Understanding a proof conceptually Let's assume that $V$ and $W$ are vector spaces over a field $\mathbb{K}$, $\lambda\in\mathbb{K}$, $\lambda\neq0$.
$S: V\rightarrow W$ and $T: W\rightarrow V$ are linear maps. Prove, that
$\lambda$ is an eigenvalue of $TS\iff\lambda$ is an eigenvalue of $ST$
What can be stated about the eigenvalues of the maps $TS$ and $ST$?
Would it also be correct if $\lambda=0$?
Proof:


*

*$\lambda$ is an eigenvalue of $TS\Rightarrow\lambda$ is an eigenvalue of $ST$ 


$TSv=\lambda v$ that is $S(TSv)=S(\lambda v)$ that is $ST(Sv)=\lambda (Sv)$


*$\lambda$ is an eigenvalue of $ST\Rightarrow\lambda$ is an eigenvalue of $TS$


$STw=\lambda w$ that is $T(STw)=T(\lambda w)$ that is $TS(Tw)=\lambda (Tw)$
I do not understand why the statement is proven by finding two eigenvectors. What would happen if we couldn't construct $Sv$ out of $v$ and $Tw$ out of $w$?
I also don't understand how to answer the latter two questions.
 A: That $\lambda$ is an eigenvalue for $T$ means that there is some vector $v$ such that $Tv = \lambda v$.
So you have that if $\lambda$ is an eigenvalue for $ST$, then there is some vector (it doesn't matter what it is) $v$ such that $(ST)v = \lambda v$. You then show that there is some other vector (it doesn't matter what it is) $w$ such that $(TS)w = \lambda w$. Hence $\lambda$ is, by definition, and eigenvalue for $TS$.
So, you have that if all eigenvalues for $ST$ are also eigenvalues for $TS$. Likewise you have the other way around by the same argument. Hence it is impossible to have an eigenvalue for one that isn't also an eigenvalue for the other. That means they have the same eigenvalues. 
Note the proof works even if $\lambda =0$.
A: The eigenvector  given/described  in 1) ,is intended to show that it is an eigenvector of the eigenvalue $\lambda$, so that $Sv$ is an eigenvector associated to $\lambda$, implying $\lambda$ is an eigenvalue associated with $ST$, showing that every eigenvalue of $TS$ is an eigenvalue of $ST$. Something similar is happening for the second eigenvalue.
