Show that a matrix has eigenvalues at zero Let $A\in\mathbb{R}^{n\times n}$, $B\in\mathbb{R}^{n\times m}$, $C\in\mathbb{R}^{m\times n}$, and define
\begin{equation}
X:=\begin{bmatrix}A&BC\\I_n&0\end{bmatrix}\in\mathbb{R}^{2n\times 2n}.
\end{equation}
Show that if $m<n$, then $X$ has $n-m$ eigenvalues at zero.
 A: Exploit the fact that there is an identity matrix in the bottom left and the zero matrix in the bottom right. By adding multiples of the bottom n rows to the top n rows, you can make A zero without affecting the rest of the matrix. Then permute columns to get the matrix into the form
$$\begin{bmatrix} BC & 0 \\ 0 & I_n \end{bmatrix} $$
From this block diagonal form you can see the rank is n plus the rank of BC.
A: Here is another way: decompose $X$ into:
$$
X =\begin{bmatrix}A & I\\ I & 0\end{bmatrix}\begin{bmatrix}I & 0\\0 & BC\end{bmatrix} = \tilde{A}\tilde{B},
$$
and observe that $\tilde{A}$ is invertible. Then you have the generalized eigenvalue problem
$$
\tilde{B} x = \lambda\tilde{A}^{-1}x.
$$
You already concluded that $BC$ has at least $n-m$ zero eigenvalues, which implies that $\tilde{B}$ also has at least $n-m$ zero eigenvalues. Let $x$ be a nonzero eigenvector corresponding to such an eigenvalue, then we have:
$$
\lambda \tilde{A}^{-1}x = 0,
$$
but this is only possible whenever $\lambda = 0$, which is an eigenvalue of $X$.
