$T,S: V\to V$, prove that $TS$ and $ST$ have the same eigenvalues hey I was trying to prove this proposition by dividing to cases and this is what I've got so far:
let's assume without loss of generality that T is invertible:
ST = [T][S]
TS = [S][T]
Pst(x) = |[T][S]-גI| = |[T][S]-גI|*|[T][T^-1]| = |[T^-1]|*|[T][S]-גI|*|[T]| =
|[S]-[T^-1]גI|*|[T]| = |[S]-ג[T^-1]|*|[T]| = |[S][T]-גI| = Pts(x)

they have the same characteristic polynomial, thus the same eigenvalues.
but what about the case when both are non-invertible?
would appreciate any kind of help.
 A: You did half the work (and you're right) : if $T$ or $S$ in an element of $\mathrm{GL}_{n}(\mathbb{C})$ (i.e. an invertible matrix), the following equality holds : $\mathrm{P}_{TS}=\mathrm{P}_{ST}$. Here, $\mathrm{P}_{TS}$ (resp. $\mathrm{P}_{ST}$) denotes the characteristic polynomial of $TS$ (resp. $ST$).
If $T$ (or $S$) is no invertible, you can use the density of $\mathcal{GL}_{n}(\mathbb{C})$ into $\mathcal{M}_{n}(\mathbb{C})$. In other words : there exist a sequence $(T_{n})_{n \geq 0}$ of invertible matrices such that $T_{n} \mathop{\longrightarrow} \limits_{n \to +\infty} T$.
For all $n$, the following holds : 
$$ \mathrm{P}_{T_{n}S} = \mathrm{P}_{ST_{n}}. $$
By continuity of the characteristic polynomial, 
$$ \mathrm{P}_{T_{n}S} \, \mathop{\longrightarrow} \limits_{n \to +\infty} \, \mathrm{P}_{TS} $$
and, as well,
$$ \mathrm{P}_{ST_{n}} \, \mathop{\longrightarrow} \limits_{n \to +\infty} \, \mathrm{P}_{ST}. $$
As a consequence : $\mathrm{P}_{TS} = \mathrm{P}_{ST}$.
A: Considering nonzero eigenvalues first, suppose $ST$ has a nonzero eigenvalue $\lambda$, so that there's some nonzero $v$ such that $$STv = \lambda v.$$
What can we say about $TS$? Since $S(Tv) = \lambda v$, let's see what happens when we apply $TS$ to $Tv$:
\begin{align*}TS(Tv) 
&= T(\lambda v)\\
&= \lambda Tv \end{align*}
where the latter holds because $T$ is linear. Note that by our assumptions, we know $Tv \neq 0$, as $S(Tv) = \lambda v \neq 0$.
Thus, if $\lambda$ is a nonzero eigenvalue of $ST$, it must be an eigenvalue of $TS$. A very similar argument will show that eigenvalues of $TS$ are also eigenvalues of $ST$.
Finally, $0$ is an eigenvalue of $TS \iff$ at least one of $S$ or $T$ is singular $\iff 0$ is an eigenvalue of $ST$.
A: Though Jibounet's answer covers what is needed, let me add something from a general group-theoretic viewpoint: In any group $G$ if $x,y\in G$ are two elements then the elements $xy$ and $yx$ are conjugate (could be the same), because $  y(xy)y^{-1}= yx$. 
Now apply this to group of non-singular matrices, then $ST$ and $TS$ are conjugate: and in matrices instead of using the word conjugate it is customary to say these transformations are similar.
Similar matrices have the same characteristic polynomials and hence the same eigen values. This is straight from the definition: $$\det(ABA^{-1}-\lambda I)= \det\big( A(B-\lambda I)A^{-1}\big)=\det A \det (B-\lambda I)\det A^{-1}= \det (B-\lambda I).t$$ Of course for singular $S,T$ one has to use the limiting argument of Jibounet. 
