# commuting matrices and simultaneous diagonalization

• Aug 7th 2013, 12:41 AM
ktotam
commuting matrices and simultaneous diagonalization
Hi,

I'm training for exam and come across a question from one of the exams:

Quote:

T and S are normal linear transformations. Also they commute TS = ST.
It is required to prove that there exists common vector basis consisting of T's (or S's) eigenvectors.
One way to prove this is to find a matrix U such that:

$\displaystyle U^{-1}[T]U = diag(\gamma_i) \mbox{ and } U^{-1}[S]U = diag(\lambda_i)$

After some manipulation I can show that TS and ST are also normal, not sure if this is useful.
I know (i searched the web to find more clues) that commute matrices should have common eigenvector, though again i didn't find simple proof for this fact.

Can anyone please give some direction of how to prove this ?

Thank you

P.S.
Our material doesn't include things like classification into groups/subgroups or representation theory so proof should use more elementary concepts like eigenvector, eigenvalues, unitary matrices
• Aug 7th 2013, 04:35 PM
Drexel28
Re: commuting matrices and simultaneous diagonalization
This is a very common theorem. See here, for example.
• Aug 8th 2013, 07:31 AM
ktotam
Re: commuting matrices and simultaneous diagonalization
Thanks for the suggesting the site.
I already seen that proof and I find it a bit wordy...
I'm thinking why to involve scalar matrices and why the need to induct on vector space dimension... (I understand that this might be also a way to prove)

Eventually I find a very nice and clear proof here: on page 8.
Thanks
• Aug 8th 2013, 09:49 PM
Drexel28
Re: commuting matrices and simultaneous diagonalization
Quote:

Originally Posted by ktotam
Thanks for the suggesting the site.
I already seen that proof and I find it a bit wordy...
I'm thinking why to involve scalar matrices and why the need to induct on vector space dimension... (I understand that this might be also a way to prove)

Eventually I find a very nice and clear proof here: on page 8.
Thanks

Never forget Keith Conrad. He usually has uncannily clear explanations of things. I'm glad you're happy!