# Show that two vectors are a basis of a subspace

1. Feb 14, 2010

### reeeky2001

Question: In R3, show that (1,-1,0) and (0,1,-1) are a basis for the subspace V={(x,y,z) $$\in$$ R3: x+y+z=0}

Attempt: By def of a basis, the vectors (1) must be linearly independent and (2) must span V.

1. For LI, show that if a(1,-1,0) + b(0,1,-1) = (0,0,0), then a=b=0.
(a,-a,0)+(0,b,-b)=(0,0,0)
(a,b-a,-b)=(0,0,0)
a=0, b-a=0, -b=0
Since the zero vector can only be expressed as a combination of a=0 and b=0, the two vectors are linearly idependent.

2. I'm not really certain how to show if the two vectors span V. If I was to guess, I'd say plug in the vectors into V, so (1,-1,0)((1,1,1)t)=0 and (0,1,-1)((1,1,1)t)=0, but I could be wrong.

2. Feb 14, 2010

### LCKurtz

To show they span the space you just need to show for any (x,y,z) with x+y+z=0 you can find a and b such that a(1,-1,0)+b(0,1,-1) = (x,y,z)

3. Feb 14, 2010

### reeeky2001

So, I show that a(1,-1,0)+b(0,1,-1)=(x,y,z) by a=x, b-a=y and -b=z. Then b=-z, a=-y-z, a=x, so x=-y-z?

Or did I misunderstand?

4. Feb 14, 2010

### LCKurtz

You have a bunch of correct equations, but it's hard to say whether you understand. Try to arrange it like this:

To get a(1,-1,0)+b(0,1,-1)=(x,y,z) you must have a = ... and b = ....
Then demonstrate those values of a and b do the job.