$\mbox{Ker} \;S$ 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.

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 $\ker(S) \ne 0$. 
