# Finding a basis of a subspace

1. Oct 11, 2009

### PenTrik

1. The problem statement, all variables and given/known data
Find a basis for each of these subspaces of R4

All vectors that are perpendicular to (1,1,0,0) and (1,0,1,1)

2. The attempt at a solution
I'm not sure how to approach this question. The only thing I can think of is that a vector that would be perpendicular to both would be where the dot product would equal zero aye?

So then that would give me

1x1 + 1x2 = 1x1 + 1x3 + 1x4 = 0.
So in which case, I'd do RREF
[[1,1,0,0]
[1,0,1,1]]

to

[[1,0,1,1]
[0,1,-1,-1]]

I get stuck here because I'm not sure how to solve for the basis at this point.
I'm also unsure if what I did for rref was correct because vectors are usually denoted by column spaces, rather than what row spaces such as what I have done.

Last edited: Oct 11, 2009
2. Oct 11, 2009

### Dick

That's the right start. But now you have to solve the linear equations. You've got 2 equations in 4 unknowns. You should be able to express the vector (a,b,c,d) in terms of two parameters by eliminating two unknowns. Can you do that?

3. Oct 11, 2009

### PenTrik

I'm not sure if this is what I'm supposed to do but

Given that I have the RREF form, that gives me
x1 + x3 + x4 = 0
x2 - x3 - x4 = 0

Given that x1 and x2 are pivots, this gives me
x1 = -x3 - x4
x2 = x3 + x4

So, if I substitute x3 and x4 with a and b, this makes my basis the span of <-1,1,1,0> and <-1,1,0,1> ?

4. Oct 11, 2009

### Dick

Well, both of those vectors are perpendicular to the given vectors. And they are linearly independent, so sure, they are a basis. The subspace they span is the perpendicular subspace, the span itself isn't a 'basis'. Those two vectors are a basis.

5. Oct 11, 2009

### PenTrik

Ah, your help was much appreciated.