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I'm dealing with the $3 \times 3$ matrix:

$$ \begin{pmatrix} 3 & 4 & 0 \\ -4 & 3 & 0 \\ 0 & 0 & 1 \end{pmatrix}. $$

I have the characteristic equation, which is $(1 - L)(L^2 - 6L + 25)$. The second term gives two complex eigenvalues: $3 \pm 4i$.

I don't understand why this matrix is not diagonalizable. There seem to be three distinct eigenvalues, so therefore there should be three linearly independent eigenvectors. Where am I going wrong?

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    $\begingroup$ It is diagonalizable, over the complex numbers. It is not diagonalizable over the reals. $\endgroup$ – levap Dec 15 '15 at 12:15
  • $\begingroup$ To rephrase levap's comment: you can diagonalize it, but only using matrices with complex entries. Apparently, this "doesn't count" for your instructor or textbook. $\endgroup$ – Omnomnomnom Dec 15 '15 at 12:37

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