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It is known that the set of eigenvalues of the following block matrix

$$ C = \begin{pmatrix} A & B \\ B & A \end{pmatrix} $$

is the union of the eigenvalues of the matrices $A + B$ and $A - B$. I am interested in the matrix of the following form

$$ C = \begin{pmatrix} A & B \\ B &-A \end{pmatrix} $$

Is there a description of the eigenvalues of $C$ in terms of $A$ and $B$?

Edit. If $AB=BA$, then we can do the following. $$ C = \begin{pmatrix} A & B \\ B &-A \end{pmatrix} \begin{pmatrix} v \\ u \end{pmatrix} =\lambda \begin{pmatrix} v \\ u \end{pmatrix} $$ implies $$ \left\{ \begin{array}{l} Av+Bu=\lambda v \\ Bv-Au=\lambda u \end{array}. \right. $$ By multiplying the first equation by $B$ and assuming $AB=BA$, we get $$ (A^2+B^2)u=\lambda^2u. $$ Therefore, $\lambda^2$ is an eigenvalue of $A^2+B^2$, what is discussed in the comments.

glS
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QMath
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    $\pm \sqrt{A^2+B^2}$, but this has sense only if $A,B$ are symmetric – Exodd Jun 13 '22 at 21:05
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    @Exodd A,B are symmetric, but what do you mean by $\sqrt{A^2+B^2}$? Eigenvalues of the matrix $\sqrt{A^2+B^2}$? – QMath Jun 13 '22 at 21:12
  • @QMath: Let $\lambda$ be an eigenvalue of $C$. Then $\det (C-\lambda I) = 0$, and expanding that gives you $\lambda^2I = A^2 + B^2$. So, $\lambda^2$ is an eigenvalue of $A^2 + B^2$. – Dan Jun 13 '22 at 21:34
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    @Dan If that were true, it would imply $A^2+B^2$ is always a scalar multiple of $I$ for any $A$ and $B$. – aschepler Jun 13 '22 at 21:40
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    @aschepler I think he meant $det(\lambda^2I - A^2 - B^2) = 0$ – Exodd Jun 14 '22 at 20:53
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    @Exodd This holds in the case $AB=BA$, but not in general. – QMath Jun 15 '22 at 11:44
  • related: https://math.stackexchange.com/q/2416817/173147, https://math.stackexchange.com/q/706513/173147, https://math.stackexchange.com/q/3264807/173147 – glS Feb 19 '23 at 17:26

1 Answers1

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Partial answer when $A$ has inverse $A^{-1}$. Use the formula for the determinant of block-partitioned matrices in terms of Schur's complement. If $(A-\lambda I)$ has an inverse then $\det(A-\lambda I)\neq 0$ \begin{align} \det \left(\begin{bmatrix} A&B \\ B& -A \end{bmatrix}-\lambda\begin{bmatrix} I&0 \\ 0& I \end{bmatrix} \right) =& \det \left(\begin{bmatrix} A-\lambda\cdot I&B \\ B& -A-\lambda\cdot I \end{bmatrix}\right) \\ =& \det(-A-\lambda I) \cdot \det\left(-(A-\lambda I) - B(A-\lambda I)^{-1} B\right) \end{align}

Thus, the eigenvalues of $\begin{bmatrix} A&B\\ B&-A \end{bmatrix} $ would be between the eigenvalues of $-A$ and between the roots of the equation $$ \det\left(-(A-\lambda I) - B(A-\lambda I)^{-1} B\right)=0 $$

Elias Costa
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