Vector Spaces: Provide a counter example to disprove

[boings]

Demonstrate with the help of a counter-example why the following is not a vector space.

1. A= ((x,y) \ni R^{2}/ x\geq0)

I have many more questions like this, but since I cannot get the first one I think I might have a chance if I understand it.

As far as an attempt at an answer, I can only grasp that vector space must be commutative and associative and I can guess that this isn't the case because as y is negative x may become negative as well which would be outside the vector space, but how might I say that if if there are no determined operations on the set of (x,y) variables?

thank you!

 

[oli4]

Hi boings
You should be able to multiply any element of your vector space by a scalar and get a new element of the same vector space
Can you think of some scalars that clearly break this rule ?
Cheers...

 

[boings]

Ok, so that would be any negative scalar which would break the rule of x\geq0?

 

[oli4]

Yes :)
multiply any element of your original 'candidate vector space' by a negative number and it clearly doesn't belong to said 'original candidate'
therefore it can't be a valid candidate :)

 

[micromass]

Some remarks (and I mean them to be constructive):

Demonstrate with the help of a counter-example why the following is not a vector space.

1. A= ((x,y) \ni R^{2}/ x\geq0)

You have the wrong math symbol here. It should be \in instead of \ni. The LaTeX code is \in

I have many more questions like this, but since I cannot get the first one I think I might have a chance if I understand it.

As far as an attempt at an answer, I can only grasp that vector space must be commutative and associative

A vector space being commutative and associate makes no sense. It is the operation + on the vector space that is commutative and associative.

 

[boings]

You have the wrong math symbol here. It should be \in instead of \ni. The LaTeX code is \in

Hah, yeah I thought that was wrong, but couldn't find the right one, thanks!

So am I correct in saying that the scalar operation on a vector space is commutative and associative?

thank you both

 

[micromass]

So am I correct in saying that the scalar operation on a vector space is commutative and associative?

Very good question.

Hmmm. I know what you mean and you are correct. But the words commutative and associative are not typically used for the scalar operation.
The scalar operation is defined as

\mathbb{R}\times V\rightarrow V:(\alpha,v)\rightarrow \alpha\cdot v

Associativity would mean that \alpha\cdot (\beta \cdot v)= (\alpha \beta)\cdot v. This law certainly holds true, but we don't use the word associative for this. The reason that we don't is that \alpha,\beta and v belong to different sets. We usually only talk about associative if the sets are the same. The word to describe the law that I have seen are "mixed associativity" and "compatibility of scalar multiplication with field multiplication". Of course, there is nothing wrong with thinking of associativity and many authors do this.

Commutativity is also a bit tricky. You want to say that \alpha\cdot v=v\cdot \alpha. But if we are rigorous, then I have to remark that v\cdot \alpha is not even defined. We defined scalar multiplication such that the scalar is always on the left of the vector. We did not define anything such that the scalar is right of the vector. Of course, we can just define \alpha\cdot v=v\cdot \alpha to be true (and many authors indeed do this). But I would still be careful in calling this relation "commutativity" (again, because \alpha and v belong to different sets).

 

[boings]

thanks a lot, that's a great reply.

So if commutativity in this case were to be proven, has it been referenced also as "mixed commutativity"?

 

[micromass]

thanks a lot, that's a great reply.

So if commutativity in this case were to be proven, has it been referenced also as "mixed commutativity"?

Commutativity can't be proven, it needs to be defined. Remember, when you have a vector space, then the only thing that is defined is something of the form \alpha\cdot v. A product where the scalar is on the right like v\cdot \alpha is not in general defined. If you want \alpha\cdot v=v\cdot \alpha, then you will need to define this to be true as the right-hand side does not make any sense until you defined it.

But anyway, even if you have \alpha \cdot v=v\cdot \alpha, then there is nothing stopping you to call this commutativity. However, I have not yet heard the term commutativity for this relationship. I don't think many authors consider this to be an interesting relationship for vector spaces in the first place, since it is just a definition and not really a very useful definition.

 

[boings]

You're right, I'm going to take your word and their word on this and not lose any sleep over whether it will be useful to me :)

thanks