Quick Subspace Question: Understanding A = [x+1,0] as a Potential Subspace

[I Like Pi]

Homework Statement

Hey, I'm trying to figure out whether A = [x+1,0] is a subspace... I know it's probably simple but what I'm confused is that...

Homework Equations

The Attempt at a Solution

u+v must be an element of A. let u = [x1+1,0] and v=[x2+1,0]. Adding them together gives you [x1+x2+2,0], where (x1+x2) = x. Therefore it wouldn't be a subspace because you now have the form [x+2,0] which is not the same?

But if I look at it, [x1+x2+2,0] should be a subspace because you can get the same value with [x+1, 0] (let x1, x2 = -1 and let x = -1, both get 0 vector, for example)
Thanks for any help!

 

[Dick]

I think your second answer is closer. You want to show [x+1,0] is a subspace of R^2 if I understand what you are asking. What things do you have show to prove A is a subspace? And yes, [x1+x2+2,0]=[(x1-1+x2)+1,0]. So it is closed under addition if I've understood you correctly.

 

[I Like Pi]

What things do you have show to prove A is a subspace? And yes, [x1+x2+2,0]=[(x1-1+x2)+1,0]. So it is closed under addition if I've understood you correctly.

Well I know the three conditions. However, what I'm not getting is that you should have your u+v in the form [x+1,0], where x is made up of x1 + x2, or am I getting the definition wrong? In this case i have (x1 + x2 + 1)...

 

[Dick]

[x1+x2+2,0]=[(x1+1+x2)+1,0] is in the form [x+1,0] where x=x1+1+x2 as you said, isn't it? Sorry, I had a typo in the previous post.

 

[I Like Pi]

[x1+x2+2,0]=[(x1+1+x2)+1,0] is in the form [x+1,0] where x=x1+1+x2, isn't it? Sorry, I had a typo in the previous post.

Oh yeah! So you can rearrange it anyway so that you get the correct form? I thought that the initial variable had to be represented by only variables (if that makes sense)...

 

[Dick]

Oh yeah! So you can rearrange it anyway so that you get the correct form? I thought that the initial variable had to be represented by only variables (if that makes sense)...

Yeah, you are probably thinking too hard. The set of vectors [x+1,0] is really the same thing as the set of vectors [x',0]. Where x'=x+1. Where both x and x' can be any real number.