# LA: Finding a basis

1. Jul 7, 2011

### Haystack

1. The problem statement, all variables and given/known data

Let S = [x y z w] $\in$ $$R^4$$ , 2x-y+2z+w=0 and 3x-z-w=0

Find a basis for S.

2. Relevant equations

3. The attempt at a solution

I started by putting the system into reduced row form:

[2 -1 2 1]
[3 0 -1 -1]

[2 -1 2 1]
[0 3 -8 -5]

[6 0 -2 -2]
[0 3 -8 -5]

[1 0 -1/3 -1/3]
[0 1 -8/3 -5/3]

Now have:

x - 1/3z - 1/3w = 0
y - 8/3z - 5/3w = 0

Letting z = s, and w = t, we get:

x = 1/3s + 1/3t
y = 8/3s + 5/3t
z = s
w = t

And this gives:

s[1/3 8/3 1 0] and t[1/3 5/3 0 1]

Where the basis vectors are:

[1/3 8/3 1 0] and [1/3 5/3 0 1]

and are linearly independent.

Did I do this correctly? I'm really struggling with these concepts and I feel like I'm missing something. Thanks in advance.

2. Jul 7, 2011

### LCKurtz

I didn't check all your arithmetic, but assuming your numbers are OK, yes. That is exactly how to work the problem.

3. Jul 8, 2011

### HallsofIvy

Personally, I would prefer this:
2x-y+2z+w=0 and 3x-z-w=0

From the second equation, w= z- 3x
Putting that into the first equation, 2x- y+ 2z+ z- 3x= -x- y+ 3z= 0 so y= 3z- x.
Having solved for y and w in terms of x and z,
[x, y, z, w]= [x, 3z- x, z, z- 3x]= [x, -x. 0, -3x]+ [0, 3z, z, z]= x[1, -1, 0, -3]+ z[0, 3, 1, 1]
so a basis is {[1, -1, 0, -3], [0, 3, 1, 1]} which is essentially what you have, multiplied by 3.

4. Jul 8, 2011

### Haystack

Thanks for the replies. Although, I'm not sure I understand your basis HallfofIvy. I can't seem to find how it's related to the one I ended up with. I understand how you got there with the substitution, but is it just another basis for S?

5. Jul 9, 2011

### HallsofIvy

It is just your method using basic algebra to solve the equations rather than "row-reduction".

Every member of this subspace is of the form <x, y, z, w> and we must have
2x-y+2z+w=0 and 3x-z-w=0

I just solved for y and w in terms of x and z and then used x and z as multipliers where you solved for x and y and used z and w as multipliers.