$\mbox{Ker} \;S$ is T-invariant, when $TS=ST$ Let $T,S:V\to S$ linear transformations, s.t: $TS=ST$, then $\ker(S)$ is $T$-invariant.
My solution:
$$\{T(v)\in V:TS(v)=0 \}=\{T(v)\in V:ST(v)=0 \}\subseteq\ker(S)$$
If its right, then why $$\{T(v)\in V:ST(v)=0 \}=\ker(S)$$?
Thank you.
 A: What you wrote is not correct. You simply have to check that if $v$ belongs to $\mbox{Ker} S$, then $Tv$ also lies in $\mbox{Ker} S$. So assume $Sv=0$. Then 
$$STv=TSv=T0=0$$
where the last equality holds because a linear transformation always sends $0$ to $0$.
Therefore $v \in \mbox{Ker} S$ implies $Tv\in \mbox{Ker} S$. In other words $$T(\mbox{Ker} S)\subseteq\mbox{Ker} S$$ that is $\mbox{Ker} S$ is invariant under $T$.
Note: if $S$ is injective, this is of course trivial. But if not, this says that that the eigenspace of $S$ with respect to the eigenvalue $0$ is invariant under $T$. More generally, every eigenspace of $S$ is invariant under $T$. When $S$ and $T$ are two cummuting diagonalizable matrices, this is the key remark when showing that they are simultaneously diagonalizable.
Note: as pointed out by Marc van Leeuwen, it is not more diffcult to show that $\mbox{Im S}$ is invariant under $T$, as $TSv=STv$. And finally, every polynomial in $S$ still commutes with $T$, so you can replace $S$ by $p(S)$ everywhere if you want.
A: To supplement the answer by julien, note that all you want to prove is
$$
T( \ker(S) ) \subseteq \ker(S).
$$
It seems you are trying to prove equality $T( \ker(S) ) = \ker(S)$, but this need not hold, think of the case when $T = 0$ (or $T = S$) and $\ker(S) \ne 0$. 
