Vector Space, (not calc i guess)

[sjmacewan]

We're working on vector spaces right now and this one problem is iving me a bit of trouble.

Is the following a vector space?

The set of all polynomials of the form n_2x^2 + n_1x + n_0
where n_0,n_1,n_2 \epsilon Z(integers)Now I'm pretty sure that this is going to end up NOT being a vector space, by using the properties that we were given (too vague and numerous to list here) But I'm not entirely sure why. I know that when it's an element of the REALS as opposed to the INTEGERS it is a vector space, but i just don't know where to start in this case :(

 

[dmoravec]

http://mathworld.wolfram.com/VectorSpace.html

here they give a list of 8 properties of vector spaces. Check and see if your polynomial set fits all eight properties.

 

[radou]

We're working on vector spaces right now and this one problem is iving me a bit of trouble.

Is the following a vector space?

The set of all polynomials of the form n_2x^2 + n_1 + n_0
where n_0,n_1,n_2 \epsilon Z(integers)


Now I'm pretty sure that this is going to end up NOT being a vector space, by using the properties that we were given (too vague and numerous to list here) But I'm not entirely sure why. I know that when it's an element of the REALS as opposed to the INTEGERS it is a vector space, but i just don't know where to start in this case :(

Did you mean the set of all polynomials of the form n_2x^2 + n_1x + n_0?
If it is a vector space, then, for any a \in \textbf{R} and p \in S, where S is your set of polynomials, and p some polynomial from this set, you have a p \in S, which is not true, because n_{0}, n_{1}, n_{2} stay integers only if a is an integer. Hm, though, I'm not sure if the scalar actually has to be a real number. Maybe there is something like a 'vector space over Z'?

 

[sjmacewan]

yes i did, sorry, i fixed it now. Thanks

And i have the 8 properties written down, I'm just sort of stumped when i try figuring them out. Using that link i would say that the first three are true:

1 it IS commutative
2 is DOES exhibit associativity of vector addition
3 there IS an additive identity (0)

but the forth one is the one that is starting to give me some troubles. I'm unsure of how to go about (dis)proving the existence of an additive inverse, it just seems like the negative values in the set of integers are going to throw it off...
edit: Why are you using a \in \textbf{R}? That's throwing me off...i understand what it is that you're saying, i just don't see where that term came from...

 

[dmoravec]

well... if you have n_2x^2 + n_1x + n_0 and you add in the negative of it -n_2x^2 + -n_1x + -n_0 you would end up with zero. Since -n_2x^2 + -n_1x + -n_0 would be in your set I think it proves that one.

EDIT: yes a vector space is closed so the scalar part doesn't work.

 

[dmoravec]

the scalar has to be part of a field,, could use real, complex and rational numbers, but not integers.

 

[radou]

yes i did, sorry, i fixed it now. Thanks

And i have the 8 properties written down, I'm just sort of stumped when i try figuring them out. Using that link i would say that the first three are true:

1 it IS commutative
2 is DOES exhibit associativity of vector addition
3 there IS an additive identity (0)

but the forth one is the one that is starting to give me some troubles. I'm unsure of how to go about (dis)proving the existence of an additive inverse, it just seems like the negative values in the set of integers are going to throw it off...



edit: Why are you using a \in \textbf{R}? That's throwing me off...i understand what it is that you're saying, i just don't see where that term came from...

Hm, better wait for someone like Halls on that one. As stated, I'm actually not sure if the scalar has to be a real number.

Well, regarding the 'fourth one'.. Why do you think anything would be 'thrown off'? If you choose a polynomial with negative integer coefficients, let's say p1, then -p1 = (-1) * p1 has positive coefficients, and you have your additive inverse.

 

[sjmacewan]

ok, that proves my doubts about that one, thanks. I would ever even have noticed that the result doesn't have to be in the set, that seems really weird to me.

 

[radou]

the scalar has to be part of a field,, could use real, complex and rational numbers, but not integers.

Thanks, that's what I thought.. So, since there is no scalar multiplication in this vector space, i.e. the space is not closed for scalar multiplying, it is no vector space. Other conditions don't have to be tested, unless I'm missing something big.

 

[sjmacewan]

hmmm, i don't think i see why there's no scalar multiplying, could you explain why?

 

[dmoravec]

if you have a polynomial n_2x^2 + n_1x + n_0
and you multiply by 2.5 the coefficients are no longer integers and it is not in the set. Therefore the set is not closed under scalar multiplication and it can't be a vector space.

 

[sjmacewan]

oooooooohh, right! That's because the scalar that you're multiplying by doesn't have to be in the set. Right? I was assuming that you COULDN'T multiply by a non-integer during scalar multiplication because they aren't included in the set...

 

[sjmacewan]

can i get a confirmation of what i just said maybe? Am i right in saying that the multipliers for scalar multiplication don't have to be a member of the set?

 

[dmoravec]

that is correct

 

[Data]

oooooooohh, right! That's because the scalar that you're multiplying by doesn't have to be in the set. Right? I was assuming that you COULDN'T multiply by a non-integer during scalar multiplication because they aren't included in the set...

Vector spaces are defined over fields. You have a set S a field F, and you have a closed addition on S, and you have multiplication between elements of F and elements of S which returns elements of S. So yes, unless S is a field the scalars must come from some different field.

You can define vector spaces over themselves: for example, \mathbb{R} is a vector space over the field \mathbb{R} using standard multiplication and addition of real numbers.

 

[matt grime]

Vector space over WHAT FIELD? Offer all the information first.

 

[sjmacewan]

it was th reals, but i got it all now and it's making perfect sense. Thanks to everyone who helped!