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Let $V$ be a finite-dimensional vector space and $T$ be a linear map from $V$ to itself (in other words, it's an operator on $V$).

I need to prove that $T$ is a scalar multiple of the identity (i.e. $T(v)=cv$ for all $v \in V$ and for some $c \in F$, the underlying field) if and only if $ST=TS$ for all operators $S$ on $V$.

Note that $ST=TS$ does not necessarily mean that they are both equal to $I$, the identity operator. The only if part is trivial. But I cannot see how to attack the if part. Any help would be greatly appreciated.

  • Consider $S(x)=vv^\top x$ for every $v\in V$. Then $v$ is an eigenvector of $T$. multiples of the identity are the only operator that have every vector as eigenvectors. – Surb Feb 13 '17 at 21:07
  • @G.Sassatelli Yes, exactly. Thanks for the link. –  Feb 13 '17 at 21:08

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