Solutions for Math 225 Assignment #5 1

Solutions for Math 225 Assignment #5 1 (1) Find a polynomial f(x) of degree at most 3 satisfying that f(0) = 2, f( 1) = 1, f(1) = 3 and f(3) = 1. Solution. By Lagrange Interpolation, ( ) (x + 1)(x 1)(x 3) f(x) = 2 (0 + 1)(0 1)(0 3) ( ) x(x 1)(x 3) + 1 ( 1)( 1 1)( 1 3) ( ) x(x + 1)(x 3) + 3 1(1 + 1)(1 3) ( ) x(x + 1)(x 1) + ( 1) 3(3 + 1)(3 1) = 2 3 (x + 1)(x 1)(x 3) 1 x(x 1)(x 3) 8 3 4 x(x + 1)(x 3) 1 x(x + 1)(x 1). 24 (2) Let T : R 3 R 3 be the linear transformation satisfying T (e 1 ) = e 2, T (e 2 ) = e 3, and rank(t 3 ) = 0. (a) Find the kernel K(T ) and the range R(T ) of T. (b) Find the matrix [T ] representing T under the standard basis. Solution. Since rank(t 3 ) = 0, T 3 = 0 and hence T 3 (e 1 ) = 0. Therefore, 0 = T 3 (e 1 ) = T (T (T (e 1 ))) = T (T (e 2 )) = T (e 3 ). It follows that T (x, y, z) = xt (e 1 ) + yt (e 2 ) + zt (e 3 ) = (0, x, y). Therefore, K(T ) = Span{e 3 } = {(x, y, z) : x = y = 0}, R(T ) = Span{T (e 1 ), T (e 2 ), T (e 3 )} = Span{e 2, e 3 } = {(x, y, z) : x = 0} and 1 http://www.math.ualberta.ca/ xichen/math22514w/hw5sol.pdf 1

2 [T ] = 0 0 0 1 0 0. 0 1 0 (3) Let V = {f(x) R[x] : deg f 3} be the vector space of real polynomials in x of degree 3 and let T 1 : V V and T 2 : V V be two linear transformations given by T 1 (f(x)) = f(x + 1) and T 2 (f(x)) = f(x 1). (a) Show that T 1 = T 1 2. (b) Find the matrix representations [T 1 ] B,B and [T 2 ] B,B of T 1 and T 2, respectively. (c) Verify that [T 1 ] B,B = [T 2 ] 1 B,B. Here B is the basis {1, x, x 2, x 3 }. Solution. Since T 1 T 2 (f(x)) = T 1 (T 2 (f(x))) = T 1 (f(x 1)) = f((x + 1) 1) = f(x) and T 2 T 1 (f(x)) = T 2 (T 1 (f(x))) = T 2 (f(x + 1)) = f((x 1) + 1) = f(x), we conclude that T 1 = T 1 2. Since T 1 (1) = 1 = 1 T 1 (x) = x + 1 = 1 + x T 1 (x 2 ) = (x + 1) 2 = 1 + 2x + x 2 T 1 (x 3 ) = (x + 1) 3 = 1 + 3x + 3x 2 + x 3 we obtain 1 1 1 1 [T 1 ] B,B = 0 1 2 3 0 0 1 3. 0 0 0 1

3 Since we obtain T 2 (1) = 1 = 1 T 2 (x) = x 1 = 1 + x T 2 (x 2 ) = (x 1) 2 = 1 2x + x 2 T 1 (x 3 ) = (x 1) 3 = 1 + 3x 3x 2 + x 3 1 1 1 1 [T 2 ] B,B = 0 1 2 3 0 0 1 3. 0 0 0 1 Finally, since 1 1 1 1 1 1 1 1 1 0 1 2 3 0 1 2 3 0 0 1 3 0 0 1 3 = 0 0 0 1 0 0 0 1 1 1 1 we conclude [T 1 ] B,B = [T 2 ] 1 B,B. (4) Which of the following statements are true and which are false? Justify your answer. (a) Let T : V W and S : U V be two linear transformations. If both T and S are onto, so is T S. Proof. True. Since S is onto, S(U) = V. So T (S(U)) = T (V ). And since T is onto, T (V ) = W. Therefore, T (S(U)) = W. Consequently, T S is onto. (b) Let T : V W and S : U V be two linear transformations. If both T and S are 1-1, so is T S. Proof. True. Suppose that T S(v 1 ) = T S(v 2 ) for some v 1, v 2 V. Then T (S(v 1 )) = T (S(v 2 )). Since T is 1-1, S(v 1 ) = S(v 2 ). And since S is 1-1, v 1 = v 2. So T S is 1-1. (c) Let T 1 : V W and T 2 : V W be two linear transformations. Then rank(t 1 + T 2 ) rank(t 1 ) + rank(t 2 ).

4 Proof. True. Since T 1 (v) R(T 1 ) and T 2 (v) R(T 2 ), (T 1 + T 2 )(v) = T 1 (v) + T 2 (v) R(T 1 ) + R(T 2 ) for all v V, R(T 1 + T 2 ) R(T 1 ) + R(T 2 ). Therefore, rank(t 1 + T 2 ) = dim R(T 1 + T 2 ) dim(r(t 1 ) + R(T 2 )) dim R(T 1 ) + dim R(T 2 ) = rank(t 1 ) + rank(t 2 ). (d) If T : V V is a linear transformation satisfying that T 2 = I, then I 2T is an isomorphism, where I is the identity map V V. Proof. True. Since (I 2T )(I + 2T ) = (I + 2T )(I 2T ) = I 4T 2 = 3I (I 2T )( 1 3 (I + 2T )) = ( 1 (I + 2T ))(I 2T ) = I 3 I 2T has an inverse and is hence bijective. (5) Do the following: (a) Find a linear transformation T : V V such that T is injective but T is not surjective. Solution. Let V = R[x]. Let T : R[x] R[x] be the map given by T (f(x)) = xf(x). It is a linear transformation because T (f(x) + cg(x)) = x(f(x) + cg(x)) = xf(x) + cxg(x) = T (f(x)) + ct (g(x)) for all f(x), g(x) R[x] and c R. Since K(T ) = {f(x) : T (f(x)) = 0} = {f(x) : xf(x) 0} = {0}, T is 1-1. And since R(T ) = Span{T (1), T (x), T (x 2 ),..., T (x n ),...} = Span{x, x 2, x 3,..., x n+1,...} R[x] T is not onto.

(b) Find a linear transformation T : V V such that T is surjective but T is not injective. Solution. Let V = R[x]. Let T : R[x] R[x] be the map given by T (f(x)) = f (x). It is a linear transformation because T (f(x) + cg(x)) = (f(x) + cg(x)) = f (x) + cg (x) = T (f(x)) + ct (g(x)) for all f(x), g(x) R[x] and c R. Since K(T ) = {f(x) : T (f(x)) = 0} = {f(x) : f (x) 0} = Span{1}, T is not 1-1. And since R(T ) = Span{T (1), T (x), T (x 2 ),..., T (x n ),...} T is onto. = Span{0, 1, 2x, 3x 2,..., nx n 1,...} = Span{1, x, x 2,..., x n,...} = R[x] (6) Find the ranks of the following linear transformations: (a) T : R 3 R 3 given by T (x, y, z) = (x y, y z, z x); 5 Solution. Since K(T ) = {(x, y, z) : x y = y z = z x = 0} = {(x, y, z) : x = y = z} = Span{(1, 1, 1)}, rank(t ) = dim R 3 dim K(T ) = 3 1 = 2. (b) T : M n n (R) M n n (R) given by T (A) = A A T ; Solution. We claim that R(T ) = {B M n n (R) : B T = B} i.e., the range of T is the subspace of skew-symmetric matrices. First, for every A M n n (R), (T (A)) T = (A A T ) T = A T A = T (A) and hence T (A) is skew-symmetric.

6 Second, for every skew-symmetric matrix B M n n (R), T ( 1 2 B) = 1 2 B 1 2 BT = 1 2 B + 1 2 B = B and hence B R(T ). Therefore, R(T ) is the subspace of skew-symmetric matrices and rank(t ) = dim R(T ) = n(n 1). 2 (c) T : R[x] R 2 given by T (f(x)) = (f(1), f (2)); Solution. Since R(T ) = Span{T (1), T (x),..., T (x n ),...} = Span{(1, 0), (1, 1),...} and {(1, 0), (1, 1)} is a basis of R 2, we conclude that R(T ) = R 2. Therefore, rank(t ) = 2. (d) T : V V given by T (f(x)) = xf (x) 2f(x), where V = {f(x) R[x] : deg f(x) 2014}. Solution. Since R(T ) = Span{T (1), T (x), T (x 2 ), T (x 3 ),..., T (x n ),..., T (x 2014 )} = Span{ 2, x, 0, x 3,..., (n 2)x n,..., 2012x 2014 } = Span{1, x, x 3,..., x n,..., x 2014 } we conclude that rank(t ) = dim R(T ) = 2014. (7) Let T : V W be an injective linear transformation. Show that T (v 1 ), T (v 2 ),..., T (v n ) are linearly independent for every linearly independent set {v 1, v 2,..., v n } in V. Proof. Suppose that T (v 1 ), T (v 2 ),..., T (v n ) are linearly dependent. Then c 1 T (v 1 ) + c 2 T (v 2 ) +... + c n T (v n ) = 0 for some c 1, c 2,..., c n R, not all zero. It follows that T (c 1 v 1 + c 2 v 2 +... + c n v n ) = c 1 T (v 1 ) + c 2 T (v 2 ) +... + c n T (v n ) = 0

7 because T is a linear transformation. That is, c 1 v 1 + c 2 v 2 +... + c n v n K(T ). Since T is 1-1, K(T ) = {0}. So c 1 v 1 + c 2 v 2 +... + c n v n = 0. Then we must have c 1 = c 2 =... = c n = 0 since v 1, v 2,..., v n are linearly independent. (8) Let A M n n (R) be an n n matrix. Show that A n = 0 if A n+1 = 0. Proof 1. Let T : R n R n be the linear transformation T (x) = Ax. Since A n+1 = 0, T n+1 = 0. We have R n = R(T 0 ) R(T 1 )... R(T n ) R(T n+1 ) = {0} where T 0 = I. If R k (T ) = R k+1 (T ) for some 0 k n, then T k (R n ) = T k+1 (R n ) T l (T k (R n )) = T l (T k+1 (R n )) for all l 0 and hence T k+l (R n ) = T k+l+1 (R n ) R(T k ) = R(T k+1 ) =... = R(T n ) = R(T n+1 ) = 0; it follows that T n = 0 and A n = 0. If R(T k ) R(T k+1 ) for all 0 k n, then rank(t k ) > rank(t k+1 ) rank(t k ) rank(t k+1 ) 1 and hence rank(t n ) = rank(t 0 ) (rank(t 0 ) rank(t 1 )) (rank(t 1 ) rank(t 2 ))... (rank(t n 1 ) rank(t n )) n } 1 1 {{... 1 } = 0. n Therefore, rank(t n ) = 0 and hence T n = 0 and A n = 0.

8 Proof 2. We claim that 0 is the only eigenvalue of A if A N = 0 for some N. Otherwise, A has an eigenvalue c 0 with a corresponding eigenvector v 0. That is, Then Av = cv. A N v = A N 1 (Av) = ca N 1 v =... = c N v. Since A N = 0, c N v = 0, which is impossible for c 0 and v 0. Therefore, all eigenvalues of A are 0 and hence the characteristic polynomial of A is f(x) = det(xi A) = x n. By Cayley-Hamilton, f(a) = 0 and hence A n = 0.