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Let $V$ be a vector space over $\mathbb F$, and Let $T,S:V\to V$ be linear transformation. Let us assume that both $S,T$ are diagonalizable.

Prove that there exists a mutual basis $B$ for which $S,T$ are both diagonal iff $ST=TS$

$$$$ Well, one direction was pretty simple:

If we assume there exists a mutual basis $v_1,\dots,v_n$ for which both $S,T$ are diagonalizable and that $\lambda_i, \mu_i$ are the eigenvalues respectively, then:

$$(ST)v_i = S(Tv_i) = S(\mu_i v_i) = \lambda_i \mu_i v_i$$

The same goes for $TS$ and this is how we prove that $ST=TS$.

$$$$ I got stuck proving the other direction.

We need to prove that there exists a basis $v_1,\dots,v_n$ for which both $S,T$ are diagonalizable. I started by taking a diagonal basis $b_1,\dots, b_n$ for $T$, and showd that $Sb_i$ is an eigenvector of $T$ for each $i$. The same can be shown for a diagonal asis for $S$. But I'm not sure if this is the right direction and how to continue from here.

Thanks.

Mickey
  • 937
  • This is a classical theorem. If you ever study the representation theory of semisimple Lie algebras this theorem allows you to consider the root space decomposition of your Lie algebra and the corresponding weight space decompositions of its finite-dimensional representations. It's fairly easy to prove (maybe not all by yourself the first time) and quite a deep theorem with many consequences. – Mathematician 42 Jul 03 '17 at 08:57

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