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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.

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    "I know that $S$ and $T$ have the same eigenvalues..." Why? The question doesn't say so, nor is there anything else to make this conclusion. – 2017-02-22

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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 $T$.

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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 $ST=TS$.