Eigenvalues of anti-triangular block matrix

Question

I have a $2n \times 2n$ real anti-triangular (skew-triangular?) block matrix of the form

$$ M = \begin{bmatrix} A & B \\ I_n & O_n \end{bmatrix} $$

where $I_n$ is the $n \times n$ identity matrix and $O_n$ is the $n \times n$ zero matrix. Note that the blocks $A$ and $B$ are also $n \times n$. Do the eigenvalues of $M$ have any specific relationships with the submatrices of $A$, $B$ (or their eigenvalues)?

Any theory or discussion would be helpful.

Answer 1

Nope as your matrix doesn't need to have a single eigenvalue in general, look at the trivial case where $n=1$ and your blockmatrix is

$$\begin{pmatrix}0 & -1 \\ 1 & 0 \\ \end{pmatrix}$$

As the characteristic polynomial is $x^2+1$ it doesn't have any eigenvalues over $\mathbb{R}$ while $A$ hast the eigenvalue $0$ and $B$ has the eigenvalue $-1$.

Answer 2

The exact relationship boils down to a matrix equation, so it does not simplify further with respect to the individual eigenvalues:

\begin{align} &\operatorname{det}\left(\matrix{xI - A & -B \\ -I & xI}\right) \\ =& \pm \operatorname{det}\left(\matrix{ -I & xI \\ xI - A & -B}\right) \quad\text{sign depending on number of row swaps}\\ =& \pm \operatorname{det}\left(\matrix{ -I & xI \\ 0 & -B + xI(xI - A)}\right) \\ =& \pm \operatorname{det}\left(B - xI(xI - A)\right) \\ =& \pm \operatorname{det}\left(B - x^2I + xI\cdot A\right) \\ \end{align}