Eigenvalues of block matrix with similar rows Let $\mathbf{A} \in \mathbb{R}^{m\times m}$, $\mathbf{B} \in \mathbb{R}^{m\times n}$ and $\mathbf{C} \in \mathbb{R}^{n\times m}$. What can be said about eigenvalues of a block matrix:
\begin{equation}
\mathbf{X} = \left[\begin{array}{cc}
\mathbf{A} & \mathbf{B}\\
\mathbf{CA} & \mathbf{CB}
\end{array}\right]
\end{equation}
I suspect there are $n$ zero eigenvalues of $\mathbf{X}$. Can somebody help me in proving this assumption. Furthermore, is there some relation to between eigenvalues of $\mathbf{X}$ and eigenvalues of its sub-blocks?
 A: It is convenient to define $Y\in\mathbb R^{m\times(m+n)}$ as
$$
Y
=
\begin{pmatrix}
A\\B
\end{pmatrix}.
$$
Your $X$ maps a vector $(x,y)\in\mathbb R^m\times\mathbb R^n$ to
$$
X(x,y)=(Ax+By,C(Ax+By))\in\mathbb R^m\times\mathbb R^n.
$$
In other words, $z\in\mathbb R^{n+m}$ is mapped to
$$
Xz=(Yz,CYz)\in\mathbb R^m\times\mathbb R^n.
$$
Let us denote the image of $Y$ by $E\subset\mathbb R^m$.
If $Y$ is surjective, then $E=\mathbb R^m$.
The image of $X$ is
$$
F=\{(w,Cw);w\in E\}.
$$
It is a relatively easy exercise to see that $\dim(F)=\dim(E)$.
(Do ask if this requires elaboration. You can also ask a separate follow-up question to settle this point.)
The dimension of $E$ is at most $m$, so the dimension of $F$ is also at most $m$.
In particular, the dimension of the image of $X$ is at most $m$, whereas the dimension of the target space is $m+n$.
This implies that the kernel of $X$ is at least $n$-dimensional.
Your guess is correct: there eigenvalue zero has at least multiplicity $n$.
To be precise, the multiplicity is $m+n-\dim(E)$.
If $A=0$ and $B=0$, then $\dim(E)=0$ and the kernel of $X$ has dimension $m+n$ — as a zero mapping should.
You also asked whether there is a way to relate the eigenvalues of $X$ to the eigenvalues of the blocks.
In general this is not possible; the only blocks that have eigenvalues are $A$ and $CB$.
Their eigenvalues are not a sufficient description, because there are also the off-diagonal blocks $CA$ and $B$.
In the simplest case $m=n=1$, the (potentially) non-zero eigenvalue of $X$ is the sum of the eigenvalues of $A$ and $CB$.
In general you cannot expect such a simple description.
For one reason, there is no natural way to pair the eigenvalues of $A$ and $CB$, because the two matrices have different sizes.
