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Suppose that $V$ is finite dimensional and $T \in L(V)$. Prove that $T$ is a scalar multiple of the identity if and only if $ST = TS$ for every $S \in L(V)$.

Suppose $T=\lambda I$ for some $\lambda \in \mathbb{F}$, showing $ST = TS$ is straightforward, but I don't know how to prove the converse.

St Vincent
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leeha
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  • I'm sure this has been answered before. – copper.hat Sep 23 '13 at 00:14
  • @VedranŠego: Do you mind me asking how you found this? I spent 5 mins. searching to no avail. – copper.hat Sep 23 '13 at 00:29
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    @copper.hat I used this: ts st is:question. It is surprising how often the duplicates are so similar to each other that most of the wording and even variables' names remain the same. However, I don't think it's the only one, as I seem to recall answering this (or a very similar) question myself, but I couldn't find that one (not that I tried too much). – Vedran Šego Sep 23 '13 at 00:32
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    @copper.hat Even though I'm not Vedran, I searched "TS = ST" and found it. – Cameron L. Williams Sep 23 '13 at 00:33
  • Thanks to both of you. It never occurred to me to search for an expression like that. I think I answered a similar one, but was unable to even find that. Oh well... – copper.hat Sep 23 '13 at 00:35

2 Answers2

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Suppose $TS=ST$ for all $S$.

Let $Tv = \lambda v$. Then $ST v = TS v = \lambda Sv$. That is, $Sv$ is an eigenvector of $T$ for all $S$. Since $S$ is arbitrary, this tells you that $Tx = \lambda x$ for all $x$. That is, $T = \lambda I$.

copper.hat
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  • copper.hat since the space id finite dimensional, only then your first line $Tx=\lambda x$ makes sense, isn't it. – Upstart Jan 30 '18 at 17:30
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    Well, sort of. By choosing $S$ appropriately we can show that $T$ is a multiple of the identity on any finite dimensional subspace from which the result follows. – copper.hat Jan 30 '18 at 18:21
  • If V is not a complex vector space, how does the assumption $Tv=\lambda v$ hold? I mean if V is not a complex vector space, then we can't be sure if eigenvalue of T exists. – Koro Nov 05 '21 at 09:39
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    @koro You are correct, I am assuming that there is at least one eigenvector. – copper.hat Nov 06 '21 at 17:16
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Look at the canonical matrices. This will help you distill what the entries of $T$ should be.

Pedro
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