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Suppose we that $V$ is finite dimensional vector space over $\mathbb{C}$ and $R, T$ are diagonalizable operators such that $RT = TR$. Show that there exists basis of $V$ such that both $R$ and $T$ are given by diagonal matrices

My attempt:

Since $T$ and $R$ are diagonalizable, then there exists matrices $P, B$ such that $T = P^{-1}TP$ and $R = B^{-1}RB$. Since $T$ and $R$ commute, $(P^{-1}TP)(B^{-1}RB) = (B^{-1}RB)(P^{-1}TP)$. I am not sure how to proceed from here.

Mike Limber
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  • We can diagonalize $R$, so we can already assume $R$ is diagonal. (And change $T$ according to the corresponding base change.) How can $T$ now look like (in order to commute with $R$)? – dan_fulea Nov 27 '19 at 13:58
  • I believe if $R$ is already diagonal and $T$ commutes with $R$, then $T$ also has to be diagonal – Mike Limber Nov 27 '19 at 14:01
  • You may start with n=2 and see how you can generalize – Weier Nov 27 '19 at 14:02
  • @Mike Limber: No. For instance, if $R$ is scalar, $T$ needn’t be. However, you can show that the eigenspaces of eg $R$ are stable under $T$, so you can diagonalize the “restrictions” of $T$ to these spaces. – Aphelli Nov 27 '19 at 14:03
  • @Mindlack How can $R$ be a scalar? It is a linear operator. – Mike Limber Nov 27 '19 at 14:08
  • @MikeLimber "...then $T$ also has to be diagonal." Here we must use the fact that $T$ is diagonal itself. Else... We can suppose that $R$ comes with "maximal blocks" of diagonal (square) submatrices, in each such block we have diagonally the same entry / eigenvalue, but this does not come any longer. A commuting $T$ must have and adapted block representation. And it is possible to have one or more non-trivial Jordan blocks "adapted / corresponding" to a block of $R$. But $T$ is also diagonal... – dan_fulea Nov 27 '19 at 14:40
  • @Mike Limber: I meant “scalar” as “scalar multiple of the identity”. – Aphelli Nov 27 '19 at 14:46

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It's been asked before. Commutative diagonalizable matrices simultaneously diagonalize. See here and here.

Tian He
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