Eigenvalues and Eigenvectors §5.2 Diagonalization
0 3 1 0 0 3 1 A: Show that A is not diagonalizable. Solution: Use Theorem 5.2.2 and show that A does not have 3 linearly independent eigenvectors. I To nd the eigenvalues, we solve det( I A) = 1 1 1 0 + 3 1 0 0 + 3 = ( 1)( +3)2 = 0: So, = 1; 3 are the only eigenvalues of A: Satya Mandal, KU Eigenvalues and Eigenvectors x5.2 Diagonalization
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