# 582hw1 - EEE 582 HW 1 SOLUTIONS Problem 1.2 Let i and pi be...

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EEE 582 HW # 1 SOLUTIONS Problem 1.2 Let ¸ i and p i be the i -th eigenvalue and eigenvector of A ,so Ap i = ¸ i p i (1) 1. A 2 p i = AAp i = A ( ¸ i p i )= ¸ 2 i p i . . . A k p i = AA k ¡ 1 p i = ¸ k i p i Hence the eigenvalues of A k are ¸ k i for i =1 ;:::;n . 2. Multiply both sides of (1) by A ¡ 1 (all eigenvalues are di®erent from zero), then A ¡ 1 Ap i = A ¡ 1 ¸ i p i ) 1 ¸ i p i = A ¡ 1 p Hence the eigenvalues of A ¡ 1 are 1 ¸ i for i =1 ;:::;n . 3. The eigenvalues of A > areg ivenbytherootso f det ¡ ¸I ¡ A > ¢ =det h ( ¸I ¡ A ) > i =det[( ¸I ¡ A )] > and for any square matrix X ,det( X )= det ( X > ), ¯nally det ¡ ¸I ¡ A > ¢ =det( ¸I ¡ A ) Hence the eigenvalues of A > are ¸ i for i =1 ;:::;n . 4. Let A H = ¹ A > (conjugate transpose), det ¡ ¸I ¡ A H ¢ =det ¡ ¸I ¡ ¹ A > ¢ =det ¡ ¸I ¡ ¹ A ¢ then det ¡ ¸I ¡ A H ¢ =det ¡ ¸I ¡ ¹ A ¢ =det h ¡ ¹ ¸I ¡ A ¢ i = det ¡ ¹ ¸I ¡ A ¢

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582hw1 - EEE 582 HW 1 SOLUTIONS Problem 1.2 Let i and pi be...

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