I saw the following theorem in a very old paper of Bendixson. Does anybody know a shorter and beautiful proof of that?
Theorem. If $A$ is a real matrix, then for each of its eigenvalues $(\lambda)$, the following inequality holds:
$ m \leq Re(\lambda) \leq M $,
where $m$ and $M$ are the minimum and maximum eigenvalues of $(A+A^{T})/2$.
Thank you!
Replacing $A$ to $A-m$ or to $M-A$ we may reduce to the following partial case: $(A+A^T)/2$ is non-negative definite, and we have to prove that $a\geqslant 0$ for any eigenvalue $\lambda=a+bi$ of $A$. We have $Av=(a+bi)v$ for a certain complex eigenvector $v$. Therefore $0\leqslant ((A+A^T)v,v)=(Av,v)+(v,Av)=(a+bi)(v,v)+(a-bi)(v,v)=2a(v,v)$ and $a\geqslant 0$ indeed.
