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?
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.