The axiom says if I multiply a scalar by a vector then rx is in the space.
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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 look like (referencing my previous post)?
So if I multiply a scalar k by vector x = (x,0) i get (kx, 0)?
Right. And does (kx,0) look like a vector (y,0) for some y?
(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?
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.
So if you add (kx, 0) + (y, 0) this is still in the space so (x, 0) is a vector space.
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.
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.
What would be the correct way to prove the last step without taking a shortcut?
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