Does Real Eigenvalues mean it is an hermitian Matrix 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?
Hint: think about the following matrix: 

$$
\left(\begin{array}{cc}
1 &amp; 1\\ 
0 &amp; 1\end{array}\right)
$$
ABSOLUTELY NOT. It is easy to construct cases with real eigenvalues, even complex coefficients, and not Hermitian.