# Finding a basis for a subspace

1. Jan 19, 2008

### Niles

[SOLVED] Finding a basis for a subspace

1. The problem statement, all variables and given/known data
We have a subspace U in R^3 defined by:

U = {(x_1 , x_2 ; x_3) | x_1 + 2*x_2 + x_3 = 0 }.

Find a basis for U.

3. The attempt at a solution

We have the following homogeneous system:

(1 2 1 | 0).

From this I find the solution to be written as a*(2,1,0)^T + b(-1,0,1)^T, where a and b are arbitrary constants. From this I find three linearly independant vectors.

Am I correct?

2. Jan 19, 2008

### Mathdope

(2,1,0) isn't a solution (but it's close). Also, if your subspace contains 3 linearly independent vectors then it's all of R^3. This is obviously not the case right?

3. Jan 19, 2008

### Niles

The solution was (-2,1,0).

Yeah, I need a condition too so it is just the subspace where x_1 + 2*x_2 + x_3 = 0 - but how do I include that?

Last edited: Jan 19, 2008
4. Jan 19, 2008

### Mathdope

Well, in the answer you gave I see two vectors (one with the correction you mentioned) that appear to be linearly independent. They must be a basis of the solution space, right?

Last edited: Jan 19, 2008
5. Jan 19, 2008

### Dick

You solved the homogeneous system and found the solution a*(-2,1,0)+b*(-1,0,1). That means (-2,1,0) and (-1,0,1) are in the subspace and they span it. They are also independent. 1) Anything in their span already automatically solves x1+2x2+x3=0, since that relation is linear. 2) What do you mean 'three' independent vectors??

6. Jan 19, 2008

### Niles

They must be basis for U. I only have two linearly independant vectors so far (not three as I wrote originally, sorry) - I need three linearly independant vectors that satisfy x_1 + 2*x_2 + x_3 = 0, right? How do I find the third?

7. Jan 19, 2008

### Niles

Doesn't U have to be spanned by three (3) vectors, since it is a subspace of R^3? If not, why?

8. Jan 19, 2008

### Mathdope

You can't find a third unless the solution space is all of R^3. You have two independent vectors that solve the equation so they must span the solution space.

9. Jan 19, 2008

### Niles

Ok, thanks to both of you.