$ST=TS$ leads to diagonalization?

Asked Oct 23 '17 at 11:41

Active Oct 23 '17 at 11:44

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Preparing for GRE math subject and I come across this problem:

$S$ and $T$ are linear transformations of a finite space $V$, $ST=TS$, each eigenspace of $S$ is one-dimensional, which of the followings are/is correct?

Each eigenvector of $S$ is also $T$'s

Each eigenvector of $T$ is also $S$'s

$T$ is diagonalizable

Since each eigenspace of $S$ is of one dimension, $S$ should be diagonalizable, but the relationship of $ST=TS$ is hard to interpret. Usually matrix is not communicative, $ST=TS$ should contain some kind of important information.

edited Oct 23 '17 at 11:44

Arnaud D.

20,884

asked Oct 23 '17 at 11:41

JP Zhang

1

Possible duplicate of Simultaneously Diagonalizable Proof – Arnaud D. Oct 23 '17 at 11:45
If $ST=TS$, and $x$ is an eigenvector of $S$, then $STx=TSx=\lambda Tx$, so $Tx$ is an eigenvector of $S$. – Oct 23 '17 at 11:52

$ST=TS$ leads to diagonalization?

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