Showing that there is no such zero vector

In summary, the conversation discusses whether the set V of ordered pairs of real numbers, defined by certain vector addition and scalar multiplication operations, is a vector space. It is shown that V fails the axiom of having a zero vector, as there are elements of V that do not have additive inverses. Therefore, V is not a vector space. It is suggested to check if other scalar multiplication axioms also fail for V.
  • #1
Mr Davis 97
1,462
44

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).
 
Last edited:
Physics news on Phys.org
  • #2
Mr Davis 97 said:

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>.
Mr Davis 97 said:
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).
 
  • #3
You just have to find a proof of what is "clearly evident".
 
  • #4
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?
 
  • #5
Mr Davis 97 said:
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.
 
  • #6
Oops, I did mess up in the original post, sorry. The rule for vector addition is (x, y) + (w, z) = (x + w, yz).
 
  • #7
Mr Davis 97 said:
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.
 
  • #8
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?
 
  • #9
Mr Davis 97 said:
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.
 
  • #10
Mark44 said:
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?
 
  • #11
Mr Davis 97 said:
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?
 
  • #12
Mark44 said:
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).
 
  • #13
Mr Davis 97 said:
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.
 

Related to Showing that there is no such zero vector

1. What does it mean to show that there is no such zero vector?

Showing that there is no such zero vector means proving that a vector space does not contain a vector that has a magnitude of zero and a direction of zero. In other words, there is no single vector that can be added to any other vector and result in the same vector.

2. How do you prove that there is no zero vector?

To prove that there is no zero vector, you can use a proof by contradiction. Assume that there is a zero vector in the vector space and then show that this leads to a contradiction. This means that the initial assumption was incorrect and there is no zero vector in the vector space.

3. Why is it important to show that there is no zero vector?

Showing that there is no zero vector is important because it is a fundamental property of vector spaces. It helps to define the structure and properties of a vector space and allows for the development of more complex mathematical concepts. It also helps to distinguish vector spaces from other mathematical structures.

4. Can a vector space have more than one zero vector?

No, a vector space can only have one zero vector. This is because the zero vector is defined as the unique vector that, when added to any other vector, results in the same vector. If there were multiple zero vectors, this property would not hold true.

5. How does showing that there is no zero vector relate to linear independence?

Showing that there is no zero vector is related to linear independence because a set of vectors is linearly independent if and only if the only solution to the linear combination that results in the zero vector is when all the coefficients are equal to zero. If there is a zero vector in the set, this property would not hold true and the vectors would not be linearly independent.

Similar threads

  • Calculus and Beyond Homework Help
Replies
0
Views
537
  • Calculus and Beyond Homework Help
Replies
6
Views
883
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
396
  • Calculus and Beyond Homework Help
Replies
26
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
24
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
686
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
Back
Top