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hw5-sol

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

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EE 221a Homework 5 Solutions 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 i =1 e λ i = 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 Thanks to Onureena Banerjee for providing part of the solution 1

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Problem 4. f ( s ) = cos ( e s ). So f ( s ) = - sin ( e s ) e s , f ( s ) = - cos ( e s ) e 2 s - sin ( e s ) e s .
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hw5-sol - EE 221a Homework 5 Solutions Fall 2007 1 Problem...

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