The range of $S$ is invariant under $T$

Question

How would I go about solving the following question...

Suppose $S,T\in L(V)$ are such that $ST=TS$. Prove that the range of $S$ is invariant under $T$.

I know that $S$ and $T$ have the same eigenvalues but I am not sure if this helps in this case. Any help would be appreciated because I am stuck on this proof.

Answer 1

You need to show that if $y\in\operatorname{Range}(S)$, then $T(y)\in\operatorname{Range}(S)$. Pick any $y\in\operatorname{Range}(S)$. By definition of range, this means that there exists some $x\in V$ such that $y=S(x)$. But then $Ty=TSx=\ldots$, and that's where you apply the given property that $ST=TS$.

Answer 2

Let $v\in V$. Then $Sv \in \operatorname*{range}(S)$ and it follows that $$ T(Sv) = S(Tv) \in \operatorname*{range}(S) $$ Since this is true for all $v\in V$, it follows that $$ Tu \in \operatorname*{range}(S) $$ for all $u \in \operatorname*{range}(S)$. Hence the range of $S$ is invariant under $T$.

Answer 3

Let v element of Range(S). This implies there exist u in V such that S(u) =v. Now T(S(u)) =T(v) S(T(u)) =T(v). Since TS=ST That is there exist T(u) in V such that S(T(u)) =T(v). Thus T(v) is there in Range of S. Therefore Range of S is invariant under T