# Problem 38

• Sep 24th 2007, 05:18 PM
ThePerfectHacker
Problem 38
Show the center of the general linear group are the proportional diagnol matrices.
Meaning, $\displaystyle Z(\mbox{GL}_n(\mathbb{R})) = \{ k I| k\in \mathbb{R}^{*} \mbox{ and }I \mbox{ identity matrix} \}$.

I try to explain this so it makes it more elementary. The set $\displaystyle \mbox{GL}_n(\mathbb{R})$ is the set of all $\displaystyle n\times n$ invertible matrices having $\displaystyle \mathbb{R}$ 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.
• Oct 25th 2007, 02:21 AM
dravid2
sorry, but i dont understand a thing
• Oct 28th 2007, 12:40 PM
putnam120
I am not sure but is this what you are saying?

Let $\displaystyle A$ be any $\displaystyle n\times n$ invertible matrix. Now if $\displaystyle B$ is an $\displaystyle n\times n$ matrix such that $\displaystyle AB=BA$ for any choice of $\displaystyle A$ then $\displaystyle B$ is a scalar multiple of the identity matrix. That is $\displaystyle B=kI$.
• Oct 28th 2007, 01:38 PM
ThePerfectHacker
Quote:

Originally Posted by putnam120
I am not sure but is this what you are saying?

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

Exactly. (Just note that $\displaystyle A,B$ are invertible).
• Mar 28th 2009, 03:18 PM
halbard
OK, let's run this past the General and see if it salutes.

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

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

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

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

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

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

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

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

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

Thus $\displaystyle Bv=kv\mbox{ for all non-zero }v\in\mathbb{R}^n$ which means that $\displaystyle B=kI$.
• Jun 16th 2009, 09:15 PM
Bruno J.
Let $\displaystyle I_{(j)}$ be the identity matrix with row j multiplied by 2. Then $\displaystyle I_{(j)}A$ is A with row j multiplied by 2, and $\displaystyle AI_{(j)}$ is A with column j multiplied by 2. From

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

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.