Thread: Determining a set is a vector space

1. The axiom says if I multiply a scalar by a vector then rx is in the space.

2. Right, but that's what you've got to prove in order to show that you've got a vector space. You're going to have to look at the specifics of your problem. What does $r\mathbf{x}$ look like (referencing my previous post)?

3. So if I multiply a scalar k by vector x = (x,0) i get (kx, 0)?

4. Right. And does (kx,0) look like a vector (y,0) for some y?

5. (y,0) does not look like (kx, 0) because y is not multiplied by a scalar.

Do I not need to multiply y by a scalar?

6. y is the same kind of number (a real number) as k, and it's the same kind of number as x. So why not let y=kx? Then it does, indeed, look like a vector in your set.

7. So if you add (kx, 0) + (y, 0) this is still in the space so (x, 0) is a vector space.

8. Originally Posted by evant8950
I understand your example, but how do I see if (x,0) is a vector space of R^2 if there is no condition like in your example such as v1+v2=0.

Thanks
Actually, there is such a condition, <v1, v2> is in this set if and only if v2= 0.

9. Originally Posted by evant8950
So if you add (kx, 0) + (y, 0) this is still in the space so (x, 0) is a vector space.
This sounds like you're using a shortcut and changing the method by which you're showing the result. Technically, I think you're right.

10. What would be the correct way to prove the last step without taking a shortcut?

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determining if something is a vector space

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