. If T is linear, then T (x y) = T (x) T (y) for all x, y V. 4. T is linear if and only if, for x 1, x 2,..., x n V and a 1, a 2,..., a n F, we have T ( a i x i = a i T (x i ). Proof.. 1. We know 0 x = 0 for all vectors x. Hence, 0 = 0 T (x) = T (0 x) = T (0) 2. Assume T is linear and c F and x, y V, then T (cx) = ct (x) and T (cx + y) = T (cx) + T (y) = ct (x) + T (y). Assume T (x + y) = T (x) + T (y) for all x, y V. Let a F and x, y V. Then T (ax) = T (ax + 0) = at (x) + T (0) = at (x) and T (x + y) = T (1 x + y) = 1 T (x) + T (y) = T (x) + T (y). 3. To start with, the definition of subtraction is: x y= x + ( y). We also wish to show: x V, 1 x = x. Given 0 x = 0 which we proved earlier, we have (1 + ( 1))x = 0 1 x + ( 1) x = 0 Since the additive inverse in unique, we conclude that ( 1) x = x. AssumeT is linear. Then T (x y) = T (x + 1 y) = T (x) + ( 1)T (y) = T (x) + ( T (y)) = T (x) T (y) for all x, y V. 4. ( ) Let n = 2, a 1 = c, and a 2 = 1. We have that T (cx 1 + x 2 ) = ct (x 1 ) + T (x 2 ). ( ) This is done by induction. Using the definition of linear transformation, we get the base case (n = 2). That is, T (a 1 x 1 + a 2 x 2 ) = T (a 1 x 1 ) + T (a 2 x 2 ) = a 1 T (x 1 ) + a 2 T (x 2 ). Now assume that n 2 and for all x 1, x 2,..., x n V and a 1, a 2,..., a n F, we have T ( a i x i ) = a i T (x i ). 1

2 2 Then for all x 1, x 2,..., x n, x n+1 V and a 1, a 2,..., a n a n+1 F, we have n+1 T ( a i x i ) = T ( a i x i + a n+1 x n+1 ) = T ( = = a i x i ) + T (a n+1 x n+1 ) a i T (x i ) + a n+1 T (x n+1 ) n+1 a i T (x i ). Defn 2. (p. 67) Let V and W be vector spaces, and let T : V W be linear. We define the null space (or kernel) N(T ) of T to be the set of all vectors x in V such that T (x) = 0; N(T ) = {x V : T (x) = 0}. We define the range (or image) R(T ) of T to be the subset W consisting of all images (under T ) of vectors in V ; that is, R(T ) = {T (x) : x V }. Theorem 2.1. Let V and W be vector spaces and T : V W be linear. Then N(T ) and R(T ) are subspaces of V and W, respectively. Proof. To prove that N(T ) is a subspace of V, we let x, y N(T ) and c F. We will show that 0 N(T ) and cx + y N(T ). We know that T (0) = 0. So, 0 N(T ). We have that T (x) = 0 and T (y) = 0 so T (cx + y) = c T (x) + T (y) = c = c 0 = 0. (The last equality is a fact that can be proved from the axioms of vector space. To prove that R(T ) is a subspace of W, assume T (x), T (y) R(T ), for some x, y V and c F. Then, ct (x)+t (y) = T (cx+y), where cx+y V, since T is a linear transformation and so, ct (x) + T (y) R(T ). Also, since 0 V and 0 = T (0), we know that 0 R(T ). Let A be a set of vectors in V. Then T (A) = {T (x) : x A}. Theorem 2.2. Let V and W be vector spaces, and let T : V W be linear. If β = {v 1, v 2,..., v n } is a basis for V, then R(T ) = span(t (β)). Proof. T (β) = {T (v 1 ), T (v 2 ),..., T (v n )}. We will show R(T ) = span(t (β)). We showed that R(T ) is a subspace of W. We know that T (β) R(T ). So, by Theorem 1.5, span(t (β)) R(T ). Let x R(T ). Then there is some x V such that T (x ) = x. β is a basis of V, so there exist scalars a 1, a 2,..., a n such that x = a 1 x 1 + a 2 x a n x n. which is in span(t (β)). Thus R(T ) span(t (β)). x = T (x ) = T (a 1 x 1 + a 2 x a n x n ) = a 1 T (x 1 ) + a 2 T (x 2 ) + + a n T (x n )

3 Defn 3. (p. 69) Let V and W be vector spaces, and let T : V W be linear. If N(T ) and R(T ) are finite-dimensional, then we define the nullity of T, denoted nullity(t ), and the rank of T, denoted rank(t ), to be the dimensions of N(T ) and R(T ), respectively. Theorem 2.3. (Dimension Theorem). Let V and W be vector spaces, and let T : V W be linear. If V is finite-dimensional, then nullity(t )+ rank(t ) = dim(v ). Proof. Suppose nullity(t ) = k and β = {n 1,..., n k } is a basis of N(T ). Since β is linearly independent in V, then β extends to a basis of V, β = {n 1,..., n k, x k+1,..., x n }, where, by Corollary 2c. to Theorem 1.10, i, x i N(T ). Notice that possibly β =. It suffices to show: β = {T (x k+1 ),..., T (x n )} is a basis for R(T ). By Theorem 2.2, T (β ) spans R(T ). But T (β ) = {0} β. So β spans R(T ). Suppose β is not linearly independent. Then there exist c 1, c 2,..., c n F, not all 0 such that 0 = c 1 T (x k+1 ) + + c n T (x n ). 3 Then and But then 0 = T (c 1 x k c n x n ) c 1 x k c n x n N(T ). c 1 x k c n x n = a 1 n 1 + a 2 n a k n k for some scalars a 1, a 2,..., a k, since β = {n 1,..., n k } is a basis of N(T ). Then 0 = a 1 n 1 + a 2 n a k n k c 1 x k+1 c n x n. The c is are not all zero, so this contradicts that β is a linearly independent set. Theorem 2.4. Let V and W be vector spaces, and let T : V W be linear. Then T is one-to-one if and only if N(T ) = {0}. Proof. Suppose T is one-to-one. Let x N(T ). Then T (x) = 0 also T (0) = 0. T being oneto-one implies that x = 0. Therefore, N(T ) {0}. It is clear that {0} N(T ). Therefore, N(T ) = {0}. Suppose N(T ) = {0}. Suppose for x, y V, T (x) = T (y). We have that T (x) T (y) = 0. T (x y) = 0. Then it must be that x y = 0. This implies that y is the additive inverse of x and x is the additive inverse of y. But also y is the additive inverse of y. By uniqueness of additive inverses we have that x = y. Thus, T is one-to-one. Theorem 2.5. Let V and W be vector spaces of equal (finite) dimension, and let T : V W be linear. Then the following are equivalent. (a) T is one-to-one. (b) T is onto. (c) rank(t ) = dim(v ).

4 4 Proof. Assume dim(v ) = dim(w ) = n. (a) (b) : To prove T is onto, we show that R(T ) = W. Since T is one-to-one, we know N(T ) = {0} and so nullity(t ) = 0. By Theorem 2.3, nullity(t ) + rank(t ) = dim(v ). So we have dim(w ) = n = rank(t ). By Theorem 1.11, R(T ) = W. (b) (c) : T is onto implies that R(T ) = W. So rank(t ) = dim(w ) = dim(v ). (c) (a) : If we assume rank(t ) = dim(v ), By Theorem 2.3 again, we have that nullity(t ) = 0. But then we know N(T ) = {0}. By Theorem 2.4, T is one-to-one. Theorem 2.6. Let V and W be vector spaces over F, and suppose that β = {v 1, v 2,..., v n } is a basis for V. For w 1, w 2,..., w n in W, there exists exactly one linear transformation T : V W such that T (v i ) = w i for i = 1, 2,..., n. Proof. Define a function T : V W by T (v i ) = w i for i = 1, 2,..., n and since β is a basis, we can express every vector in V as a linear combination of vectors in β. For x V, x = a 1 v 1 + a 2 v a n v n, we define T (x) = a 1 w 1 + a 2 w a 2 w n. We will show that T is linear. Let c F and x, y V. Assume x = a 1 v 1 +a 2 v 2 + +a n v n and y = b 1 v 1 + b 2 v b n v n. Then cx + y = (ca i + b i )v i (1) T (cx + y) = T ( (ca i + b i )v i ) = (ca i + b i )w i = c a i w i + b i w i = ct (x) + T (y) Thus, T is linear. To show it is unique, we need to show that if F : V W is a linear transformation such that F (v i ) = w i for all i [n], then for all x V, F (x) = T (x). Assume x = a 1 v 1 + a 2 v a n v n. Then T (x) = a 1 w 1 + a 2 w a n w n = a 1 F (v 1 ) + a 2 F (v 2 ) + + a n F (v n ) = F (x) Cor 1. Let V and W be vector spaces, and suppose that V has a finite basis {v 1, v 2,..., v n }. If U, T : V W are linear and U(v i ) = T (v i ) for i = 1, 2,..., n, then U = T. Defn 4. (p. 80) Given a finite dimensional vector space V, an ordered basis is a basis where the vectors are listed in a particular order, indicated by subscripts. For F n, we have {e 1, e 2,..., e n } where e i has a 1 in position i and zeros elsewhere. This is known as the standard ordered basis and e i is the i th characteristic vector. Let β = {u 1, u 2,..., u n } be an ordered basis for a finite-dimensional vector space V. For x V, let a 1, a 2,..., a n be the

5 unique scalars such that x = [x] β = a i u i We define the coordinate vector of x relative to β, denoted [x] β, by a 1 a 2 Defn 5. (p. 80) Let V and W be finite-dimensional vector spaces with ordered bases β = {v 1, v 2,..., v n } and γ = {w 1, w 2,..., w n }, respectively. Let T : V W be linear. Then for each j, 1 j n, there exists unique scalars a i,j F, a i m, such that m T (v j ) = t i,j w i 1 j n. So that [T (v j )] γ = We call the m n matrix A defined by A i,j = a i,j the matrix representation of T in the ordered bases β and γ and write A = [T ] γ β. If V = W and β = γ, then we write A = [T ] β.. a n. t 1,j t 2,j. t m,j. Fact 2. [a 1 x 1 + a 2 x a n x n ] B = a 1 [x 1 ] B + a 2 [x 2 ] B + + a n [x n ] B Proof. Let β = {v 1, v 2,..., v n }. Say x j = b j,1 v 1 + b j,2 v b j,n v n. The LHS: a 1 x 1 + a 2 x a n x n = a 1 (b 1,1 v 1 + b 1,2 v b 1,n v n ) +a 2 (b 2,1 v 1 + b 2,2 v b 2,n v n ) + +a n (b n,1 v 1 + b n,2 v b n,n v n ) = (a 1 b 1,1 + a 2 b 2,1 + + a n b n,1 )v 1 +(a 1 b 1,2 + a 2 b 2,2 + + a n b n,2 )v 2 + The i th position of the LHS is the coefficient c i of v i in: +(a 1 b 1,n + a 2 b 2,n + + a n b n,n )v n a 1 x 1 + a 2 x a n x n = c 1 v 1 + c 2 v c n v n And so, c i = a 1 b 1,i + a 2 b 2,i + + a n b n,i. The RHS: The i th position of the RHS is the sum of the i th position of each a j [x j ] β, which is the coefficient of v i in a j x j = a j (b j,1 v 1 + b j,2 v b j,n v n ) and thus, a j b j,i. We have the i th position of the RHS is a 1 b 1,i + a 2 b 2,i + + a n b n,i. 5

6 6 Fact 2. (a) Let V and W be finite dimensional vector spaces with bases, β and γ, respectively. [T ] γ β [x] β = [T (x)] γ. Proof. Let β = {v 1, v 2,..., v n }. Let x = a 1 v 1 + a 2 v a n v n. Then [T (x)] γ = [T (α 1 v 1 + α 2 v α n v n )] γ = [α 1 T (v 1 ) + α 2 T (v 2 ) + + α n T (v n ))] γ = α 1 [T (v 1 )] γ + α 2 [T (v 2 )] γ + + α n [T (v n )] γ by fact 2 t 1,1 t 1,n α 1... = t n,1 t n,n α n = [T ] γ β [x] γ Fact 3. Let V be a vector space of dimension n with basis β. {x 1, x 2,..., x k } is linearly independent in V if and only if {[x 1 ] β, [x 2 ] β,..., [x k ] β } is linearly independent in F n. Proof. {x 1, x 2,..., x k } is linearly dependent a 1, a 2,..., a k, not all zero, such that a 1 x 1 + a 2 x 2 + a k x k = 0 a 1, a 2,..., a k, not all zero, such that [a 1 x 1 + a 2 x 2 + a k x k ] β = [0] β = 0 since the only way to represent 0( V ) in any basis is 0( F n ) a 1, a 2,..., a k, not all zero, such that a 1 [x 1 ] β + a 2 [x 2 ] β + a k [x k ] β = 0 {[x 1 ] β, [x 2 ] β,..., [x k ] β }is a linearly dependent set Defn 6. (p. 99) Let V and W be vector spaces, and let T : V W be linear. A function U : W V is said to be an inverse of T if T U = I W and UT = I V. If T has an inverse, then T is said to be invertible. Fact 4. v and W are vector spaces with T : V W. If T is invertible, then the inverse of T is unique. Proof. Suppose U : W V is such that T U = I W and UT = I V and X : W V is such that T X = I W and XT = I V. To show U = X, we must show that w W, U(w) = X(w). We know I W (w) = w and I V (X(w)) = X(w). U(w) = U(I W (w)) = UT X(w) = I V X(w) = X(w). Defn 7. We denote the inverse of T by T 1. Theorem Let V and W be vector spaces, and let T : V W be linear and invertible. Then T 1 : W V is linear.

7 Proof. By the definition of invertible, we have: w W, T (T 1 (w)) = w. Let x, y W. Then, (2) (3) (4) (5) T 1 (cx + y) = T 1 (ct (T 1 (x)) + T (T 1 (y))) = T 1 (T (ct 1 (x) + T 1 y)) = T 1 T (ct 1 (x) + T 1 y) = T 1 (x) + T 1 y 7 Fact 5. If T is a linear transformation between vector spaces of equal (finite) dimension, then the conditions of being a.) invertible, b.) one-to-one, and c.) onto are all equivalent. Proof. We start with a) c). We have T T 1 = I V. To show onto, let v V then for x = T 1 (v), we have: T (x) = T (T 1 (v)) = I V (v) = v. Therefore, T is onto. By Theorem 2.5, we have b) c). We will show b) a). Let {v 1, v 2,..., v n } be a basis of V. Then {T (v 1 ), T (v 2 ),..., T (v n )} spans R(T ) by Theorem 2.2. One of the corollaries to the Replacement Theorem implies that {x 1 = T (v 1 ), x 2 = T (v 2 ),..., x n = T (v n )} is a basis for V. We define U : V V by, i, U(x i ) = v i. By Theorem 2.6, this is a well-defined linear transformation. Let x V, x = a 1 v 1 + a 2 v a n v n for some scalars a 1, a 2,..., a n. UT (x) = UT (a 1 v 1 + a 2 v a n v n ) = U(a 1 T (v 1 ) + a 2 T (v 2 ) + + a n T (v n ) = U(a 1 x 1 + a 2 x a n x n ) = a 1 U(x 1 ) + a 2 U(x 2 ) + + a n U(x n ) = a 1 v 1 + a 2 v a n v n = x Let x V, x = b 1 x 1 + b 2 x b n x n for some scalars b 1, b 2,..., b n. T U(x) = T U(b 1 x 1 + b 2 x b n x n ) = T (b 1 U(x 1 ) + b 2 T (x 2 ) + + b n T (x n ) = T (b 1 v 1 + b 2 v b n v n ) = b 1 T (v 1 ) + b 2 T (v 2 ) + + b n T (v n ) = b 1 x 1 + b 2 x b n x n = x Defn 8. (p. 100) Let A be an n n matrix. Then A is invertible if there exits an n n matrix B such that AB = BA = I. Fact 6. If A is invertible, then the inverse of A is unique. Proof. A M n (F ) and AB = BA = I n. Suppose AC = CA = I n. Then we have B = BI n = BAC = I n C = C.

8 8 Defn 9. We denote the inverse of A by A 1. Lemma 1. (p. 101) Let T be an invertible linear transformation from V to W. Then V is finite-dimensional if and only if W is finite-dimensional. In this case, dim(v ) = dim(w ). Proof. ( ) T : V W is invertible. Then T T 1 = I W and T 1 T = I V. Assume dim V = n and {v 1, v 2,..., v n } is a basis of V. We know T is onto since, if y W then for x = T 1 (y), we have T (x) = y. We know R(T ) = W since T is onto and {T (v 1 ), T (v 2 ),..., T (v n )} spans W, by Theorem 2.2. So W is finite dimensional and dim W n = dim V. ( ) Assume dim W = m. T 1 : W V is invertible. Applying the same argument as in the last case, we obtain that V is finite dimensional and dim V dim W. We see that when either of V or W is finite dimensional then so is the other and so dim W dim V and dim V dim W imply that dim V = dim W. Theorem Let V and W be finite-dimensional vector spaces with ordered bases β and γ, respectively. Let T : V W be linear. Then T is invertible if and only if [T ] γ β is invertible. Furthermore, [T 1 ] β γ = ([T ] γ β ) 1. Proof. Assume T is invertible. Let dim V = n and dim W = m. By the lemma, n = m. Let β = { 1, 2,..., n } and γ = {w 1, w 2,..., w n }. We have T 1 such that T T 1 = I W and T 1 T = I V. We will show that [T 1 ] β γ is the inverse of [T ] γ β. We will show the following 2 matrix equations: (6) (7) [T ] γ β [T 1 ] β γ = I n [T 1 ] β γ[t ] γ β = I n To prove (1) we will take the approach of showing that the i th column of [T ] γ β [T 1 ] β γ is e i, the i th characteristic vector. Notice that for any matrix A, Ae i is the i th column of A. Consider [T ] γ β [T 1 ] β γe i. By repeated use of Fact 2a, we have: [T ] γ β [T 1 ] β γe i = [T ] γ β [T 1 ] β γ[w i ] γ = [T ] γ β [T 1 (w i )] β = [T T 1 (w i )] γ = [w i ] γ = e i To prove (7) the proof is very similar. We will show that the i th column of [T 1 ] β γ[t ] γ β is e i, the i th.

9 9 Consider [T 1 ] β γ[t ] γ β e i. By repeated use of Fact 2a, we have: [T 1 ] β γ[t ] γ β e i = [T 1 ] β γ[t ] γ β [v i] β = [T 1 ] β γ[t (v i )] γ = [T 1 (T (v i ))] β = [v i ] β = e i Thus we have ( ) and the Furthermore part of the statement. Now we assume A = [T ] γ β is invertible. Call its inverse A 1. We know that it is square, thus n = m. Let β = { 1, 2,..., n } and γ = {w 1, w 2,..., w n }. Notice that the i th column of A is A i = (t 1,i, t 2,i,..., t n,i ) where T (v i ) = t 1,i w 1 + t 2,i w t n,i w n To show T is invertible, we will define a function U and prove that it is the inverse of T. Let the i th column of A 1 be C i = (c 1,i, c 2,i,..., c n,i ). We define U : W V by for all i [n], U(w i ) = c 1,i 1 +c 2,i c n,i n. Since we have defined U for the basis vectors γ, we know from Theorem 2.6 that U is a linear transformation. We wish to show that T U is the identity transformation in W and UT is the identity transformation in V. So, we show (1) T U(x) = x, x W and (2) UT (x) = x, x V. Starting with (1.). First lets see why, i [n], T (U(w i )) = w i. T (U(w i )) = T (c 1,i 1 +c 2,i c n,i n ) = c 1,i (t 1,1 w 1 + t 2,1 w t n,1 w n ) +c 2,i (t 1,2 w 1 + t 2,2 w t n,2 w n ) Gathering coefficients of w 1, w 2,..., w n, we have: + + c n,i (t 1,n w 1 + t 2,n w t n,n w n ) T (U(w i )) = (c 1,i t 1,1 + c 2,i t 1,2 + + c n,i t 1,n )w 1 +(c 1,i t 2,1 + c 2,i t 2,2 + + c n,i t 2,n )w (c 1,i t n,1 + c 2,i t n,2 + + c n,i t n,n )w n We see that the coefficient of w j in the above expression is Row j of A dot Column i of A 1, which is always equal to 1 if and only if i = j. Thus we have that T (U(w i )) = w i. Now we see that for x W then x = b 1 w 1 +b 2 w 2 + +b n w n for some scalars b 1, b 2,..., b n. T U(x) = T U(b 1 w 1 + b 2 w b n w n ) = T (b 1 U(w 1 ) + b 2 U(w 2 ) + + b n U(w n )) = b 1 T (U(w 1 )) + b 2 T (U(w 2 )) + + b n T (U(w n )) = b 1 w 1 + b 2 w b n w n = x

10 10 Similarly, UT (x) = x for all x in V. Cor 1. Let V be a finite-dimensional vector space with an ordered basis β, and let T : V V be linear. Then T is invertible if and only if [T ] β is invertible. Furthermore, [T 1 ] β = ([T ] β ) 1. Proof. Clear. Defn 10. Let A M m n (F ). Then the function L A : F n F m where for x F n, L A (x) = Ax and is called left-multiplication by A. Fact 6. (a) L A is a linear transformation and for β = {e 1, e 2,..., e n }, [L A ] β = A. Proof. We showed in class. Cor 2. Let A be an n n matrix. Furthermore, (L A ) 1 = L A 1. Then A is invertible if and only if L A is invertible. Proof. Let V = F n. Then L A : V V. Apply Corollary 1 with β = {e 1, e 2,..., e n } and [L A ] β = A, we have: L A is invertible if and only if A is invertible. The furthermore part of this corollary says that (L A ) 1 is left multiplication by A 1. We will show L A L A 1 = L A 1L A = I V. So we must show L A L A 1(x) = x, x V and L A 1L A (x) = x, x V. We have L A L A 1(x) = AA 1 x = x and L A 1L A (x) = A 1 Ax = x. Defn 11. Let V and W be vector spaces. We say that V is isomorphic to W if there exists a linear transformation T : V W that is invertible. Such a linear transformation is called an isomorphism from V onto W. Theorem Let V and W be finite-dimensional vector spaces (over the same field). Then V is isomorphic to W if and only if dim(v ) = dim(w ). Proof. ( ) Let T : V W be an isomorphism. Then T is invertible and dim V = dim W by the Lemma. ( ) Assume dim V = dim W. Let β = {v 1,..., v k } and γ = {w 1, w 2,..., w k } be bases for V and W, respectively. Define T : V W where v i w i for all i. Then [T ] γ β = I which is invertible and Theorem 2.18 implies that T is invertible and thus, an isomorphism. Fact 7. Let P M n (F ) be invertible. W is a subspace of F n implies L P (W ) is a subspace of F n and dim(l P (W )) = dim(w ). Proof. Let x, y L P (W ), a F, then there exist x and y such that P x = x and P y = y. So P (ax + y ) = ap x + P y = ax + y. So ax + y L P (W ). Also, we know 0 W and L P (0) = 0 since L P is linear, so we have that 0 L P (W ). Therefore, L P (W ) is a subspace. Let {x 1, x 2,..., x k } be a basis of W. a 1 P (x 1 ) + a 2 P (x 2 ) + a k P (x k ) = 0 P (a 1 x 1 + a 2 x 2 + a k x k ) = 0 a 1 x 1 + a 2 x 2 + a k x k = 0 a 1 = a 2 = = a k = 0 {P (x 1 ), P (x 2 ),..., P (x k )} is linearly independent

11 To show that {P (x 1 ), P (x 2 ),..., P (x k )} spans L P (W ), we let z L P (W ). Then there is some w W, such that L P (W ) = z. If w = a 1 x 1 + a 2 x a k x k, we have z = P w = P (a 1 x 1 + a 2 x a k x k ) = a 1 P (x 1 ) + a 2 P (x 2 ) + + a k P (x k ). And with have that {P (x 1 ), P (x 2 ),..., P (x k )} is a basis of L P (W ) and dim(w ) = dim(l P (W )). 11 So far, we did not define the rank of a matrix. We have: Defn 12. Let A M m n (F ). Then, Rank(A)= Rank(L A ). And the range of a matrix, R(A) is the same as R(L A ). Fact 8. Let S M n (F ). If S is invertible, then R(S) = F n. Proof. L S : F n F n. We know S is invertible implies L S is invertible. We already know that, since L S is onto, R(L S ) = F n. By the definition of R(S), we have R(S) = F n. Fact 9. S, T M n (F ), S invertible and T invertible imply ST is invertible and its inverse is T 1 S 1. Proof. This is the oldest proof in the book. Theorem Let V and W be finite-dimensional vector spaces over F of dimensions n and m, respectively, and let β and γ be ordered bases for V and W, respectively. Then the function Φ : L(V, W ) M m n (F ) defined by Φ(T ) = [T ] γ β for T L(V, W ), is an isomorphism. Proof. To show Φ is an isomorphism, we must show that it is invertible. Let β = {v 1, v 2,..., v n } and γ = {w 1, w 2,..., w n }. We define U : M m n (F ) L(V, W ) as follows. Let A M m n (F ) and have i th column: a 1,i a 2,i. a n,i So U maps matrices to transformations. U(A) is a transformation in L(V, W ). We describe the action of the linear transformation U(A) on v i and realize that this uniquely defines a linear transformation in L(V, W ). i [n], U(A)(v i ) = a 1,i w 1 + a 2,i w a n,i w n. Then A = [U(A)] γ β. Thus, ΦU(A) = A. To verify that U(Φ)(T ) = T, we see what the action is on v i. U(Φ(T ))(v i ) = U([T ] γ β )(v i) = t 1,i w 1 + t 2,i w t n,i w n = T (v i ) Cor 1. Let V and W be finite-dimensional vector spaces over F of dimensions n and m, respectively. Then L(V, W ) is finite-dimensional of dimension mn.

12 12 Proof. This by Theorem Theorem Let β and γ be two ordered bases for a finite-dimensional vector space V, and let Q = [I V ] γ β. Then (a) Q is invertible. (b) For any v V, [v] γ = Q[v] β. Proof. We see that Let β = {v 1,..., v n } and γ = {w 1,..., w n } We claim the inverse of Q is [I V ] β γ. We see that the i th column of [I V ] γ β [I V ] β γ is [I V ] γ β [I V ] β γe i. We have [I V ] γ β [I V ] β γe i = [I V ] γ β [I V ] β γ[w i ] γ = [I V ] γ β [I V (w i )] β = [I V (I V (w i ))] γ = [w i ] γ = e i and [I V ] β γ[i V ] γ β e i = [I V ] β γ[i V ] γ β [v i] β = [I V ] β γ[i V (v i )] γ = [I V (I V (v i ))] β = [v i ] β = e i We know, [v] γ = [I V (v)] γ = [I V ] γ β [v] β = Q[v] β. Defn 13. (p. 112) The matrix Q = [I V ] γ β defined in Theorem 2.22 is called a change of coordinate matrix. Because of part (b) of the theorem, we say that Q changes β- coordinates into γ-coordinates. Notice that Q 1 = [I V ] β γ. Defn 14. (p. 112) A linear transformation T : V V is called a linear operator. Theorem Let T be a linear operator on a finite-dimensional vector space V, and let β and γ be ordered bases for V. Suppose that Q is the change of coordinate matrix that changes β-coordinates into γ-coordinates. Then Proof. The statement is short for: [T ] β = Q 1 [T ] γ Q. [T ] β β = Q 1 [T ] γ γq.

13 13 Let β = {v 1, v 2,..., v n }. As usual, we look at the i th column of each side. We have The i th column of [T ] β β. Q 1 [T ] γ γqe i = [I V ] β γ[t ] γ γ[i V ] γ β [v i] β = [I V ] β γ[t ] γ γ[i V (v i )] γ = [I V ] β γ[t ] γ γ[v i ] γ = [I V ] β γ[t (v i )] γ = [I V (T (v i ))] β = [T (v i )] β = [T ] β β [v i] β = [T ] β β e i Cor 1. Let A M n n (F ), and let γ be an ordered basis for F n. Then [L A ] γ = Q 1 AQ, where Q is the n n matrix whose j th column is the j th vector of γ. Proof. Let β be the standard ordered basis for F n. Then A = [L A ] β. So by the theorem, [L A ] γ = Q 1 [L A ] β Q. Defn 15. (p. 115) Let A and B be in M n (F ). We say that B is similar to A if there exists an invertible matrix Q such that B = Q 1 AQ. Defn 16. (p. 119) For a vector space V over F, we define the dual space of V to be the vector space L(V, F ), denoted by V. Let β = {x 1, x 2,..., x n } be an ordered basis for V. For each i [n], we define f i (x) = a i where a i is the i th coordinate of [x] β. Then f i is in V called the i th coordinate function with respect to the basis β. Theorem Suppose that V is a finite-dimensional vector space with the ordered basis β = {x 1, x 2,..., x n }. Let f i (1 i n) be the ith coordinate function with respect to β and let β = {f 1, f 2,..., f n }. Then β is an ordered basis for V, and, for any f V, we have f = f(x i )f i. Proof. Let f V I. Since dim V = n, we need only show that f = f(x i )f i, from which it follows that β generates V, and hence is a basis by a Corollary to the Replacement Theorem. Let g = f(x i )f i. For 1 j n, we have ( ) g(x j ) = f(x i )f i (x j ) = f(x i )f i (x j ) = f(x i )δ i,j = f(x j ).

14 14 Therefore, f = g.

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