# Proving a set is a subspace

1. Aug 22, 2011

### NewtonianAlch

1. The problem statement, all variables and given/known data
Show that the set:

S = {x $\in$ R$^{4}$| x = $\lambda$(2,0,1,-1)$^{T}$ for some $\lambda$ $\in$ R

is a subspace of R$^{4}$

3. The attempt at a solution

For the subspace theorem to hold, 3 conditions must be met:

1) The zero vector must exist
3) Closed under scalar multiplication

1) If $\lambda$ = 0, the vector becomes (0,0,0,0)$^{T}$ - therefore that's the zero vector.

If we define two new vectors, u and v and two scalars $\alpha$ and $\beta$ respectively.

u + v = ?

3) For closure under multiplication, isn't this obviously already closed? Heck it's being multiplied by a scalar quantity already.

2. Aug 22, 2011

### Dick

Well, if u=a*(2,0,1,-1)^T and v=b*(2,0,1,-1)^T can you show u+v has the form something*(2,0,1,-1)^T? What's the something? Sure 3) is kind of obvious, but you still have to say why.

3. Aug 22, 2011

### NewtonianAlch

The scalar would be (a + b)

So (a + b)*(2,0,1,-1)$^{T}$

Shouldn't it be (a + b)*(4,0,2,-2)$^{T}$ though ? Because if we do u + v, the vectors would add as well as the scalars.

4. Aug 22, 2011

### Dick

The vectors DO NOT add as well as the scalars. a*(2,0,1,-1)^T=(2a,0,a,-a)^T. b*(2,0,1,-1)^T=(2b,0,b,-b)^T. Add them. You don't get (a+b)*(4,0,2,-2)^T, do you? This is the distributive rule with the vector part constant.

5. Aug 22, 2011

### NewtonianAlch

Dang, just when I thought I was getting the hang of this stuff. No, it doesn't become (4,0,2,-2)$^{T}$

It becomes (2a + 2b, 0, a+b, -a - b)$^{T}$

So that would actually be (a + b)*(2,0,1,-1)$^{T}$ when you factor it out.

Interesting...I have to go over the notes again.

Thanks a lot for your help Dick.

6. Aug 23, 2011

### HallsofIvy

Staff Emeritus
No, that's not even true for numbers: av+ bv= (a+b)v whether v is a vector or a number.