6 ms·

As a bumbling idiot who never did university level math, I don’t understand the question or the answer. (I did understand the probability one and knew the right

As a bumbling idiot who never did university level math, I don’t understand the question or the answer. (I did understand the probability one and knew the right answer to that at least.) I’m trying to catch up. Would you be kind enough to explain it?

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.
throwawaymath 8y ago
Sure. Operations in applied linear algebra - such as matrix multiplication and solving systems of linear equations - are formalized by the theory of vector spaces, much like calculus is formalized through the theory of analysis. Vector spaces are algebraic structures which axiomatize the linearity you need to carry out these operations. If you can establish your equations exist in a vector space, you can prove that they admit linear relations and are thus solvable as linear systems. More precisely, a vector space V is any set S defined over a field F which is closed under both vector addition and scalar multiplication, where the elements of S are called vectors and the elements of the underlying field F are called scalars. The term "closed" means that for every pair of vectors x, y in V there exists a vector x + y in V, and for every scalar c in F and vector x in V there exists a vector cx in V. There are eight axioms in total, for things like associativity and commutativity, but those aren't germane to this particular example. What's important is that vector spaces are what allow you to form linear combinations of things, which is the scaffolding you need to prove things like linear dependence and independence; whether or not a system of linear equations has no solutions, one solution or infinitely many solutions, etc. Fields are the algebraic structures which formalize the elementary arithmetic you're already familiar with over sets like the the complex numbers, the real numbers, the rationals, etc. They are sets which are closed under "regular" addition and multiplication. Notably, integers do not comprise a field because integers do not have multiplicative inverses. Multiplicative inverses are the axiomatic way of establishing that in any field, division must be possible. So concretely, there is no multiplicative inverse 1/n for any integer n. There is in the set of rationals, but not the set of integers. Therefore integers are not closed under multiplication, and they cannot comprise a field. Since the integers do not comprise a field, you cannot define a vector space over the integers, because the scalars used to define scalar multiplication in vector spaces are just elements of the underlying field. If you try to define a vector space over a set without multiplicative closure, the vector space cannot be closed under scalar multiplication. Among other things, linear combinations stop being invertible (or even possible in general), and linear relations don't exist. So circling back to the specific question: it's asking which of the given sets comprises a vector space. You can make all kinds of abstract vector spaces (e.g. the set of all polynomials over a field, the set of all polynomials with degree at most n over a field, the set of all continuous functions, etc). But if you stick with the definition of a vector space, you don't need to tediously test each of the given sets for the eight axioms. You just have to remember the integers don't comprise a field, so the set of all triples of integers can't be a vector space either. Hopefully that's clear, let me know if you'd like me to clarify anything.
archgoon 8y ago
A vector space is a set of elements that are closed under addition and scalar multiplication (multiplying by a real or complex number). The precise definition of what we mean by that is covered by the vector space axioms. http://mathworld.wolfram.com/VectorSpace.html http://mathworld.wolfram.com/VectorSpace.html ' A. The set of all complex numbers. You can add two complex numbers, and multiply them by a real (or a complex number in this case) and still have a complex number. So it's a vector space. ' B. The set of all functions from (0,1) to \mathbb R that are twice differentiable. Two twice differentiable functions can be added together to yield a twice differentiable function (f(x)+g(x))''=f''(x)+g''(x) and scaled by a scalar (cf(x))''=c(f''(x)) ' C. The set of all polynomials in x with real coefficients that have x^2+x+1 as a factor. Given two polynomials P,Q, which have x^2+x+1 as a factor they can be expressed as P(x) = f(x)(x^2+x+1) Q(x) = g(x)(x^2+x+1) So P(x)+Q(x) = (f(x)+g(x))(x^2+x+1) And of course c*P(x) = cf(x)(x^2+x+1) Which is also a polynomial with real coefficients. ' D. The set of all triples (a,b,c) of integers. Almost works, except multiplying by a scalar would yield a tuple of real numbers. Kinda silly. In programming, if you attempted to write a function function List<Int> multiply(List<Int> list, Real c) { return list.map((x) => x * c); } You'd get a compiler exception since the return type wouldn't match (or you'd get some warning about shortening the precision). ' E. The set of all sequences (x_1,\dots,x_n)\in\mathbb R^n such that x_1+\dots+x_n=0 and x_1+2x_2+\dots+nx_n=0. Adding two sequences (x_1,...x_n) + (y_1...,y_n) gets you (x_1+y_1...,x_n+y_n) which plugging into the above two equations, and rearranging, will show that you'd get (x1+y1) + (x2+y2)... + (xn+yn) = (x1+x2+..xn) + (y1+...+yn) = (0) + (0) and the same for the second equation. Also, scalar scaling works fine. ' So only D has any issues. The short explanation as to why we insist on scalars being real or complex is that a major goal of linear algebra is to provide ways of solving equations. And you really want to perform division to solve equations, which real and complex numbers let you do. Integers aren't closed over division, so they aren't good for solving linear equations. For example, if your vector space is integer triplets then 2 * x = (1,0,0) Wouldn't have any solutions in the set of integer triplets. This is a linear equation, and the goal of linear algebra is to provide solutions, so it's better to have the framework yield the answer of x = (.5,0,0) and then say "Oh, the answer lies outside of the original set, so it wasn't a vector space to start with." Well, that's kinda the extrinsic view of it, the intrinsic view would throw an exception :).

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.

throwawaymath 8y ago

Sure. Operations in applied linear algebra - such as matrix multiplication and solving systems of linear equations - are formalized by the theory of vector spaces, much like calculus is formalized through the theory of analysis. Vector spaces are algebraic structures which axiomatize the linearity you need to carry out these operations. If you can establish your equations exist in a vector space, you can prove that they admit linear relations and are thus solvable as linear systems. More precisely, a vector space V is any set S defined over a field F which is closed under both vector addition and scalar multiplication, where the elements of S are called vectors and the elements of the underlying field F are called scalars. The term "closed" means that for every pair of vectors x, y in V there exists a vector x + y in V, and for every scalar c in F and vector x in V there exists a vector cx in V. There are eight axioms in total, for things like associativity and commutativity, but those aren't germane to this particular example. What's important is that vector spaces are what allow you to form linear combinations of things, which is the scaffolding you need to prove things like linear dependence and independence; whether or not a system of linear equations has no solutions, one solution or infinitely many solutions, etc. Fields are the algebraic structures which formalize the elementary arithmetic you're already familiar with over sets like the the complex numbers, the real numbers, the rationals, etc. They are sets which are closed under "regular" addition and multiplication. Notably, integers do not comprise a field because integers do not have multiplicative inverses. Multiplicative inverses are the axiomatic way of establishing that in any field, division must be possible. So concretely, there is no multiplicative inverse 1/n for any integer n. There is in the set of rationals, but not the set of integers. Therefore integers are not closed under multiplication, and they cannot comprise a field. Since the integers do not comprise a field, you cannot define a vector space over the integers, because the scalars used to define scalar multiplication in vector spaces are just elements of the underlying field. If you try to define a vector space over a set without multiplicative closure, the vector space cannot be closed under scalar multiplication. Among other things, linear combinations stop being invertible (or even possible in general), and linear relations don't exist. So circling back to the specific question: it's asking which of the given sets comprises a vector space. You can make all kinds of abstract vector spaces (e.g. the set of all polynomials over a field, the set of all polynomials with degree at most n over a field, the set of all continuous functions, etc). But if you stick with the definition of a vector space, you don't need to tediously test each of the given sets for the eight axioms. You just have to remember the integers don't comprise a field, so the set of all triples of integers can't be a vector space either. Hopefully that's clear, let me know if you'd like me to clarify anything.

archgoon 8y ago

A vector space is a set of elements that are closed under addition and scalar multiplication (multiplying by a real or complex number). The precise definition of what we mean by that is covered by the vector space axioms. http://mathworld.wolfram.com/VectorSpace.html http://mathworld.wolfram.com/VectorSpace.html ' A. The set of all complex numbers. You can add two complex numbers, and multiply them by a real (or a complex number in this case) and still have a complex number. So it's a vector space. ' B. The set of all functions from (0,1) to \mathbb R that are twice differentiable. Two twice differentiable functions can be added together to yield a twice differentiable function (f(x)+g(x))''=f''(x)+g''(x) and scaled by a scalar (cf(x))''=c(f''(x)) ' C. The set of all polynomials in x with real coefficients that have x^2+x+1 as a factor. Given two polynomials P,Q, which have x^2+x+1 as a factor they can be expressed as P(x) = f(x)(x^2+x+1) Q(x) = g(x)(x^2+x+1) So P(x)+Q(x) = (f(x)+g(x))(x^2+x+1) And of course c*P(x) = cf(x)(x^2+x+1) Which is also a polynomial with real coefficients. ' D. The set of all triples (a,b,c) of integers. Almost works, except multiplying by a scalar would yield a tuple of real numbers. Kinda silly. In programming, if you attempted to write a function function List<Int> multiply(List<Int> list, Real c) { return list.map((x) => x * c); } You'd get a compiler exception since the return type wouldn't match (or you'd get some warning about shortening the precision). ' E. The set of all sequences (x_1,\dots,x_n)\in\mathbb R^n such that x_1+\dots+x_n=0 and x_1+2x_2+\dots+nx_n=0. Adding two sequences (x_1,...x_n) + (y_1...,y_n) gets you (x_1+y_1...,x_n+y_n) which plugging into the above two equations, and rearranging, will show that you'd get (x1+y1) + (x2+y2)... + (xn+yn) = (x1+x2+..xn) + (y1+...+yn) = (0) + (0) and the same for the second equation. Also, scalar scaling works fine. ' So only D has any issues. The short explanation as to why we insist on scalars being real or complex is that a major goal of linear algebra is to provide ways of solving equations. And you really want to perform division to solve equations, which real and complex numbers let you do. Integers aren't closed over division, so they aren't good for solving linear equations. For example, if your vector space is integer triplets then 2 * x = (1,0,0) Wouldn't have any solutions in the set of integer triplets. This is a linear equation, and the goal of linear algebra is to provide solutions, so it's better to have the framework yield the answer of x = (.5,0,0) and then say "Oh, the answer lies outside of the original set, so it wasn't a vector space to start with." Well, that's kinda the extrinsic view of it, the intrinsic view would throw an exception :).