7_1 - Section 7.1: 1 TFFTFFTT Section 7.1: 2b The...

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Section 7.1: 1 TFFTFFTT Section 7.1: 2b The characteristic polynomial is ( t - 4)( t - 1), so the eigenvalues are distinct, and the matrix is diagonalizable. The basis of eigenvectors can be chosen as (2 , 3) t , and (1 , - 1) t , leading to a Jordan Form of ± 4 0 0 - 1 ² Section 7.1: 2c The characteristic polynomial is - ( t + 1)( t - 2) 2 . For λ = - 1, the eigenvector can be chosen to be (1 , 3 , 0) t . For λ = 2, E 2 has dimension 1. K 2 must therefore have dimension 2, and we can pick any vector in K 2 which is not in E 2 to generate a cycle. To calculate K 2 find the nullspace of ( T - 2 I ) 2 . We can use x 0 = (1 , - 1 , 0) t , and then ( T - 2 I ) x 0 = (1 , 1 , 1) t . Thus we can use for a basis β = { (1 , 3 , 0) t , (1 , 1 , 1) t , (0 , 1 , - 1) t } , and with respect to β , the matrix of T is - 1 0 0 0 2 1 0 0 2 Section 7.1: 3a Using the standard basis for the polynomials of degree less than or equal to 2, the matrix of T is 2 - 1 0 0 2 - 2 0 0 2 . the dimension of the nullspace of T - 2 I is one, of ( T - 2 I ) 2 is 2, and of ( T - 2 I ) 3 is 3. Thus there is a single 3 cycle. We can generate it from x 2
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This note was uploaded on 02/02/2012 for the course MATH 115A 262398211 taught by Professor Fuckhead during the Spring '10 term at UCLA.

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7_1 - Section 7.1: 1 TFFTFFTT Section 7.1: 2b The...

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