In a previous question here, I asked the question below for block matrices and received an answer showing the question is true if $\mathcal B$ is hermitian and false, in general if $\mathcal B$ is non-hermitian. However, numerical experiments suggest it is still true if we are talking only about matrices rather than block matrices and this is the content of this question.
We consider matrices
$$\mathcal A = \begin{pmatrix} 0 & a\\\bar a& 0 \end{pmatrix}$$ and
$$\mathcal B = \begin{pmatrix} 0 & b\\c & 0 \end{pmatrix}$$
with $a,b,c \in \mathbb C.$
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 you have any questions.
Best Answer
The characteristic polynomial is even in both $X$ and $t$ : $P_t(X)=Q(X^2,t^2)$ where $$Q(Y,s)=(Y-s)^2-(2|a|^2+|b|^2+|c|^2)(Y-s)-4|a|^2s+|a^2-b\bar c|^2.$$ The variation of $s\mapsto Y(s)$ is given by the derivative $$2(Y-s)(Y'-1)-(2|a|^2+|b|^2+|c|^2)(Y'-1)-4|a|^2=0.$$ The sign of $Y'$ changes when $s$ crosses the value $$s_0=\frac1{4|a|^2}(2|a|^2+|b|^2+|c|^2)(2|a|^2-|b|^2-|c|^2).$$ This value is admissible for $t_0^2$ if it is positive, that is is $2|a|^2>|b|^2+|c|^2$.
In conclusion, the absolute values of the eigenvalues are monotonous functions of $s$ whenever $2|a|^2\le|b|^2+|c|^2$. If on the contrary $2|a|^2>|b|^2+|c|^2$, then one of the eigenvalues is not monotone, and the change happens when $t$ crosses $$t_0=\frac1{2|a|}\sqrt{(2|a|^2+|b|^2+|c|^2)(2|a|^2-|b|^2-|c|^2)}.$$