# Find the basis and it's dimension

• May 3rd 2009, 10:22 AM
orendacl
Find the basis and it's dimension
Find a basis for

W = {|a b|
_____|c d|
:a+d=b+c} and state the dimension of W.

Showing my work:
I thought one way to solve this would be to exhange values in the matrix:

Then establish our vector set. Determine if the set spans for W. We then show linear independence...and therefore our basis.

Problem is:
I set up matrices like this:

a |1 0| +b|0 0| c|0 0| +d|0 1|
...|1 0|....|0 0|...|0 0|....|0 1|

I then took a+b=c+d
and said: a-c=-d-b

Anyway, some years ago I did some similar math and thought of this:

Wy not replace
|a b|
|c d|

with:
|a a|
|e e|

I used e is epsilon. Please feel free to trash my "thought" process and do not follow my example to solve it as it need not be solved this way, that is: I am just trying what I remember from previous years.

You could say I am kind of grapsing at straws here guys (and gals) and need some help!

Thanks.
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• May 3rd 2009, 11:01 AM
HallsofIvy
Okay, here's the "trashing": Your "all 0" matrices make no sense. The "0 matrix" can't be a basis vector since any set of matrices containing the 0 matrix can't be independent.

a+ d= b+ c gives one condition on the four numbers. You can solve for one of the numbers in terms of the other three. For example, you can write d= b+ c- a. I have no idea what you mean by "e= epsilon". What is epsilon?

Here's how I would do the problem:

Taking a= 1, b= c= 0, d= -1 so $\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}$ is in the space.

Taking b= 1, a= c= 0, d= 1 so $\begin{bmatrix}0 & 1 \\ 0 & 1\end{bmatrix}$ is in the space.

Taking c= 1, a= b= 0, d= 1 so $\begin{bmatrix}0 & 0 \\ 1 & 1\end{bmatrix}$ is in the space.

The space has dimension 3 and those three matrices form a basis.

Of course, you could also solve for b, c, or d and get a different, but equally correct, basis.