Let us say I know that a given $N\times N$ matrix has all its eigenvalues as real, does it mean, it is hermitian. How do I prove (or disprove) that?
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Hint: think about the following matrix:
$$ \left(\begin{array}{cc} 1 & 1\\ 0 & 1\end{array}\right) $$
icurays1
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1+1, though it might be useful to ponder a matrix like $$\begin{pmatrix} 2 & 1 \ 0 & 1 \end{pmatrix}$$ as well. – Branimir Ćaćić May 08 '13 at 09:20
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ABSOLUTELY NOT. It is easy to construct cases with real eigenvalues, even complex coefficients, and not Hermitian.
