Showing that there is no such zero vector

[Mr Davis 97]

Homework Statement

Let V denote the set of ordered pairs of real numbers. If (x, y) and (w, z) are elements of V and a is a real scalar, define (x, y) + (w, z) = (x + w, yz) and a(x, y) = (ax, y). Is V a vector space?

Homework Equations

The Attempt at a Solution

Going through the vector space axioms everything is fine until we need to show that a zero vector exists.
Specifically, it is clearly evident that no such zero vector exists. However, how do we prove that no such zero vector exists, and hence it is not a vector space? For example, we can easily show that (0, 0) and (1, 1) do not work. However, this does not show that now pair will work (although clearly no pair will work).

 

[Mark44]

Homework Statement

Let V denote the set of ordered pairs of real numbers. If (x, y) and (w, z) are elements of V and a is a real scalar, define (x, y) + (w, z) = (x + y, wz) and a(x, y) = (ax, y). Is V a vector space?

Homework Equations

The Attempt at a Solution

Going through the vector space axioms everything is fine until we need to show that a zero vector exists.
Specifically, it is clearly evident that no such zero vector exists. However, how do we prove that no such zero vector exists, and hence it is not a vector space?

How is it clearly evident that no such zero vector exists?

I think that I would start by assuming that <u, v> is a zero vector in V. Then for an arbitrary vector <x, y> it must be true that <x, y> + <u, v> = <x, y>.

For example, we can easily show that (0, 0) and (1, 1) do not work. However, this does not show that now pair will work (although clearly no pair will work).

 

[PeroK]

You just have to find a proof of what is "clearly evident".

 

[Mr Davis 97]

So (x, y) + (u, v) = (x, y) ==> (x + u, vy) = (x, y)

therefore, u = 0, v = 1. So would (0, 1) be the zero vector in this vector space?

 

[PeroK]

So (x, y) + (u, v) = (x, y) ==> (x + u, vy) = (x, y)

therefore, u = 0, v = 1. So would (0, 1) be the zero vector in this vector space?

Can you confirm the rule for vector addition as this is very different from what you have in the original post.

 

[Mr Davis 97]

Oops, I did mess up in the original post, sorry. The rule for vector addition is (x, y) + (w, z) = (x + w, yz).

 

[Mark44]

So (x, y) + (u, v) = (x, y) ==> (x + u, vy) = (x, y)

therefore, u = 0, v = 1. So would (0, 1) be the zero vector in this vector space?

Looks like you found a zero vector, even though it was "clearly evident" that no such vector exists.

 

[Mr Davis 97]

Yes yes, I have learned my lesson. However, now let's look at if an inverse vector always exists.

$(x, y) + (u, v) = (0, 1) \implies (x + u, yv) = (0, 1) \implies u = -x,~~ v = 1/y$

Does this show that an inverse vector always exists?

 

[Mark44]

Yes yes, I have learned my lesson. However, now let's look at if an inverse vector always exists.

$(x, y) + (u, v) = (0, 1) \implies (x + u, yv) = (0, 1) \implies u = -x,~~ v = 1/y$

Does this show that an inverse vector always exists?

No, it doesn't. There are lots of (i.e., an infinite number of) vectors that don't have additive inverses.

 

[Mr Davis 97]

No, it doesn't. There are lots of (i.e., an infinite number of) vectors that don't have additive inverses.

So any vector of the form (a, 0) doesn't have an inverse, so thus this is not a vector space?

 

[Mark44]

So any vector of the form (a, 0) doesn't have an inverse, so thus this is not a vector space?

What do you think? Does this particular axiom allow for exceptions?

 

[Mr Davis 97]

What do you think? Does this particular axiom allow for exceptions?

No, since there must exist an inverse for each element in the vector space, and obviously (a, 0) is an element (as it is a 2-tuple).

 

[Mark44]

No, since there must exist an inverse for each element in the vector space, and obviously (a, 0) is an element (as it is a 2-tuple).

There you go. And since V fails this axiom, it is not a vector space. It's possible that there are other axioms among the scalar multiplication bunch that fail, but since you found that some elements of V don't have additive inverses, that's enough to say that V isn't a vector space. It might be useful, though, to see if some of the scalar multiplcation axioms fail -- might be good practice for a test.