If $S,T \in L(V)$, why does $S(T(x))$ having $0$ as an eigenvalue imply $T(S(x))$ has $0$ as an eigenvalue also? I am trying to prove that if $S,T \in L(V)$, then $S(T(x))$ having $0$ as an eigenvalue implies $T(S(x))$ has $0$ as an eigenvalue also.
I know there are two cases but I am having difficulty understanding the second case.
Let us assume that $S(T(x))=0*x$ for $x \neq0$.
First case, $T(x) \neq 0 $. In this case, S has a kernel so if $S(T(x))=0$, then $TS(T(x))=0$, hence $0$ is an eigenvalue of $T(S(x))$.
Now, the case I dont completely get: $T(x)=0$. In this case, my book says that $S$ has no kernel then $S$ is invertible. Hence, $T(x)=0$ means $TS(S^{-1}x)=0$.
Hence, no matter if $T(x) \neq 0 $ or $T(x) = 0 $, then $S(T(x))$ having $0$ as an eigenvalue implies $T(S(x))$ has $0$ as an eigenvalue also.
What I dont get:
What does it mean for S to have a kernel or not have a kernel? That the nullspace is trivial? (only $0$ gets sent to $0$)?
Also, in the second case, why is S invertible? I thought that being invertible here would mean S is also injective and hence its nullspace or kernel is trivial. So basically, why does the fact that $T(x)=0$ mean that S has a trivial nullspace? It seems like S is being forced to only take on $0$ as a value, it has no other choice of elements to send, so how do I know S wont send some other element to 0? Thanks!!!
 A: Here's how I would prove your point:
Because $0$ is an eigenvalue for $S\circ T$, there exists such an $x\neq 0$ that $S(T(x))=0$. Now, you have $2$ cases:


*

*If $S$ is invertible, then, because $S(T(x))=0$, this means that $T(x)=S^{-1}(0)=0$. From $T(x)=0$, it's simple to show that $T(S(S^{-1}(x)))=0$.

*If $S$ is not invertible, then there exists a vector $y\neq0$ in the kernel of $S$, meaning that $S(y)=0$. Therefore, $T(S(y))=T(0)=0$, meaning $0$ is an eigenvalue of $S\circ T$

A: A linear operator has $0$ as an eigenvalue if and only if it fails to be injective, which in finite dimension means its determinant is zero. So your result follows from $\det(S\circ T)=\det(S)\det(T)=\det(T\circ S)$.
A: If $ST$ has $0$ as an eigenvalue, then $ST$ is not one-to-one. 
Thus, at least one of $S$ and $T$ is not one-to-one, for if they were both one-to-one so would be their product. 
Case I. $S$ is not one-to-one. Then $Sx=0$, for some $x\ne 0$, and hence $TSx=T0=0$, and thus $0$ is an eigenvalue of $TS$.
Case II. $S$ is one-to-one and $T$ is not one-to-one. Then $S$ is onto as well, as $V$ is finite dimensional, and hence $S$ possesses an inverse $S^{-1}\in{\mathcal L}(V)$. So as $T$ is not one-to-one, there exists a $x\ne 0$, with $Tx=0$ and if $y=S^{-1}x$, then
$$
(TS)y=TS(S^{-1}x)=Tx=0,
$$ 
and thus $0$ is an eigenvalue of $TS$.
A: If $V$ is finite dimensional, you know that, for $S,T\in L(V)$,
$$
\def\rk{\operatorname{rk}}
\rk ST\le\min\{\rk S,\rk T\}
$$
where $\rk S$ denotes the rank. Moreover, $\lambda$ is an eigenvalue of an endomorphism $F$ if and only if $F-\lambda I$ is not invertible. Because of the rank-nullity theorem, this is equivalent to $F-\lambda I$ not being surjective.
Thus, with $\dim V=n$, saying that $0$ is not an eigenvalue of $TS$ is equivalent to $TS$ having rank $n$; therefore
$$
n=\rk TS\le\min\{\rk T,\rk S\}
$$
which implies $\rk T=\rk S=n$. Therefore both $T$ and $S$ are invertible and $ST$ is invertible too, so $0$ is not an eigenvalue of $ST$.
