Let $K$ be a field. Suppose, we have linear maps $f: K^n \mapsto K^n$, $g: K^m \mapsto K^m$ and $h: K^n \mapsto K^m$ such that $h$ is surjective and $h \circ f = g\circ h$. Then every eigenvalue of $g$ is also an eigenvalue of $f$.
I tried a few things but wasn't successful.
Let $\lambda \in K$. We want to prove that if $g-\lambda \text{id}$ is not surjective, then $f-\lambda \text{id}$ is not surjective.
Equivalently; if $f-\lambda \text{id}$ is surjective, then so is $g-\lambda \text{id}$.
Indeed, this follows from $$ h(f-\lambda \text{id}) = hf-\lambda h = gh-\lambda h = (g-\lambda \text{id})h $$ and the fact that for two functions $A \stackrel k \to B \stackrel l \to C$; if $lk$ is surjective, then so is $l$.