<s>[INST] What are the eigenvalues of $\begin{pmatrix} A & B \\ B &-A \end{pmatrix}$ in terms of $A$ and $B$? 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.
 [/INST] 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
$$
\det\left(-(A-\lambda I) - B(A-\lambda I)^{-1} B\right)=0
$$
</s>