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 ∀S∈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, ∀v∈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≠ST, might be the way to go, but haven’t made much progress.

Thanks in advance!

**Answer**

For a basis-free answer, consider S∈L(V) given by Sx=f(x)v for some vector v and some linear functional f on V. Then TSx=f(x)Tv=STx=f(Tx)v for any x. In particular, as long as a nontrivial linear functional f on V exists, there is x

such that f(x)≠0, and then Tv=αv for all v, where α=f(Tx)/f(x).

This works even for infinite-dimensional spaces, although I think in general you need the Axiom of Choice to get a nontrivial linear functional on a vector space.

**Attribution***Source : Link , Question Author : abeln , Answer Author : Robert Israel*