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Unformatted text preview: EE221A Problem Set 7 Solutions  Fall 2011 Problem 1. With the given information, we can determine the Jordan form J = TAT 1 of A to be, J = 1 1 1 1 1 1 1 2 1 2 1 2 . Thus, cos ( e J ) = cos e 1 e 1 sin e 1 cos e 1 cos e 1 e 1 sin e 1 cos e 1 cos e 1 cos e 2 e 2 sin e 2 1 2 ( e 2 sin e 2 + e 2 2 cos e 2 ) cos e 2 e 2 sin e 2 cos e 2 , and cos ( e A ) = T 1 ( cos ( e J )) T . Problem 2. We know that there is a single eigenvalue = 0 with multiplicity 6, and that the size of the largest Jordan block is 3. We know that rank( A ) = rank ( T 1 JT ) = rank( J ) since T is full rank (apply Sylvesters inequality). Then J must have rank of at least 2, arising from the 1s in the superdiagonal in the Jordan block of size 3. If all the other Jordan blocks were size 1, then there would be no additional 1s on the superdiagonal, so the lower bound on rank( A ) is 2. Now the most 1s on the superdiagonal that this matrix could have is 4, which would be the case if there were two Jordan blocks of size 3. So rank( A ) 4...
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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.
 Fall '10
 ClaireTomlin

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