sorry, but i dont understand a thing

Results 1 to 6 of 6

- September 24th 2007, 06:18 PM #1

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

## Problem 38

Show the center of the general linear group are the proportional diagnol matrices.

Meaning, .

I try to explain this so it makes it more elementary. The set is the set of all invertible matrices having 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.

- October 25th 2007, 03:21 AM #2

- Joined
- Oct 2007
- Posts
- 27

- October 28th 2007, 01:40 PM #3

- Joined
- Nov 2006
- From
- Florida
- Posts
- 228

- October 28th 2007, 02:38 PM #4

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

- March 28th 2009, 04:18 PM #5

- Joined
- Mar 2009
- Posts
- 91

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

Suppose that commutes with all . Assume that and choose any with . Clearly .

Claim: The set is linearly dependent.

Assume for contradiction purposes that and are linearly independent. Then so are and .

The linearly independent sets can each be extended to a basis.

Therefore there exists an invertible matrix with and .

But then which shows that and are linearly dependent after all, contradiction.

Therefore is l.d. and there must exist a non-zero scalar such that .

For any non-zero there is an invertible matrix such that .

Therefore .

Thus which means that .

- June 16th 2009, 10:15 PM #6
Let be the identity matrix with row j multiplied by 2. Then is A with row j multiplied by 2, and is A with column j multiplied by 2. From

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.