Problem 38

ThePerfectHacker · #1 $Old$ September 24th, 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 $Old$ October 25th, 2007, 04:21 AM

sorry, but i dont understand a thing

putnam120 · #3 $Old$ October 28th, 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 $Old$ October 28th, 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).

halbard · #5 $Old$ March 28th, 2009, 05:18 PM

OK, let's run this past the General and see if it salutes.

Suppose that $B$ commutes with all $A\in\mbox{GL}_n(\mathbb{R})$ . Assume that $B\neq0$ and choose any $u\in\mathbb{R}^n$ with $Bu\neq0$ . Clearly $u\neq0$ .

Claim: The set $\{u, Bu\}$ is linearly dependent.

Assume for contradiction purposes that $u$ and $Bu$ are linearly independent. Then so are $u$ and $u-Bu$ .

The linearly independent sets $\{u, Bu\}\mbox{ and }\{u, u-Bu\}$ can each be extended to a basis.

Therefore there exists an invertible matrix $A$ with $Au=u$ and $A(Bu)=u-Bu$ .

But then $u-Bu=ABu=BAu=Bu$ which shows that $u$ and $Bu$ are linearly dependent after all, contradiction.

Therefore $\{u, Bu\}$ is l.d. and there must exist a non-zero scalar $k$ such that $Bu=ku$ .

For any non-zero $v\in\mathbb{R}^n$ there is an invertible matrix $C$ such that $v=Cu$ .

Therefore $Bv=B(Cu)=BCu=CBu=C(ku)=kCu=kv$ .

Thus $Bv=kv\mbox{ for all non-zero }v\in\mathbb{R}^n$ which means that $B=kI$ .

Bruno J. · #6 $Old$ June 16th, 2009, 11:15 PM

Let $I_{(j)}$ be the identity matrix with row j multiplied by 2. Then $I_{(j)}A$ is A with row j multiplied by 2, and $AI_{(j)}$ is A with column j multiplied by 2. From

$I_{(j)}A=AI_{(j)}$

we conclude that A has a diagonal entry and zeroes everywhere else on the jth row and the jth column. In this fashion we conclude that A is diagonal.

It is then easy to show that the diagonal entries are all equal.