m21b_worksheet18-sols.pdf - Diagonalization a1 0 0 b1 0 0 1...

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Diagonalization 1. Warmup. Calculate a 1 0 0 0 a 2 0 0 0 a 3 b 1 0 0 0 b 2 0 0 0 b 3 . Solution. a 1 0 0 0 a 2 0 0 0 a 3 b 1 0 0 0 b 2 0 0 0 b 3 = a 1 b 1 0 0 0 a 2 b 2 0 0 0 a 3 b 3 . 2. Recap. If we have a discrete dynamical system ~x ( t + 1) = A~x ( t ) , where A is an n × n matrix, and we’re given an initial condition ~x (0) , what’s our strategy for finding ~x ( t ) in closed form? 3. You are given the following information about a 3 × 3 matrix A : ~v 1 = 1 0 5 is an eigenvector of A with eigenvalue 2, and ~v 3 = 0 2 7 is an eigenvector
of A with eigenvalue - 2. (a) Explain why B = ( ~v 1 ,~v 2 ,~v 3 ) is a basis of R 3 .
(b) If T ( ~x ) = A~x , find the B -matrix of T .
(Remember that saying [ T ( ~v 1 )] B = 3 0 0 means exactly that T ( ~v 1 ) = 3 ~v 1 + 0 ~v 2 + 0 ~v 3 .) Therefore, the B -matrix of T is the diagonal matrix D = 3 0 0 0 2 0 0 0 - 2 .

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