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# Math 219 Homework 3 Solutions 1. Let A = 1 1 . Find the eigenvalues of A and their associated eigenvectors.

2 4

## Solution: 1 1 = 0 = (1 )(4 ) + 2 = 0 = 1 = 2 and 2 = 3. 2 4 For 1 = 2, 1 1 2 4 x1 x = 1 2 = x1 x2 = 0 = x1 = x2 . x2 x2 r for some r R. r

Thus, all eigenvectors associated with 1 = 2 are of the form In particular, x1 = For 2 = 3, 1 1 2 4 1 . 1

## Thus, all eigenvectors associated with 1 = 2 are of the form In particular, x2 = 1 . 2

2. Let A be an n n matrix for which In A is nonsingular. If is an eigenvalue of A, show that (1 )1 is an eigenvalue of (In A)1 . Solution: Let be an eigenvalue of A. Then Ax = x = x Ax = x x = (In A)x = x(1 ). Thus, 1 is an eigenvalue of In A. Now pre-multiply both sides by (In A)1 : (In A)1 (In A)x = (In A)1 x(1 ) x = (In A)1 x(1 ) 1 x = (In A)1 x 1 Thus, (1 )1 is an eigenvalue of (In A)1 . 3. (Exercise 1.30, p. 31) If A and B conform under multiplication, show that the nonzero eigenvalues of AB coincide with the nonzero eigenvalues of BA. Solution: Let be a nonzero eigenvalue of AB. Then, for some vector x = 0, ABx = x = BABx = Bx. Let y = Bx. Then is an eigenvalue of BA with associated eigenvector y. 1

4. Let P be an n n idempotent matrix (i.e., P2 = P). Prove that the matrix In P is also idempotent. Solution: Since P2 = P, (In P)(In P) = In P P + P2 = In P P + P = In P. Thus, In P is idempotent. 5. (Exercise 2.1, p. 70) Let P = A B , where P is an n n orthogonal matrix. Show that A A is an idempotent matrix. Solution: Let A be n m where m n. Then B is n (n m). Since P is orthogonal, P P = In . Therefore, A B A B = AA AB BA BB = Im 0(nm)m 0m(nm) . Inm

Thus, A A = Im . Since Im is idempotent, A A is idempotent. 6. Let X be an n p matrix with full column rank. Show that P = X(X X)1 X is symmetric and idempotent. Solution: Show P is symmetric: P = (X(X X)1 X ) = (X ) [(X X)1 ] X = X[(X X) ]1 X = X(X X)1 X = P. Show P is idempotent: P2 = X(X X)1 X X(X X)1 X = XIp (X X)1 X = P. 7. Solution: See Microsoft Word document. For the explanation to part (j): We know y = y + . Thus, y ||y||2 = || + ||2 = ( + ) ( + ) = ( + )( + ) = y y+ + y+ y y y y y since y and are orthogonal. = || ||2 +|| ||2 y
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