Eigenvalues of a block matrix

Question

For $X=\left(\begin{array}{cc} A & B\\ C & 0\end{array}\right)$, how are eigenvalues of $X$ related to the eigenvalues of $A$?

Answer 1

Not much can be said. However, if $A$ is square and $X$ is Hermitian (hence $A$ is Hermitian and $C=B^\ast$) and $\lambda_1(M)\le\lambda_2(M)\le\lambda_3(M)\cdots$ denote the eigenvalues of a Hermitian matrix $M$ arranged in increasing order, we have the following interlacing inequality: $$ \lambda_k(X)\le\lambda_k(A)\le\lambda_{k+n-r}(X) $$ for $1\le k\le r$, when $A$ is $r\times r$ and $X$ is $n\times n$.

Also, if the four blocks have equal sizes, the characteristic polynomial of $X$ can be simplified as follows: $$ \det\pmatrix{xI-A&-B\\ -C&xI} = \det(x^2I - xA - BC). $$ Such simplification is valid because the two blocks at the bottom of the LHS commute. You can see that $\det(x^2I - xA - BC)$ has little resemblance to $\det(\lambda I-A)$ and we don't expect any relationship between the eigenvalues of $X$ and $A$ in general.

Answer 2

Say that for some v, $ Av = \lambda v $ and $ C v = 0 $. Then letting $ w = \left(\begin{array}{c} v \\ 0 \end{array}\right) $, what is $ X w $? Similarly if $ w = \left(\begin{array}{c} x \\ y \end{array}\right) $ is an eigenvalue of $ X $ and $ B y = 0 $, then what is $ A x$?