# Homework Help: Determine if the space is a subspace testing both closure axioms

1. Jun 9, 2009

### jackxxny

1. The problem statement, all variables and given/known data
determine if the space is a subspace testing both closure axioms.

in R^2 the set of vectors (a,b) where ab=0

2. Relevant equations

3. The attempt at a solution

i just used the sum and product which are the closure axioms.

But at the end how do you tell if the resulting vector is a subspace?

(a,b) + (c,d) = (a+c, b+d)

then ???

x=constant

x(a,b) = (xa,xb) then abx^2 =0

then?????

2. Jun 9, 2009

### CompuChip

Re: Subspace

How about you look at (1, 1) and (1, -1) ?

3. Jun 9, 2009

### Cyosis

Re: Subspace

If (a,b) and (c,d) are in the set. For it to be a subspace (a+c,b+d) needs to be in the set. To be in the set (a+c).(b+d)=0 needs to be satisfied. Proof that the equality holds. Then do the same for scalar multiplication. If both equalities hold you can conclude it is a subspace.

Last edited: Jun 9, 2009
4. Jun 9, 2009

### jackxxny

Re: Subspace

so if i set
a=1 , b=1 , c=1 , d=-1

i will have 1+1-1-1 = 0

Is that the way to prove it?

is isn't the roduct beetween a and b has to be equal to 0. I understand by that that one of the 2 terms has to be 0 right?

5. Jun 9, 2009

### jackxxny

Re: Subspace

then if i set (0,12) and (45,0) both are in the set but when i add them

(45,12) they are not in the space anymore right?

45*12 is not equal to 0 right?

6. Jun 9, 2009

### Cyosis

Re: Subspace

What you say in post 4 is not correct. The sum of both may be zero but separately they aren't even in the subspace, since (1,1), 1*1=1!=0.

What you say in post 5 is correct, because 0*12=45*0=0.

That said is this really the problem? As you've stated the problem it could be read that a and b are vectors in R^2 and they are a subspace when a.b=0 (inner product). I think CompuChip interpreted your question like this.

Last edited: Jun 9, 2009
7. Jun 9, 2009

### Pengwuino

Re: Subspace

Yes, that is an example of how it works. I'm not sure why one would use (1,1), (1,-1). They aren't in the subspace.

As Cyosis began, you can use two arbitrary vectors, X=(a,b) and Y=(c,d) where ab=0 and cd=0. However, investigating closure under addition, X+Y = (a+c,b+d). Let's let e=a+c, f=b+d meaning X + Y = (e,f). For closure to be true, ef=0. Plug back in what e and f are and you get (a+c)(b+d). Does this equal 0? That is, is it in the subspace for arbitrary a,b,c,d?

8. Jun 9, 2009

### jackxxny

Re: Subspace

so under addition does not work.... by it does work under product....

in the end is not a subspace.....

it's so good learning the material like that.....thanks to all of you....

9. Jun 11, 2009

### CompuChip

Re: Subspace

Ah, I must apologize.
I misread the condition of "ab equal to 0" as "ab not equal to 0".