Relation Between Eigenvalues of a Matrix and its Real Part Let $A$ be an $n \times n$ complex matrix. The real part of $A$ is $\frac{A^H + A}{2}$. What is the relation between the eigenvalues of $A$ and $\frac{A^H + A}{2}$? I know that the eigenvalues of $A^H$ are the complex conjugates of $A$'s eigenvalues.
In fact there are some inequalities. We know that eigenvalues of hermite matrices are real. So $A_r=\frac{1}{2}(A+A^H),A_i=\frac{1}{2\mathrm{i}}(A+A^H)$ are hermite matrices. Write their eigenvalues as $ x_1\le x_2\le\cdots, \le x_n;y_1\le y_2,\le\ldots\le y_n$;  Our homework is that all eigenvalues $a+b\mathrm{i}$ of $A$ satisfies $ x_1\le a\le x_n,y_1 \le b \le y_n$.How can I prove it?

If $A$ is normal, the eigenvalues of $\frac{A + A^{*}}{2}$ are the real parts of the eigenvalues of $A$. If $A$ is not normal, there isn't any nice relation. For example, if
$$ A = \begin{pmatrix} 0 & 2\varepsilon \\ 0 & 0 \end{pmatrix} $$
then the eigenvalues of $A$ are zero but the eigenvalues of 
$$ \frac{A + A^{*}}{2} = \begin{pmatrix} 0 & \varepsilon \\ \overline{\varepsilon} & 0 \end{pmatrix} $$
are $\pm |\varepsilon|$.
