Selected Solutions to Linear Algebra Done Right by Sheldon Axler

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1 Selected Solutions to Linear Algebra Done Right by Sheldon Axler Contents Matt Rosenzweig 1 Chapter Chapter Chapter Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercises 17, Exercise Exercise Exercise Exercise Exercise

2 1 Chapter 1 2 Chapter 2 3 Chapter 3 Exercise 1 Let V be a F-vector space with dim(v ) = 1, and let T : V V be a linear operator. theorem, we have that By the rank-nullity 1 = dim(v ) = dim(ker(t )) + dim(im(t )) If T = 0, then T v = 0 v v V ; if T 0, then dim(im(t )) 1. Hence, dim(im(t )) = 1 and since Im(T ) V is a subspace, we conclude that V = Im(T ). Hence, for some v 0 V, T v = λv, where λ F. Consider the linear operator T λi : V V. Since v ker(t ), we see that Im(T λi) = {0}. 3.1 Exercise 2 Consider the function f : R 2 R defined by f(x, y) := ( x 3 + y 3) 1 3, (x, y) R 2 Clearly, f(1, 0) + f(0, 1) = 2, but f(1, 1) = for λ R, f(1, 0) + f(0, 1), which shows that f is not additive. However, 3.2 Exercise Exercise 4 f(λx, λy) = ( λ 3 x 3 + λ 3 y 3) 1 3 = ( λ 3 (x 3 + y 3 ) ) 1 3 = λ ( x 3 + y 3) 1 3 = λf(x, y) Let T Hom F (V, F), and suppose that u V \ ker(t ). 3.4 Exercise 5 Let T : V W be an injective linear map, and let v 1,, v n V be a linearly independent list in V. For α 1,, α n F, we have by linearity that 0 = α 1 T v α n T v n = T (α 1 v α n v n ) α 1 v α n v n ker(t ) = {0}, where the last equality follows by injectivity. We conclude that T v 1,, T v n is a linearly independent list in W. 3.5 Exercise 6 Let V 0, V 1,, V n be F-vector spaces and S j Hom F (V j, V j 1 ) be injective. We prove that S 1 S n : V n V 0 is injective by induction on n. The case n = 1 is true by hypothesis, so suppose S 1 S n 1 : V n 1 V 0 is injective. Observe that ker(s 1 S n ) = {v V 0 : S n v ker(s 1 S n 1 )} ker(s n ) = {v V 0 : S n v ker(s 1 S n 1 )} By our induction hypothesis, S n v ker(s 1 S n 1 ) S n v = 0 v = 0. We conclude that ker(s 1 S n ) = {0}. 3.6 Exercise 7 Suppose v 1,, v n V span V and T Hom F (V, W ) is surjective. Let w W, such that w = T v, for v V. Then v = a 1 v a n v n, for scalars a 1,, a n F, so that w = T v = T (a 1 v a n v n ) = a 1 T v a n T v n span F {T v 1,, T v n } 2

3 3.7 Exercise Exercise 9 Let T : F 4 F 2 be a linear map such that ker(t ) = { (x 1, x 2, x 3, x 4 ) F 4 : x 1 = 5x 2, x 3 = 7x 4 } To show that T is surjective, it suffices by the rank-nullity theorem to show that dim(ker(t )) 2. I claim that the vectors (5, 1, 0, 0) and (0, 0, 7, 1) form a basis for ker(t ). Clearly, these are linearly independent elements in ker(t ). Suppose that (x 1, x 2, x 3, x 4 ) ker(t ). Then x 1 = 5x 2, x 3 = 7x 4, so 3.9 Exercise 10 (x 1, x 2, x 3, x 4 ) = (5x 2, x 2, 7x 4, x 4 ) = x 2 (5, 1, 0, 0) + x 4 (0, 0, 7, 1) Suppose there exists a linear map T : F 5 F 2 with ker(t ) = { (x 1,, x 5 ) F 5 : x 1 = 3x 2, x 3 = x 4 = x 5 } Clearly, T 0 since the element (1, 0, 0, 0, 0) / ker(t ). By the rank-nullity theorem, it suffices show that dim(ker(t )) 2 to arrive at a contradiction. But this is evident since (1, 3, 0, 0, 0) and (0, 0, 1, 1, 1) ker(t ) are linearly independent Exercise 11 Let V, W be a F-vector spaces and T Hom F (V, W ) such that both ker(t ) and Im(T ) are finite-dimensional. Let {v 1,, v n } be a basis set for ker(t ) and {w 1,, w m : w j = T u j, u j V, 1 j m} be a basis set for Im(T ). Observe that u 1,, u m are linearly independent in V. I claim that span F {v 1, v n, u 1,, u m } = V. Assume the contrary: there is some v V \span F {v 1,, v n, u 1,, u m }. T v Im(T ) \ {0}, so there are scalars a 1,, a m, not identically zero, such that 3.11 Exercise 12 T v = a 1 w a m w m = a 1 T u a m T u m = T (a 1 u 1 + a m u m ) Let V and W be F-vector spaces with n = dim(v ) <, m = dim(w ) <. Suppose there exists a surjective linear map T : V W. Then by the rank-nullity theorem, n = dim(ker(t )) + dim(im(t )) = dim(ker(t )) + dim(w ) m Now suppose m n. Let {v 1,, v n } be a basis set for V and {w 1,, w m } be a basis set for W. Let T : V W be the unique linear operator defined by T v j := w j, for 1 j m, and T v j := 0, for m < j n. Clearly, T is surjective Exercise 13 Let V and W be F-vector spaces with n = dim(v ) <, m = dim(w ) <, and suppose U V is a subspace. Suppose there exists T Hom F (V, W ) such that ker(t ) = U. Then by the rank-nullity theorem, dim(v ) = dim(ker(t )) + dim(im(t )) = dim(u) + dim(im(t )) dim(u) = dim(v ) dim(im(t )) dim(u) dim(v ) dim(w ) Now suppose that dim(u) dim(v ) dim(w ) Exercise 14 Let V, W be F-vector spaces with dim(w ) = n <. Suppose that T is injective. Let {w 1, cdots, w n } be a basis set for W. Now suppose there exists S Hom F (W, V ) such that ST = I : V V. 3

4 3.14 Exercise Exercise 16 Suppose U and V are finite-dimensional F-vector spaces and S : V W and T : U V are linear operators. Observe that u ker(st ) T u ker(s) or u ker(t ). Let u 1,, u n be a basis for ker(st ), where n = dim(ker(st )). Set m = dim(ker(s)) and l = dim(ker(t )). Consider the disjoint subsets of indices E 1 := {j : u j ker(t )} and E 2 := {j : T u j 0, T u j ker(s)}. Clearly, E 1 dim(ker(t )). By Exercise 3.5, the T u j, for j E 2, form a linearly independent list in V. Hence, E 2 dim(ker(s)), 3.16 Exercises 17, 18 dim(ker(st )) = n = E 1 + E 2 dim(ker(t )) + dim(ker(s)) The solutions to these exercises follow from the result that the matrix of the composition of two linear operators is the product their respective matrices in corresponding order Exercise 19 Let T Hom F (F n, F m ), and let a 1,1 a 1,n A := M(T ) =.. a m,1 a m,n be the matrix of T with respect to the standard bases of F n and F m. For any (x 1,, x n ) F n, T (x 1,, x n ) = x 1 T (1, 0,, 0) + + x n T (0,, 0, 1) = x 1 (a 1,1,, a m,1 ) + + x n (a 1,n,, a m,n ) = (x 1 a 1,1 + + x n a 1,n,, x 1 a m,1 + + x n a m,n ) 3.18 Exercise 22 We first prove the direction. Let S 1, T 1 denote the inverse operators of S and T, respectively. Then T 1 S 1 : V V is a linear operator and (T 1 S 1 )(ST ) = T 1 (S 1 S)T = T 1 IT = T 1 T = I, (ST )(T 1 S 1 ) = S(T T 1 )S 1 = SS 1 = I which shows that T 1 S 1 is the inverse of ST. We now prove the direction. Since ker(t ) ker(st ), we see that T is injective, hence invertible. Any w ker(s) can be written as w = T v for some v V, which implies that ker(s) = {0}, hence T is injective and therefore invertible Exercise 23 Let V be a finite-dimensional F-vector space, and set n := dim(v ). It suffices to show ST = I T S = I, since the argument for the reverse implication is the same. Since ker(t ) ker(st ) and n <, we have that T is invertible. Since ker(s) V = Im(T ), we see that S is also invertible. Hence, so that ST = I = T S Exercise 24 T = IT = (S 1 S)T = S 1 (ST ) = S 1 I = S 1 T 1 = S Let V be al F-vector space with n = dim(v ) <, and let T Hom F (V ). We first prove. Let λ F such that T = λi. Then for any S Hom F (V ), ST = S(λI) = λ(si) = (λi)s = T S We now prove. Suppose that ST = T S S Hom F (V ). Let {v 1,, v n } be a basis set for V. Define n linear operators S 1,, S n : V V by S j (v k ) = δ jk v j, 1 k n 4

5 For each 1 j n, we have T v j = a 1,j v a n,j v n Observe that for 1 j, k n, δ kj (a 1,j v a n,j v n ) = T (δ kj v j ) = (T S k )v j = (S k T )v j = S k (a 1,j v a n,j v n ) = a k,j v k We see that if k j, then a k,j = 0, and if k = j, then a j,j v j = a 1,j v a n,j v n Since {v 1,, v n } is a basis set, the preceding equality implies that a i,j = 0 for i {1,, n} \ {j}. To complete the proof, we must show that a 1,1 = = a n,n. This follows from observing that for any distinct i, j {1,, n}, we can define the operator S i,j by S i,j v i := v j, S i,j v j := v i and S i,j v k := 0 for k / {i, j} Exercise 25 a i,i v i = T v i = T S i,j v j = S i,j T v j = S i,j (a j,j v j ) = a j,j v i a i,i = a j,j Let V be a finite-dimensional F-vector space with dim(v ) = n > 1. Let v 1,, v n be a basis for V. Let T, S : V V be the unique linear operatos respectively defined by T v 1 = v 1, T v j = 0 2 j n, Sv 1 = 0, Sv j = v j 2 j n Clearly, T and S are nonvertible since they are not injective; however, T + S = I, which shows that the set of noninvertible operators on V is not a subspace of Hom F (V ). 5