the commuting algebra of a matrix [closed]

Actually I'm not sure about the term "commuting algebra". I might have used it wrongly.

Let $A$ be a non-scalar $2 \times 2$ matrix. Consider the ring of all $2 \times 2$ matrices which commute with $A$. We can show this ring is commutative!

This result is related to this question: http://mathoverflow.net/questions/13349/tate-module-of-cm-elliptic-curves

I just did this exercise by down-to-earth linear algebra computation. I'm wondering if there is a more conceptual proof, and is it still true for $n>2$?

edited Jan 31 2010 at 22:11

Pete L. Clark
40k●4●103●227

asked Jan 30 2010 at 9:46

natura
892●3●13

the centralisator of $(1,0,0;0,1,0;1,0,0)$ is not commutative. – Martin Brandenburg Jan 30 2010 at 9:54

The centralizer (or commutant) of any nonscalar idempotent will give you a counterexample when n>2. – Jonas Meyer Jan 30 2010 at 10:00

ah, yes, I just realized the proof of the 2 dimensional case can be simplified if we consider the Jordan matrix, which is really simple in 2 dim case...so yeah, shouldn't generalize to n>3... – natura Jan 30 2010 at 10:11

Someone help me to close this question... – natura Jan 30 2010 at 10:13

Closed upon request of the questioner. – Pete L. Clark Jan 30 2010 at 16:58