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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.

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    $\begingroup$ Try to use $A^2 = \begin{bmatrix}A^*A & 0\\0 & A^*A\end{bmatrix}$ and $Ax=\lambda x \Longrightarrow A^2 x = \lambda^2 x$ $\endgroup$ – Michael Galuza Jul 8 '15 at 13:19
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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.

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  • $\begingroup$ @ Joonas Ilmavirta , you show that $(spectrum(\mathcal{A}))^2=$two copies of $spectrum(AA^*)$. Let $\sigma$ be a singular value of $A$ of multiplicity $k$; then $\mathcal{A}$ admits the eigenvalues $\sigma$, with multiplicity $p$ and $-\sigma$, with multiplicity $q$ with $p+q=2k$. Yet you do not prove that $p=q$. $\endgroup$ – loup blanc Jul 9 '15 at 8:39
  • $\begingroup$ @loupblanc, true. I only found the squared eigenvalues of $\mathcal A$ with multiplicity, not the multiplicities of the eigenvalues themselves. I believe this is what I wrote, although not very emphatically. $\endgroup$ – Joonas Ilmavirta Jul 9 '15 at 8:48
  • $\begingroup$ Yes Joonas, I agree. $\endgroup$ – loup blanc Jul 9 '15 at 8:52
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$\det(\mathcal{A}-\lambda I_{2n})=\det(\lambda ^2 I_n-AA^*)$. Then the eigenvalues of $\mathcal{A}$ are $\pm$ the singular values of $A$ (with multiplicity).

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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.

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    $\begingroup$ If $A$ is not hermitean, you should conjugate transpose some of your $A$s. $\endgroup$ – Joonas Ilmavirta Jul 8 '15 at 13:35

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