# Linear Algebra Finding Basis for space.

1. Nov 20, 2009

### jordan123

1. The problem statement, all variables and given/known data
Says, The set W = {(x,y,z,w) : x+z=0, 2y+w=0} is a subspace of R^4. Find a basis for W, and state the dimension.

3. The attempt at a solution

What I did:

W= {(-z,-w/2,z,w): z,w are in R}
= {z(-1,0,1,0) + w(0,-1/2, 0, 1)}
= span {(-1,0,1,0), (0,-1/2,0,1)}

(-1,0,1,0), (0,-1/2,0,1) is LI, because not a scalar multiple. Basis has dimension 2.

I got this wrong.

What is wrong? I guess I could have used the other variables, x and y, instead of z,w Would that have been correct? Or should I go to my teacher and get some marks, brought me down 7 percent?

Thanks

Last edited: Nov 20, 2009
2. Nov 20, 2009

### fireb

This seems right.
I used w= -2y and got [ 0 1 0 -2] which is essentially the same basis. i guess u used y=-w/2.
And ur first base is right

3. Nov 21, 2009

### jordan123

Yeah, I just used x=-z and y=-(w/2)

4. Nov 21, 2009

### Staff: Mentor

The vectors you got are a basis for W, and the dimension of this subspace of R^4 is 2. You should definitely talk to your teacher and ask why you lost points for this problem. I got exactly the same vectors you did.

5. Nov 21, 2009

### jordan123

Awesome, thanks! I will. And the odd thing is, on the midterm it is marked right and then crossed out and marked wrong.

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