# Linear Algebra - Subspace

1. Oct 21, 2010

### lkh1986

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

Show that S = {(a+1,b,0)|a,b are real numbers} is NOT a subspace of R^3.

2. Relevant equations

3. The attempt at a solution

I take a specific counter example:

Let k = 0 inside real, and u = (1+1,1,0) inside S
ku = 0(1+1,1,0) = (0,0,0) not inside S
So, S is not a subspace.
[I can let k = 0 right? Because 0 is also a real number]

Another counter example:
Let u = v = (-1+1,1,0) inside S
u + v = (0,2,0) not inside S
So, S is not a subspace.

Which of the counter examples should I use? It seems that the first one makes more sense to me. The second one is really weird.

Thanks.

2. Oct 21, 2010

### Dick

I don't think you should use either one. (0,0,0) is the case a=(-1) b=0. (0,2,0) is a=(-1) b=2. They are both is S. In fact, S IS a subspace. It's the same as the set {(a,b,0)}.

3. Oct 21, 2010

### lkh1986

Yes, that's what I think as well. S is a subspace, because since the unknown a is a real number, so why don't I replace a+1 by another unknown, let say c, where c is also a real number.

The thing is, the answer provided says S is NOT a subspace. So it has me confused.

So, here's my steps to show S is a subspace.

(i) When a = -1, b = 0, (0,0,0) inside S.
(ii) Let vector u = (u1+1,u2,0) and vector v = (v1+1,v2,0), both inside S.
So u + v = ((u1+v1+1)+1, (u2+v2),0) also inside S.
(iii) Let k inside real and u = (u1+1,u2,0)
ku = k(u1+1,u2,0) = (ku1+k,ku2,0)= ((ku1+k-1)+1,ku2,0) also inside S.

The last part, i.e. (ku1+k,ku2,0)= ((ku1+k-1)+1,ku2,0) is a bit strange but somehow it seems okay for me.

4. Oct 21, 2010

### Dick

Your proof looks good to me. I suspect somebody just wasn't thinking clearly when they wrote the problem.

5. Oct 21, 2010

### lkh1986

Thanks very much for the help. Greatly appreciated. :)