Problem 38

ThePerfectHacker · #1 $Old$ 09-24-2007, 07:18 PM

Show the center of the general linear group are the proportional diagnol matrices.
Meaning, $Z(\mbox{GL}_n(\mathbb{R})) = \{ k I| k\in \mathbb{R}^{*} \mbox{ and }I \mbox{ identity matrix} \}$ .

I try to explain this so it makes it more elementary. The set $\mbox{GL}_n(\mathbb{R})$ is the set of all $n\times n$ invertible matrices having $\mathbb{R}$ as their entries. This set is called the "general linear group". The "center" of this set are all the matrices which commute with everything else. So you need to show if a matrix commute with all invertible matrices then it must be a proportional diagnol matrix.

dravid2 · #2 $Old$ 10-25-2007, 04:21 AM

sorry, but i dont understand a thing

putnam120 · #3 $Old$ 10-28-2007, 02:40 PM

I am not sure but is this what you are saying?

Let $A$ be any $n\times n$ invertible matrix. Now if $B$ is an $n\times n$ matrix such that $AB=BA$ for any choice of $A$ then $B$ is a scalar multiple of the identity matrix. That is $B=kI$ .

ThePerfectHacker · #4 $Old$ 10-28-2007, 03:38 PM

Quote:

Originally Posted by putnam120 $View Post$

I am not sure but is this what you are saying?

Let $A$ be any $n\times n$ invertible matrix. Now if $B$ is an $n\times n$ matrix such that $AB=BA$ for any choice of $A$ then $B$ is a scalar multiple of the identity matrix. That is $B=kI$ .

Exactly. (Just note that $A,B$ are invertible).