hw5-sol - EE 221a Homework 5 Solutions Fall 2007 1 Problem...

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1 Fall 2007 Problem 1. Since f ( s ) = e s is an analytic function, the eigen-values of e A are e λ i , i = 1 , 2 , . . . , n , where λ i ’s are eigen-values of A . So, det( e A ) = n Y i =1 e λ i 6 = 0 Problem 2. v R n , Av N ( A ), that is, A 2 v = θ, v . So, A 2 = 0. The statement is wrong. A 1 = 0 is a solution of A 2 = 0, and A 2 = 0 1 . . . 0 0 0 . . . 0 . . . . . . . . . . . . 0 0 . . . 0 is also a solution. Even with change of basis, A 1 is not A 2 . Problem 3. The minimum polynomial is s 3 , so the eigen-values of A are all zeros, and the largest Jordan block has size 3. Write A = V JV - 1 , where J is in Jordan form. Clearly rank ( A ) = rank ( J ). The J which has the lowest rank is J = 0 1 0 1 0 0 0 0 The J which has the highest rank is J = 0 1 0 1 0 0 1 0 1 0 So, 2 rank ( J ) 4. 1
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This note was uploaded on 12/09/2011 for the course EE 221A taught by Professor Clairetomlin during the Fall '10 term at University of California, Berkeley.

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hw5-sol - EE 221a Homework 5 Solutions Fall 2007 1 Problem...

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