# $T$ is a diagonalizable linear operator on a finite dimensional vector space $V$. Is every linear operator commuting with $T$ a polynomial of $T$?

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.

## 1 Answer

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 $id$.

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