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I'm a little confused on the wording of this statement in my book:

A is orthogonally diagonalizable and has real eigenvalues if and only if A is symmetric.

They're not saying that only symmetric matrices can have all real eigenvalues, right?

Can matrices that are not symmetric have all real eigenvalues?

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  • $\begingroup$ The adverb "orthogonally" is the crucial part here. $\endgroup$ – user228113 Mar 11 '17 at 23:02
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    $\begingroup$ consider $\begin{pmatrix} a & b \\ 0 & c\end{pmatrix}$ with $a,b,c\in \Bbb R$. This matrix is not symmetric if $b \neq 0$ but its eigenvalues are $a,c$. $\endgroup$ – Surb Mar 11 '17 at 23:04
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They're not saying that only symmetric matrices can have all real eigenvalues, right?

No, they say that only a symmetric matrix can: have all real eigenvalues and be orthogonally diagonalizable.

Can matrices that are not symmetric have all real eigenvalues?

Sure. You can just pick a triangular (but not diagonal) matrix with real entries on the main diagonal.

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