TITLE: Eigenvalues and eigenvectors of antidiagonal block matrix
QUESTION [6 upvotes]: 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!

REPLY [5 votes]: $\det(\lambda I-C)=\det\pmatrix{\lambda I&-A\\ -B&\lambda I}$. 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.
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