I have a complex-valued block matrix $N=\begin{bmatrix} A & B \\ C & -A^* \end{bmatrix}$, where $A$ is diagonal, $B=B^*$, and $C$ is rank-1 but not Hermitian. If $C$ were Hermitian, $N$ would be a complex-Hamiltonian matrix and have the property that $\lambda \in \sigma(H) \implies -\bar\lambda \in \sigma(H)$, i.e., its eigenvalues will be symmetric about the imaginary axis.
Now it turns out, numerically I found for the specific $N$ I have, $N$ has same eigenvalues as the Hamiltonian matrix $H=\begin{bmatrix} A & B \\ (C+C^*)/2 & -A^* \end{bmatrix}$. Hence, $N$ also has eigenvalues symmetric w.r.t imaginary axis.
I am looking to ascertain under what additional conditions (on various blocks of $N$) will this be the case ? In other words, under what conditions is the spectrum of $N$ only dependent on the symmetric part of $C$ ?
I was able to prove that without any additional hypothesis, $\lambda \in \sigma(N) \implies -\bar\lambda \in \sigma(M)$, where $M=\begin{bmatrix} A & B \\ C^* & -A^* \end{bmatrix}$, but not the above result regarding equivalence of spectrum of $H$ and $N$.
**Some more details
If $H$ is a Hamiltonian, it satisfies $JHJ-H^*=0$, where $J=\begin{bmatrix} 0 & I_n \\ -I_n & 0 \end{bmatrix}$. The near-Hamiltonian $N$ yields $JNJ-N^*=\begin{bmatrix} C-C^* & 0 \\ 0 & 0 \end{bmatrix}$
