Eigenvalues of Hermitian dilation of any square matrix 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.
 A: If $A$ is hermitean and its eigenvalues are $\{\lambda_1,\dots,\lambda_n\}$, then $\mathcal A$ is also hermitean and its eigenvalues are $\{-\lambda_1,+\lambda_1,\dots,-\lambda_n,+\lambda_n\}$.
(This is true also allowing repeated eigenvalues; the multiplicities behave as you would guess.)
If $v\in\mathbb C^n$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $(v,\pm v)\in\mathbb C^{2n}$ is an eigenvector of $\mathcal A$ with eigenvalue $\pm\lambda$.
Counting dimensions shows that these form an eigenbasis for $\mathbb C^{2n}$.
Consider then a more general $A$.
If $w\in\mathbb C^{2n}$ satisfies $\mathcal Aw=\mu w$ for some $\mu\in\mathbb R$, then $\mathcal A^2w=\mu^2w$.
The reason for considering the square is that
$$
\mathcal A^2
=
\begin{pmatrix}
AA^*&0\\
0&A^*A
\end{pmatrix}.
$$
The matrices $AA^*$ and $A^*A$ are positive semidefinite and hermitean, so their eigenvalues are positive real numbers.
The squared eigenvalues of $\mathcal A$ are thus the eigenvalues of $AA^*$ and $A^*A$ (squared singular values of $A$), counted with multiplicity.
Since $\mathcal A$ is traceless and hermitean, its eigenvalues must add up to zero (with multiplicity).
This strongly suggests (but does not fully prove) that if $\lambda$ is an eigenvalue of $\mathcal A$, then $-\lambda$ is an eigenvalue with the same multiplicity.
A: $\det(\mathcal{A}-\lambda I_{2n})=\det(\lambda ^2 I_n-AA^*)$. Then the eigenvalues of $\mathcal{A}$ are $\pm$ the singular values of $A$ (with multiplicity).
A: 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.
