How do I show that some eigenvalues of two square matrices of different dimensions are the same? I have three matrices in a field $F$:
$X \in F^{a,a}, Y \in F^{b,b}, Z \in F^{a,b}$, where $a,b \in \mathbb{N}, a \geq b$ and $\text{rank}(Z) = b$. The following term describes their relation:
$X \cdot Z = Z \cdot Y$
I want to show that all eigenvalues of $Y$ are eigenvalues of $X$.
 A: Note: Not all of the determinants you wrote down are well-defined, since $XZ$ and $ZY$ are not quadratic if $a \neq b$.
Also, as stated, this does not seem quite correct. Set $a = 2, b = 1$ and
$$X = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}, \quad
Y = (1), \quad
Z = \begin{pmatrix} 1 \\ 0 \end{pmatrix},$$
then $a \geq b$, the rank of $Z$ is equal to $b = 1$, and
$$XZ = \begin{pmatrix} 1 \\ 0 \end{pmatrix} = ZY.$$
Then $X$ has the eigenvalues 1 and 2, but $Y$ has the eigenvalue 1. Note that you can replace the 2 in $X$ with any other value.

Generally, with problems like these it's a good idea to start by comparing if you can somehow make eigenvectors of one of the matrices into an eigenvector of the other one, and here this gives us some insight, as well.
If $x \neq 0$ is an eigenvector of $Y$ to the value $\lambda$, then
$$\lambda Zx = Z(\lambda x) = Z Yx = XZx,$$
hence, if $Zx \neq 0$, then it is an eigenvector of $X$ to the value $\lambda$. But the full rank condition on $Z$ and $a \geq b$ means that its kernel is empty, so if $x \neq 0$ then also $Zx \neq 0$. So every eigenvalue of $Y$ is an eigenvalue of $X$.
But, looking at the counterexample above, that's all that you can show from the given requirements. X might have a bunch of other eigenvalues that you do not know anything about from the given data.
