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 SL(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, vV. 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 TSST, might be the way to go, but haven’t made much progress.

Thanks in advance!

Answer

For a basis-free answer, consider SL(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

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