Solution08_121 - Solutions to PS 8 (Math 121) J. Scholtz...

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Solutions to PS 8 (Math 121) J. Scholtz January 6, 2007 Question 1: 7.1/2(a) We are to find the basis of the generalazied eigenvectors and the corresponding Jordan form of matrix A : A = ± 1 1 - 1 3 ² This matrix gives us a double eigenvalue of λ = 2. Hence we need to find all the vectors such that: A = ± - 1 1 - 1 1 ²± a b ² = 0 This is satisfied by a (1 , 1). Therefore we need to search for another generalized eigenvector. So we need to find a vector such that ( A - 2 I ) 2 v = 0 but ( A - 2 I ) v 6 = 0. The solution to the first equation is any vector since ( A - 2 I ) 2 = 0, therefore we just need to find a vector such that ( A - 2 I ) 6 = 0, hence anything linearly independent from a (1 , 1), pick v = ( - 1 , 0), so that ( A - 2 I ) v = (1 , 1). Then the Jordan form is: J = ± 2 1 0 2 ² Question 2: 7.1/3(a) First, by action on the standard basis we can establish that: A = 2 - 1 0 0 2 - 2 0 0 2 The only (triple) eigenvalue is again λ = 2. We can see that the rank( A - 2 I ) = 2 and so there must be a three-cycle of generalized eigenvectors, such that ( A - 2 I ) 3 v = 0 but ( A - 2 I ) 2 v 6 = 0. But we know that ( A - 2 I ) 3
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This note was uploaded on 07/03/2011 for the course MATH 110 taught by Professor Gurevitch during the Spring '08 term at University of California, Berkeley.

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Solution08_121 - Solutions to PS 8 (Math 121) J. Scholtz...

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