I am looking for a proof check, as well as a bit of help.
Let $X$ be a finite-dimensional vector space over $\mathbb{C}$. Let $R$ and $S$ be linear operators on $X$.
(a) Prove that $SR$ and $RS$ have the same set of nonzero eigenvalues.
(b) Let $\lambda$ be a nonzero eigenvalue of $SR$. Let $U$ be the generalized eigenspace of $SR$ corresponding to $\lambda$. Let $W$ be the generalized eigenspace of $RS$ corresponding to $\lambda$. Show that $S$ maps $U$ into $W$ injectively.
(a) Let $\lambda$ be an eigenvalue of $SR$. There are two cases. If $\lambda \neq 0$, let $x$ be an eigenvector corresponding to $\lambda$. Now $SRx = \lambda x$. Since $\lambda \neq 0$, $Rx \neq 0$, so $RS(Rx) = R\lambda x = \lambda R(x)$. So $\lambda$ is an eigenvalue of $RS$ as well.
Suppose instead that $\lambda = 0$. Then, by the Invertible Matrix Theorem, $\det(SR) = 0$. But, $\det(SR) = \det(RS) = 0$, so $RS$ is also not invertible, hence it has 0 as an eigenvalue as well.
I would appreciate any corrections or feedback for part (a). For part (b), I am not sure how to proceed. I am aware that if $v \in U$, it can be written as a unique linear combination of generalized eigenvectors of $SR$, then I'm trying to show that the image of such vectors under $S$ are in $W$, but I'm not quite sure how to show this, let alone that this is an injective mapping.