Eigenvalues and eigenvectors of antidiagonal block matrix

Question

I have $2$ square matrices $A_m$ and $B_m$ which are symmetric and of size $m\times m$. And the 3rd matrix is

$$C = \begin{bmatrix} 0 & A \\ B & 0\end{bmatrix}$$

Now, I would like to calculate the eigenvalues and eigenvectors of matrix $C$. How can I get it? Or how does it related to the eigenvalues and eigenvectors of $A$ and $B$?

Thank you very much in advance!

Answer 1

@A.G. idea is a good start:

If we assume that $A,B$ commute, then we get an easy result.

$C^2$ is symmetric, so it is diagonalisable. Let $P$ be a matrix which is the change to the diagonal basis.

$(PC^2 P^{-1}) = (PCP^{-1})^2 = D$, where $D$ some diagonal matrix, so $PCP^{-1} = D^{1/2}$, which you can compute by taking the square roots of the entries of $D$.

Answer 2

$\det(\lambda I-C)=\det\pmatrix{\lambda I&-A\\ -B&\lambda I}$. Since all square subblocks have the same sizes and the two subblocks at bottom commute, the determinant is equal to $\det(\lambda^2 I - AB)$. Therefore, the eigenvalues of $C$ are the square roots of eigenvalues of $AB$. That is, for each eigenvalue $t$ of $AB$, the two roots of $\lambda^2-t=0$ are eigenvalues of $C$.

As pointed out in a comment, we have $\det(C)=\det(-AB)$ and hence there is some relation between the product of the eigenvalues of $C$ and the products of the eigenvalues of $A$ and $B$, but besides that, very few about the spectrum or the eigenvectors of $AB$ can be said even if the spectra and eigenvectors of $A$ and $B$ are fully known. When both $A$ and $B$ are positive definite, we do have some bounds for the eigenvalues of $AB$. See "Evaluating eigenvalues of a product of two positive definite matrices" on this site or "Eigenvalues of product of two symmetric matrices" on MO.