# Homework Help: Subspace question

1. Sep 26, 2011

### Nope

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

[PLAIN]http://img683.imageshack.us/img683/4530/unledkw.jpg [Broken]
My textbook did not explain very well, but I understand this can be zero vector and it is closed under scalar multiplication.
thanks!

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 5, 2017
2. Sep 26, 2011

### Dick

[x,y]=[1,-1] satisfies x^2=y^2. So does [1,1]. Does their sum?

Last edited: Sep 26, 2011
3. Sep 26, 2011

### Nope

oh ,
(1+1, 1-1)
(2,0) not in (x,y)
am i correct?
or can you explain to me in word?

Last edited: Sep 26, 2011
4. Sep 26, 2011

### Dick

Do you agree [1,-1] and [1,1] are in your subspace? What's the sum [1,-1]+[1,1]? It's vector addition, right?

5. Sep 26, 2011

### Nope

yes, (2,0)

6. Sep 26, 2011

### Dick

Is [2,0] in your set where x^2+y^2=1?

7. Sep 26, 2011

### Dick

Right! (2,0) is not in (x,y) such that x^2=y^2. So your set is NOT closed under addition.

8. Sep 26, 2011

### Nope

not in the set, but why is x^2+y^2 equal to one?

9. Sep 26, 2011

### Dick

It isn't. I meant to write x^2=y^2 or x^2-y^2=0. My mistake.

10. Sep 26, 2011

### Nope

is there another way to prove it? like using x1 or y2

11. Sep 26, 2011

### Dick

The most direct way to prove a set is NOT closed under addition is to find two elements in the set whose sum is NOT in the set. I'm not sure why you would want another way.

12. Sep 26, 2011

### Nope

I got it, thanks!