Lecture 8.2- Similarity and Diagonalization

# Lecture 8.2- Similarity and Diagonalization - 8.2...

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8.2 Similarity and Diagonalization MATH232 D100 2016-3 Paul Tupper SFU Burnaby

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Powers of a Matrix Suppose that A = - 7 18 - 3 8 and we want to calculate A 5 1 0
This is tedious A 1 0 = - 7 18 - 3 8 1 0 = - 7 - 3 A 2 1 0 = - 7 18 - 3 8 - 7 3 = - 5 - 3 A 3 1 0 = - 7 18 - 3 8 - 5 - 3 = - 19 - 9 A 4 1 0 = - 7 18 - 3 8 - 19 - 9 = - 29 - 15 A 5 1 0 = - 7 18 - 3 8 - 29 - 15 = - 67 - 33 but there is a better way.

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Review of Eigenvalues Recall that if v is an eigenvector of A with eigenvalue λ , then A k v is much easier to compute. Unfortunately But we will try to express 1 0 as a linear combination of eigenvectors.
We obtain the eigenvectors of A by For λ = 2 , the eigenspace is For λ = - 1 , the eigenspace is

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Now note that each eigenspace is 1 -dimensional (a line), and so has a single vector as its basis If we put these two vectors together, we get a basis for R 2 . This basis is called an eigenbasis of R 2 with respect to the matrix A .
Now, our strategy for computing A 5 1 0 is this: We express v = 1 0 in terms of our eigenbasis B = { (2 , 1) , (3 , 1) } , that is Then the effect of multiplying by A 5 is

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But we can go much further now!
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