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.
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 $$ \det\left(-(A-\lambda I) - B(A-\lambda I)^{-1} B\right)=0 $$