We consider block matrices
$$\mathcal A = \begin{pmatrix} 0 & A\\A^* & 0 \end{pmatrix}$$ and
$$\mathcal B = \begin{pmatrix} 0 & B\\C & 0 \end{pmatrix}.$$
Then we define the new matrix
$$T(t) = \begin{pmatrix} \mathcal A+t & \mathcal B \\ \mathcal B^* & \mathcal A-t\end{pmatrix}.$$
Numerical experiments seem to show that the eigenvalues of $[0,\infty) \ni t\mapsto T(t)$ have the property that their absolute values are monotonically increasing in $t \ge 0.$ However, I do not have a proof of this, does anybody know how this follows? (The eigenvalues of $T(t)$ seem to come in pairs $\pm \lambda$ with $\lambda = \lambda(t) \ge 0$, i.e. $+\lambda(t)$ is increasing, while $-\lambda(t)$ is decreasing.
To illustrate the effect, consider
$$T(t)=\begin{pmatrix}
t & 1& 0& 2\\
1 & t & 0& 0\\
0 & 0& -t & 1\\
2& 0 & 1 & -t
\end{pmatrix},$$ then the eigenvalues of $T(t)$ are
$$ \pm 1 \mp \sqrt{2+t^2}.$$
Please let me know if anything is unclear.
Best Answer
The idea is to apply a unitary congruence $$U=\dfrac{1}{\sqrt{2}}\begin{pmatrix}I&-I\\I&I\end{pmatrix}.$$ I consider here $\mathcal{B}$ to be hermitian, $\mathcal{B}=\mathcal{B}^*$, whereas the general case may be 'different'.
So $$R=UT(t)U^*=\begin{pmatrix}\mathcal{A}-\mathcal{B}&tI\\tI&\mathcal{A}+\mathcal{B}\end{pmatrix}.$$ Similarly $$ R-\lambda I=\begin{pmatrix}-\lambda I&(A-B)&tI&0\\(A^*-C)&-\lambda I&0&tI\\tI&0&-\lambda I &(A+B)\\0&tI&(A^*+C)&-\lambda I\end{pmatrix}. $$ Using the well known determinant formula for block matrices with a commuting off-diagonal block, you obtain that the eigenvalues $\lambda$ satisfy $$\det\left(\begin{pmatrix}-\lambda I&F\\G&-\lambda I\end{pmatrix}\begin{pmatrix}-\lambda I&2A-F\\2A^*-G&-\lambda I\end{pmatrix}-t^2I\right)=0.$$ Equivalently, $$\det\begin{pmatrix}(\lambda^2-t^2)I +F(2A^*-G)&-2\lambda A\\-2\lambda A^*&(\lambda^2-t^2)I+G(2A-F)\end{pmatrix}=0.$$
The monotonicity of the eigenvalues follows and $-\lambda$ is also an eigenvalue since $$\begin{pmatrix}Q&X\\Y&Z\end{pmatrix}$$ is unitarily congruent to $$\begin{pmatrix}Q&-X\\-Y&Z\end{pmatrix}.$$
If $\mathcal{B}$ is not hermitian the block matrix $$T(t)=\begin{pmatrix}tI&A&0&B\\A^*&tI&C&0\\0&C^*&-tI&A\\B^*&0&A^*&-tI\end{pmatrix}$$ may not be similar to $-T(t)$ for $2\times 2$ blocks, $t=2$, $A=\begin{pmatrix}1&0\\0&1\end{pmatrix}, B=\begin{pmatrix}0&0\\0&5\end{pmatrix}$ and $C= \begin{pmatrix}0&0\\3&0\end{pmatrix}.$
The property seems to hold for $T(t)\in \mathbb{M}_4(\mathbb{C})$, consider the characteristic polynomial of $T(t)$ and $-T(t)$.