Now lets nd an eigenvector for 4 lets reduce the

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Unformatted text preview: s. 15. a) Let’s find the eigenvectors for the eigenvalue 1. We need to solve (A − I )X = 0. The three equations in the system are the same: x + y + z = 0, so the solution set is two dimensional and has basis {[1, 0, − 1], [0, 1, − 1]}. Now let’s find an eigenvector for 4. Let’s reduce the matrix A − 4I : −2 1 1 1 −2 1 1 − 2 1 R3 + R2 1 − 2 1 R2 + 2R1 1 − 2 1 R1 ↔ R2 − 2 1 1 0 −3 3 1 0 1 −1 R3 − R1 1 1 −2 1 1 −2 0 3 − 3 − 3 R2 0 0 0 which means that the solution space is given by z = t, y = t, x = t, so [1, 1, 1] is a basis for it. 1 01 Therefore, writing eigenvectors as columns, P = 0 1 1 , and we’ll have our −1 −1 1 100 P − 1A P = 0 1 0 . 004 301 b) Let’s look for eigenvector for the eigenvalue 1: the matrix B − I = 0 0 1 is al 000 ready row-reduced, so we can easily see that [0, 1, 0] is a basis for the null space. But then we only have one eigenvector corresponding to the eigenvalue 1 which has algebraic multiplicity two. Therefore B is not diagonalizable. c) He’s right: if they were similar, since A is diagonalizable, B would be too. 2...
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