Let $V$ be a vector space, $\dim V=n$. Let $T,S:V \to V$ be linear transformations.
Suppose $\lambda=0$ is an eigenvalue of $T \circ S$, prove that $Ker(S \circ T) \ne 0$.
Let $B$ be a basis for $V$. Note that $[T \circ S]_B^B=[T]_B^B[S]_B^B$
0 is an eigenvalue of $T \circ S$, so it it an eigenvalue of $[T \circ S]_B^B$, so $\Bbb{nul}[T \circ S]_B^B= \Bbb{nul}[T]_B^B[S]_B^B \ne 0$.
So $\Bbb{rank}([T]_B^B[S]_B^B)<n$.
What can I say about the rank of $[S]_B^B[T]_B^B$ in that case? Is it also less then $n$? Why?
Thanks!