If $m_1,m_2$ are the minimal polynomials of $ST$ and $TS$ prove $m_2(x)=x^im_1(x)$ where $i=-1,0$ or $1$ Let V be a finite dimensional vector space, and let $S,T:V \rightarrow V$ be linear transformations. Let $m_1,m_2$ denote the minimal polynomials of $ST$ and $TS$ respectively how would you prove:
$m_2(x)=x^im_1(x)$ where $i=-1,0$ or $1$ and $a$ is an eigenvalue of $ST$ iff a is an eigenvalue of $TS$?
 A: Suppose that $a$ is an eigenvalue of $ST$. If $\mathbf{v}\in E_a$, the eigenspace of $ST$ associated to $a$, then
$$TS\Bigl(T(\mathbf{v})\Bigr) = T\Bigl(ST(\mathbf{v})\Bigr) = T(a\mathbf{v}) = aT(\mathbf{v}),$$
so either $T(\mathbf{v})=\mathbf{0}$, or else $a$ is also an eigenvalue of $TS$. 
So you have two cases: if $T(\mathbf{v})=\mathbf{0}$ for every $\mathbf{v}\in E_a$, then $T$ is not invertible; what can you conclude $a$ and about $TS$ in that case? 
The other case is that there exists $\mathbf{v}\in E_a$ with $T(\mathbf{v})\neq\mathbf{0}$. What can you conclude about $TS$ in that case?
Can you apply the same argument starting with $TS$ instead of with $ST$?
For the minimal polynomials, notice that for any $n\gt 0$,
$$T(ST)^nS = (TS)^{n+1}.$$
If $m_1(x) = x^k + a_{k-1}x^{k-1}+\cdots + a_1x + a_0$ is the minimal polynomial of $ST$, then
$$\begin{align*}
(ST)^k + a_{k-1}(ST)^{k-1}+\cdots + a_1(ST) + a_0I &= 0\\
T\Bigl((ST)^k + a_{k-1}(ST)^{k-1}+\cdots + a_1(ST) + a_0I\Bigr)S &= 0\\
T(ST)^kS + a_{k-1}T(ST)^{k-1}S + \cdots + a_1T(ST)S + a_0TS &= 0\\
(TS)^{k+1} + a_{k-1}(TS)^k + \cdots + a_1(TS)^2 + a_0(TS) & = 0,
\end{align*}$$
so $TS$ satisfies $xm_1(x)$. Hence $m_2(x)$ divides $xm_1(x)$.  By a symmetric argument, $m_2(x)$ divides $xm_1(x)$.
If $0$ is not an eigenvalue of $ST$ nor of $TS$, then $x$ cannot be a factor of the minimal polynomials; what can you conclude then?
If $0$ is an eigenvalue of $ST$ and of $TS$, then you cannot simply "take them out"; but it does tell you that every irreducible factor, except perhaps for the irreducible factor $x$, must be the same (and raised to the same degree) in $m_1(x)$ and in $m_2(x)$. Now you just need to deal with $x$, and show that the degree to which it shows in $m_1(x)$ and in $m_2(x)$ can differ by at most $1$, which will give you the result you want.
