# Linear Algebra - Subspace Checks

1. Feb 7, 2013

### cowmoo32

1. The problem statement, all variables and given/known data
Determine whether the following sets form subspaces of R$^{2}$

A){($x_{1}$,$x_{2}$)$^{T}$ | $x_{1}$$x_{2}$=0}
B){($x_{1}$,$x_{2}$)$^{T}$ | $x_{1}$=3$x_{2}$}
2. Relevant equations
checks:
Does zero vector exist?
Is the space closed under addition?
Is the space closed under scalar multiplication?

3. The attempt at a solution
I know B is a subspace, but I'm not sure why. I can check the zero vector and the scalar, but I'm not 100% sure how to define closed under addition.

Also, I know A is not a subspace. Again, I know how to check for the zero vector, but I'm lost on addition and scalar multiplication, at least as a general form.

Last edited: Feb 7, 2013
2. Feb 7, 2013

### Dick

Give me some examples of vectors in A.

3. Feb 7, 2013

### cowmoo32

[1,0] [0,1] [n,0] [0,n].... one value has to be zero for x1*x2=0 I had the answers backwards. B is a subspace and A is not.

4. Feb 7, 2013

### Dick

Good! So check addition [1,0]+[0,1]=[1,1]. Is [1,1] in A?

5. Feb 7, 2013

### Dick

To prove closure under addition take [x,y] in B, i.e. x=3y and [u,v] in B, so u=3v. You want to show [x,y]+[u,v]=[x+u,y+v] is in B. Can you show that?

6. Feb 8, 2013

### HallsofIvy

Not "does the zero vector exist" but rather "is the zero vector in the subspace"
The zero vector, <0, 0>, clearly exists!

If <x, y> and <u, v> are in A then we know that xy= 0 and uv= 0. Their sum is <x+u, y+ v> so we need to look at (x+ u)(y+ v)= xy+ xv+ uy+ uv= xv+ uy. Is that necessarily 0?
If <x, y> and <u, v> are in B then we know that x= 3y and u= 3v. Their sum is <x+u, y+ v> so we need to compare x+ u and 3(y+ v). Are they necessarily equal?