1
$\begingroup$

I'm trying to answer this question

True or false? $T$ is a diagonalizable linear operator on a finite dimensional vector space $V$. Then every linear operator that commutes with $T$ is a polynomial of $T$.

I think the statement false but unable to find a counter example. So any help is appreciated.

$\endgroup$
0

1 Answer 1

Reset to default
4
$\begingroup$

This is false, at least when $\dim V > 1$: The identity transformation $id$ on $V$ (which is diagonalizable) commutes with every linear operator $V \to V$, but every polynomial in $id$ is a multiple of $id$; in particular, linear operators that are not multiples of $id$ commute with $id$ but are not polynomials in $id$.

$\endgroup$
3
  • $\begingroup$ I wonder why does it take this long to upvote a simple, correct answer like this one. +1 $\endgroup$
    – Timbuc
    Dec 14, 2014 at 13:56
  • $\begingroup$ @Travis thank you for your help $\endgroup$
    – SSH
    Dec 15, 2014 at 4:39
  • $\begingroup$ You're welcome. $\endgroup$
    – Travis Willse
    Dec 15, 2014 at 11:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .