# Subspace of R2 involving absolute value

1. Feb 10, 2013

### Mdhiggenz

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

{(x1,x2)T| |x2|=|x2|}

So my first thought is we would have to check for both cases (x1,x1) and (-x1,-x1)

a=(x1,x1)T b=(v1,v1)T

βa=(βx1,βx1)T for the case where a<0 βa(-βx1,-βx1)T

thus it is closed under scalar multiplication.

a+b=(x1+v1,x1+v1) for the case where a and b<0 [-(x1+v1,x1+v1)]T

I would think this would be a subset, however the book says this is false. Where did I go wrong?

2. Relevant equations

3. The attempt at a solution

2. Feb 10, 2013

### Dick

You are basically ignoring the absolute value. Are (1,1) and (1,-1) in your subspace? How about the sum?

3. Feb 10, 2013

### Mdhiggenz

No because a+b would give you (2,0) and x1 and x2 have to be equal?

4. Feb 10, 2013

### Dick

No, because |2| is not equal |0|. You don't have to have x1=x2 to be in the subspace. x1=(-x2) also works. But (2,0) doesn't fit either one.

5. Feb 10, 2013

### SammyS

Staff Emeritus
Did you really mean |x2| = |x2| ?

6. Feb 10, 2013

### Mdhiggenz

Oops that was a typo it was suppose to be |x1|=|x2|.

Also Dick, I'm a bit confused on why you chose (1,1) (1,-1)

7. Feb 10, 2013

### Dick

I was just trying to find two vectors in your set which didn't sum to be something in the set. Choosing one that satisfies x1=x2 and another that satisfies x1=(-x2) will do it. Nothing special about the 1's.

8. Feb 10, 2013

### Mdhiggenz

But they must be equal? for instance we could use (2,2) and (2,-2) or (-2,2) and (2,2)

9. Feb 10, 2013

### Dick

(1,1) and (3,-3) work fine too. Their sum isn't in the set either.