# A linear operator commuting with all such operators is a scalar multiple of the identity.

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 $\forall S \in {\cal L}(V), TS = ST$. Here, ${\cal L}(V)$ 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 $w$, 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.

For a basis-free answer, consider $S \in L(V)$ given by $S x = f(x) v$ for some vector $v$ and some linear functional $f$ on V. Then $T S x = f(x) T v = S T x = f(T x) v$ for any x. In particular, as long as a nontrivial linear functional $f$ on $V$ exists, there is $x$
such that $f(x) \ne 0$, and then $T v = \alpha v$ for all $v$, where $\alpha = f(T x)/f(x)$.