Prove that $T,S$ are simultaneously diagonalizable iff $TS=ST$.

Question

Definition: We say that $S,T$ are simultaneously diagonalizable if there's a basis, $B$ which composed by eigen-vectores of both $T$ and $S$

Show that $S,T$ are simultaneously diagonalizable iff $ST=TS$.

I tried both directions, but couldn't get much further.

I'd be glad for help.

Thanks.

You should try searching the site first, as a matter of course — Ben Grossmann, Dec 12 '14 at 22:06

score 5 · Answer 1 · answered Dec 12 '14 at 22:14

5

"If" is not true. A counterexample is $S=({}^{0\;1}_{0\;0})$ and $T=({}^{1\;0}_{0\;1})$. They trivially commute, but $S$ is not diagonalizable at all, and so in particular not simultaneously with $T$.

answered Dec 12 '14 at 22:14

hmakholm left over Monica

291,611

Cute matrices. ${}$ – copper.hat Dec 12 '14 at 22:18

mookid · Accepted Answer · 2014-12-12T22:04:13.480

3

hint: If $S = P^{-1} D_S P$ and $T = P^{-1} D_T P$ with $D_{S,T}$ diagonal then what are the values of $ST$ and $TS$?

edited Dec 12 '14 at 22:04

answered Dec 12 '14 at 21:58

mookid

28,795

Prove that $T,S$ are simultaneously diagonalizable iff $TS=ST$.

2 Answers2