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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.

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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 $-T(t)$.

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  • $\begingroup$ Thanks, so indeed it seems to hinge on $B$ being hermitian?-Even though the result might not... $\endgroup$
    – Sascha
    Commented Mar 16, 2022 at 17:08
  • $\begingroup$ Welcome. Yes this is for $\mathcal{B}$ hermitien, (in general no clue). You could give an example in the general case maybe this helps. $\endgroup$
    – Toni Mhax
    Commented Mar 16, 2022 at 17:22
  • $\begingroup$ thanks, yes, i added a slightly more general example. $\endgroup$
    – Sascha
    Commented Mar 16, 2022 at 17:35
  • $\begingroup$ First your problem seems for $n\times n$ block matrices so you may want to keep the initial statement and give the example in $2 \times 2$ like you did. $\endgroup$
    – Toni Mhax
    Commented Mar 16, 2022 at 17:51
  • $\begingroup$ thanks, indeed the initial statement was more general, but maybe this scenario sheds some light on what we are missing? :) $\endgroup$
    – Sascha
    Commented Mar 16, 2022 at 18:16

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