Subspace Question: Why Not Closed Under Addition?

[Nope]

Homework Statement

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

Homework Equations

The Attempt at a Solution

 

[Dick]

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

 

[Nope]

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

 

[Dick]

can you show me?, because I don't get the closed under addition part at all..

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?

 

[Nope]

yes, (2,0)

 

[Dick]

yes, (2,0)

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

 

[Dick]

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

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

 

[Nope]

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

 

[Dick]

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

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

 

[Nope]

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

 

[Dick]

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

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.

 

[Nope]

I got it, thanks!