Whether a matrix of the kind $A+\sqrt{-1}\cdot B$ is invertible

Question

Let $A_0=(a_{ij})_{N\times N}$ be a real symmetric matrix. Now denote the following two real symmetric matrices by $$ A=\left(\cos{\frac{(i+j)\pi}{2m}}\cdot a_{ij}\right)_{N\times N},\,\,\,B=\left(\sin{\frac{(i+j)\pi}{2m}}\cdot a_{ij}\right)_{N\times N}, $$ where $m\in\mathbb{Z}^+$.

Suppose that both $A$ and $B$ are invertible. Can we prove that the complex matrix $A+\sqrt{-1}\cdot B$ is also invertible?

Since $B$ is invertible, thus it's equvivalent to prove that $\sqrt{-1}\cdot I+AB^{-1}$ is invertible. If $AB^{-1}$ is symmetic (which is equvivalent to $AB=BA$ ), then its eigenvalues are all real, and the proof ends. However, in our case we don't have $AB=BA$.

Answer 1

Here is a counterexample. Let $$ A=\begin{pmatrix} -1 & 0 \cr 0 & 1\end{pmatrix},\; B=\begin{pmatrix} 0 & 1 \cr 1 & 0\end{pmatrix}. $$ Both $A$ and $B$ are real, symmetric, invertible matrices. However,
$$ A+iB=\begin{pmatrix} -1 & i \cr i & 1\end{pmatrix} $$ has determinant zero, so is not invertible.

Answer 2

No. E.g. consider $m=1$, $$ \begin{aligned} A_0&=\pmatrix{1&1\\ 1&1},\\ C&=\left(\cos\frac{(i+j)\pi}{2}\right)_{i,j\in\{1,2\}}=\pmatrix{-1&0\\ 0&1},\\ S&=\left(\sin\frac{(i+j)\pi}{2}\right)_{i,j\in\{1,2\}}=\pmatrix{0&-1\\ -1&0},\\ A&=C\circ A_0=C,\\ B&=S\circ A_0=S,\\ A+\mathbf iB&=C+\mathbf i S=\pmatrix{-1&-\mathbf i\\ -\mathbf i&1}=\pmatrix{1\\ \mathbf i}\pmatrix{-1&-\mathbf i}.\\ \end{aligned} $$