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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 $\lambda$ satifyy $$\det\left(\begin{pmatrix}-\lambda I&F\\G&-\lambda I\end{pmatrix}\begin{pmatrix}-\lambda I&2A-F\\2A^*-G&-\lambda I\end{pmatrix}-t^2I\right)=0.$$ 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 $-T(t)$.

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 $-T(t)$.

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 $\lambda$ satifyy $$\det\left(\begin{pmatrix}-\lambda I&F\\G&-\lambda I\end{pmatrix}\begin{pmatrix}-\lambda I&2A-F\\2A^*-G&-\lambda I\end{pmatrix}-t^2I\right)=0.$$ 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 $$\begin{pmatrix}Q&-X\\-Y&Z\end{pmatrix}.$$

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 $\lambda$ satifyy $$\det\left(\begin{pmatrix}-\lambda I&F\\G&-\lambda I\end{pmatrix}\begin{pmatrix}-\lambda I&2A-F\\2A^*-G&-\lambda I\end{pmatrix}-t^2I\right)=0.$$ 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 $-T(t)$.

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 $\mathcal{B}=\mathcal{B}^*$, wether the general case may be 'different'. So $R=UT(t)U^*=\begin{pmatrix}\mathcal{A}-\mathcal{B}&tI\\tI&\mathcal{A}+\mathcal{B}\end{pmatrix}.$ Also $R-\lambda I=\begin{pmatrix}-\lambda I&(A-B)&tI&0\\(A^*-C)&-\lambda I&0&tI\\tI&0&-\lambda I &(A+B)\\0&tI&(A^*+C)&-\lambda I\end{pmatrix}.$ Using the well known determinant for block matrices with a commuting off-diagonal block you get the eigenvalues $\lambda$ verify $\det(\begin{pmatrix}-\lambda I&F\\G&-\lambda I\end{pmatrix}\begin{pmatrix}-\lambda I&2A-F\\2A^*-G&-\lambda I\end{pmatrix}-t^2I)=0.$ 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 $\begin{pmatrix}Q&-X\\-Y&Z\end{pmatrix}$.

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 $\lambda$ satifyy $$\det\left(\begin{pmatrix}-\lambda I&F\\G&-\lambda I\end{pmatrix}\begin{pmatrix}-\lambda I&2A-F\\2A^*-G&-\lambda I\end{pmatrix}-t^2I\right)=0.$$ 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 $$\begin{pmatrix}Q&-X\\-Y&Z\end{pmatrix}.$$