6 ms·

you need closure on scalar mult. if (2,3,4) is a valid int triple & 1/3 is your scalar then (2/3,1,4/3) throws you out of the group so there goes your closure.

by dxbydt 8y ago

you need closure on scalar mult. if (2,3,4) is a valid int triple & 1/3 is your scalar then (2/3,1,4/3) throws you out of the group so there goes your closure. unlike op, you don’t really need to know about fields to solve this.

zazen 8y ago
But if you don't know about fields, you could use the integers instead of the reals for your scalars, and imagine you've made a perfectly good closed vector space that way.
throwawaymath 8y ago
While that's the right idea, I'd push back against not needing to know about fields since the scalars are just field elements. If you try to define a vector space over the integers, it's more accurate to say you can't choose 1/n as a scalar, because 1/n doesn't exist in your underlying field. Your closure ends before you even get to choose the element. For students it might not be immediately obvious why that's a problem for vector spaces, but yes it does mean scalar multiplication won't be closed in the vector space. And more practically speaking, if you tried to solve a system of equations without invertible linear combinations, you'd have no linearity whatsoever. Elementary row operations likewise cease to be invertible, so matrix reduction isn't possible...the whole thing breaks down really.
- ajross 8y ago
  > I'd push back against not needing to know about fields since the scalars are just field elements. The point was that you don't need to know the jargon of "field" and the full set of implications. It's enough to know that multiplying integers by non-integer scalars can give non-integers, which means that "scalar multiplication" can produce a thing that is not a "triple of integers". So it's not a well defined vector space operation. No need for "field" or "closure" or any other jargon not in the question as posed.
  - zazen 8y ago
    > It's enough to know that multiplying integers by non-integer scalars can give non-integers Not quite: you also need to know that multiplying by integer scalars instead isn't an option. The question as posed asked as to use the "obvious" choice of scalar multiplication, and to a student who hasn't yet taken the "field" part on board, it might seem obvious to achieve closure by using the integers for scalars.
    - dxbydt 8y ago
      This isn’t quite right. When I personally learnt these things in an undergrad program in math in the US, we learnt monoids. Then we learnt semigroups. Then groups. Then abelian groups. Then vector spaces. Then on the midterm we got questions exactly like the one we are debating here - is this guy a vector space, is that guy a semigroup, is that guy abelian etc. At that point, none of us knew what a ring was, what a field was etc. In the US you learn things like cosets and Lagrange’s theorem way before you even get to fields. That’s why I said you don’t need fields. If you have (2,3,4) and want to navigate to (5,6,7) who is also in your space and you have scalar mult as your tool of choice then mult with 2 gets you to (4,6,8) but then you are stuck. Soon you realize no matter what you do you can’t navigate that space without fractions. A working definition of a space might be - you have a member in that space, you can get to every other member by just scalar mult. Addition is just freebie because you can rephrase it as bunch of scalar mults.
      - throwawaymath 8y ago
        You learned cosets and Lagrange's theorem before you learned fields? Did you take a course in abstract algebra before you took analysis? If so that seems a little unconventional to me, but I don't see another explanation since fields are taught in analysis.
        dxbydt 8y ago
        yeah i took abstract algebra before real analysis. we did hit rings & fields in algebra but by that time it was finals week & they got minimal coverage. we used Herstein, that’s the order in that book.
gizmo686 8y ago
If the triple is defined over the integers, why would you allow 1/3 as a scalar in the first place. By this logic, should R not be a vector space, as it is not closed under scalar multiplication by i, or Q not be a vector space, as it is not closed under scalar multiplication by sqrt(2)?
- skh 8y ago
  Something being sidestepped in the post you responded to is that what is being talked about is a valid algebraic object with lots of structure to it. It’s called a module which you can think of as a sort of vector space. It’s just that the scalars may not have the property that they have multiplicative inverses. (I’m deliberately focusing on rings that are integral domains for the nitpickers.). When talking about these objects you have to include the underlying scalar set. For instance the real numbers are a vector space over the rationals. They are a different vector space over the reals. They are not a vector space over the complex numbers and are not a vector space over the integers. But they are a module over the integers. But not a module over the complex numbers.
  - throwawaymath 8y ago
    Everything you've said is true, but circling back to the example given, we still can't choose a scalar 1/n for integral n. Yes the integers are a ring, and yes you can define a module over a ring which generalizes a vector space. But the point being spoken to here is that the explanation is backwards: you can't choose 1/n from Z. Therefore you can't use it as a scalar, so you'd never even break closure in the vector space. The hypothesis doesn't work before you can engage that contradiction.
    - skh 8y ago
      You are correct and I wasn't trying to criticize what you wrote. gizmo686's post (the one I responded to) indicated a sense of insight into these issues. I wanted gizmo686 to feel justified in his/her thoughts. Namely, that what we call modules are natural objects and they look at feel like vectors spaces on the surface.
ska 8y ago
You are doing this the wrong way around, as your scalars are in Z and so you can't just "pick" 1/n, any more than you can pick pi. In fact, closure under scalar multiplication is there. Pick d in Z and d(a,b,c) = (da,db,dc) is fine. The real problem is, I need an inverse. So if that exists, we have : e(da,db,dc) = (a,b,c) and e must exist in the set for (a,b,c) != (0,0,0). Now you are trying to find e that behaves like 1/d , but you've left the set - no good.

by dxbydt 8y ago

zazen 8y ago

But if you don't know about fields, you could use the integers instead of the reals for your scalars, and imagine you've made a perfectly good closed vector space that way.

throwawaymath 8y ago

While that's the right idea, I'd push back against not needing to know about fields since the scalars are just field elements. If you try to define a vector space over the integers, it's more accurate to say you can't choose 1/n as a scalar, because 1/n doesn't exist in your underlying field. Your closure ends before you even get to choose the element. For students it might not be immediately obvious why that's a problem for vector spaces, but yes it does mean scalar multiplication won't be closed in the vector space. And more practically speaking, if you tried to solve a system of equations without invertible linear combinations, you'd have no linearity whatsoever. Elementary row operations likewise cease to be invertible, so matrix reduction isn't possible...the whole thing breaks down really.

ajross 8y ago
> I'd push back against not needing to know about fields since the scalars are just field elements. The point was that you don't need to know the jargon of "field" and the full set of implications. It's enough to know that multiplying integers by non-integer scalars can give non-integers, which means that "scalar multiplication" can produce a thing that is not a "triple of integers". So it's not a well defined vector space operation. No need for "field" or "closure" or any other jargon not in the question as posed.
- zazen 8y ago
  > It's enough to know that multiplying integers by non-integer scalars can give non-integers Not quite: you also need to know that multiplying by integer scalars instead isn't an option. The question as posed asked as to use the "obvious" choice of scalar multiplication, and to a student who hasn't yet taken the "field" part on board, it might seem obvious to achieve closure by using the integers for scalars.
  - dxbydt 8y ago
    This isn’t quite right. When I personally learnt these things in an undergrad program in math in the US, we learnt monoids. Then we learnt semigroups. Then groups. Then abelian groups. Then vector spaces. Then on the midterm we got questions exactly like the one we are debating here - is this guy a vector space, is that guy a semigroup, is that guy abelian etc. At that point, none of us knew what a ring was, what a field was etc. In the US you learn things like cosets and Lagrange’s theorem way before you even get to fields. That’s why I said you don’t need fields. If you have (2,3,4) and want to navigate to (5,6,7) who is also in your space and you have scalar mult as your tool of choice then mult with 2 gets you to (4,6,8) but then you are stuck. Soon you realize no matter what you do you can’t navigate that space without fractions. A working definition of a space might be - you have a member in that space, you can get to every other member by just scalar mult. Addition is just freebie because you can rephrase it as bunch of scalar mults.
    - throwawaymath 8y ago
      You learned cosets and Lagrange's theorem before you learned fields? Did you take a course in abstract algebra before you took analysis? If so that seems a little unconventional to me, but I don't see another explanation since fields are taught in analysis.
      - dxbydt 8y ago
        yeah i took abstract algebra before real analysis. we did hit rings & fields in algebra but by that time it was finals week & they got minimal coverage. we used Herstein, that’s the order in that book.

gizmo686 8y ago

If the triple is defined over the integers, why would you allow 1/3 as a scalar in the first place. By this logic, should R not be a vector space, as it is not closed under scalar multiplication by i, or Q not be a vector space, as it is not closed under scalar multiplication by sqrt(2)?

skh 8y ago
Something being sidestepped in the post you responded to is that what is being talked about is a valid algebraic object with lots of structure to it. It’s called a module which you can think of as a sort of vector space. It’s just that the scalars may not have the property that they have multiplicative inverses. (I’m deliberately focusing on rings that are integral domains for the nitpickers.). When talking about these objects you have to include the underlying scalar set. For instance the real numbers are a vector space over the rationals. They are a different vector space over the reals. They are not a vector space over the complex numbers and are not a vector space over the integers. But they are a module over the integers. But not a module over the complex numbers.
- throwawaymath 8y ago
  Everything you've said is true, but circling back to the example given, we still can't choose a scalar 1/n for integral n. Yes the integers are a ring, and yes you can define a module over a ring which generalizes a vector space. But the point being spoken to here is that the explanation is backwards: you can't choose 1/n from Z. Therefore you can't use it as a scalar, so you'd never even break closure in the vector space. The hypothesis doesn't work before you can engage that contradiction.
  - skh 8y ago
    You are correct and I wasn't trying to criticize what you wrote. gizmo686's post (the one I responded to) indicated a sense of insight into these issues. I wanted gizmo686 to feel justified in his/her thoughts. Namely, that what we call modules are natural objects and they look at feel like vectors spaces on the surface.

ska 8y ago

You are doing this the wrong way around, as your scalars are in Z and so you can't just "pick" 1/n, any more than you can pick pi. In fact, closure under scalar multiplication is there. Pick d in Z and d(a,b,c) = (da,db,dc) is fine. The real problem is, I need an inverse. So if that exists, we have : e(da,db,dc) = (a,b,c) and e must exist in the set for (a,b,c) != (0,0,0). Now you are trying to find e that behaves like 1/d , but you've left the set - no good.