TITLE: Eigenvalues of Hermitian dilation of any square matrix
QUESTION [5 upvotes]: Let us consider any $n \times n$ matrix $A$. My question is, what are the eigenvalues of 
\begin{equation}
\mathcal{A} = \begin{bmatrix} 0 & A \\ A^* & 0 \end{bmatrix}. 
\end{equation}
Of course since $\mathcal{A}$ is traceless Hermitian, if $\lambda$ is an eigenvalue, $-\lambda$ is also an eigenvalue.
Motivation & further questions: If $A$ is Hermitian with eigenvalues $\lambda_j$, then the eigenvalues of $\mathcal{A}$ are $\pm \lambda_j$. Can we say something similar when $A$ is not Hermitian? Granted, $A$ may not be diagonalizable, can we say something in terms of singular values? What if, for a simple case, all eigenvalues of $A$ be real, though $A$ may not be Hermitian? Advanced thanks for any help/ suggestions.

REPLY [2 votes]: This matrix can be diagonalized by $\mathcal{U} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$ gives $\mathcal{U} \mathcal{A} \mathcal{U}^{-1} = \begin{bmatrix} A & 0 \\ 0 & -A \end{bmatrix}$, so the eigenvalues are $\pm$ the eigenvalues of $A$, just as you said, and that's all of them.