Proof that eigenvalues of a conjugated block matrix are complex conjugated pairs

Question

I am trying to solve the set of differential equations $\frac{d\mathbf{X}(t)}{dt}=\mathbf{C}\mathbf{X}(t)+\mathbf{d}$, where $\mathbf{X}(t)$ and $\mathbf{d}$ are column vectors, and $\mathbf{C}=\begin{bmatrix} A & B\\B^* &A^*\end{bmatrix}$, where A and B are $N\times N$ matrices. If I can prove that $\mathbf{C}$ has $2N$ eigenvalues which are $N$ complex conjugated pairs, I can solve the set of equations by converting it to a single differential equation and then using Laplace transform to solve it.

So I start with the eigenvalue equation of a block matrix as $\det(\mathbf{C}-\lambda I)=\det(A-\lambda I)\det\{(A^*-\lambda I)-B^*(A-\lambda I)^{-1}B\}=0$.

I am not sure how to go further. Could someone please help me? Thanks a lot

Answer 1

Hint: $$\mathbf C^*=\begin{bmatrix}A^* & B\\B^* & A\end{bmatrix}.$$ Can you show that this matrix has the same eigenvalues as $\mathbf C$?

Answer 2

Thanks Carl for the hint. I can show this $\mathbf{C}^*=\begin{bmatrix} A^* & B^*\\B & A\end{bmatrix}$ is the similar matrix of $\mathbf{C}$ because there exists a matrix $P=\begin{bmatrix}\mathbf{0} &J\\ J& \mathbf{0}\end{bmatrix}$, where J is anti-diagonal identity matrix, that satisfies $\mathbf{C}^* =P\mathbf{C}P^{-1}$. Since two similar matrices have the same eigenvalues, I have proved the problem. By the way, is there any way to directly solve the system of differential equations $\frac{d\mathbf{X}(t)}{dt}= \mathbf{CX}(t)+\mathbf{d}$ ?