The question is from Axler's "Linear Algebra Done Right", which I'm using for self-study.
We are given a linear operator $T$ over a finite dimensional vector space $V$. We have to show that $T$ is a scalar multiple of the identity iff $\forall S \in L(V), TS = ST$. Here, $L(V)$ denotes the set of all linear operators over $V$.
One direction is easy to prove. If $T$ is a scalar multiple of the identity, then there exists a scalar $a$ such that $Tv = av$, $\forall v \in V$. Hence, given an arbitrary vector $w$, $$TS(w) = T(Sw) = a(Sw) = S(aw) = S(Tw) = ST(w)$$ where the third equality is possible because $S$ is a linear operator. Then, it follows that $TS = ST$, as required.
I am, however, at a loss as to how to tackle the other direction. I thought that a proof by contradiction, ultimately constructing a linear operator $S$ for which $TS \neq ST$, might be the way to go, but haven't made much progress.
Thanks in advance!