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Can I determine if these matrices are are diagonilizable without any calculations? I know I can determine it by calculating the eigenvalues and eigenvectors, but I am not sure if it can be done without the calculations. Thanks

$$ \begin{bmatrix} -2 && 1 && 1 \\ 1 && -2 && 1 \\ 1 && 1 && -2 \end{bmatrix} $$

I think that this one can be diagonalized based on the theorem that any real symmetric matrix can be diagonalized (correct me if I'm wrong). But what about this one:

$$ \begin{bmatrix} 0 && 0 && 1 \\ 1 && 0 && 0 \\ 0 && 1 && 0 \end{bmatrix} $$

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  • $\begingroup$ Not without any calculations. It requires calculations to determine symmetricity for example. $\endgroup$ – mathreadler May 19 '17 at 17:18
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The second example is a permutation matrix. Every permutation matrix is diagonalisable over $\Bbb C$, since each permutation matrix satisfies $A^m=I$ for some $m$, and $x^m-1=0$ has $m$ distinct roots over $\Bbb C$.

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  • $\begingroup$ could you explain a bit more what $x^m - 1 = 0$ is? and is there a way to prove that $A^m = I$? I am still kind of confused.. thanks $\endgroup$ – ivana14 May 19 '17 at 17:43
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    $\begingroup$ @ivana14 In your example, $A^3=I$. $\endgroup$ – Angina Seng May 19 '17 at 17:44

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