How Craig Barton wishes he’d taught maths

My understanding of learning is that it both takes a longer time to actually learn something useful and that we're inefficient in learning or retaining anything because we rushed from one topic to another in our courses.

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.

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.

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

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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.

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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))

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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.

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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).

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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.

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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 :).

You can generalize vector spaces to work over arbitrary rings (in which case it is a module). I did have to do a double take on that question, because in my mind when I see someone talking about a vector space over a ring, I just silently translate it to a module (because 90% of the time, the person just misspoke, 9% of the time, even if they meant vectorspace, the immediate follow up would be "no, but does your point still stand of we consider it a module?", and maybe 1% of the time they are actually being adversarial.

As I remember, most first courses in linear algebra don't deal with the subject as anything like abstract algebra and aren't going to be targeting students who can distinguish a ring from a field. Essentially, until you can past the calculus sequence, most math in the US is centered on calculation with only a few forays into proofs and definitions.

I'm a high school math and science teacher, and I completely agree this is a problem. Several of my students struggled greatly with simple arithmetic until I set down and gave them reasons for why a negative times a negative was a positive (which meant I also had to explain to them why multiplication works like it does).

I personally think part of the problem is how elementary education is structured. At least in my state, elementary school teachers are expected to be extreme generalists, and only have to take a math class or two -- and then nothing above simple college algebra (which is fine). But they always complain about how difficult it is, and they don't understand math, often taking the state's required exam multiple times because they can't pass math. Yet these are the people we allow to teach kids math; it's a huge issue when they're being taught math by people who don't understand why it works, only the algorithms they've memorized.

Coincidentally, this is also why the 'new math' was so lambasted -- these teachers (and often, parents) don't understand how numbers work, thus they think it's useless to teach kids to subtract 20 and add 2 instead of subtracting 18. Despite the fact that one is much easier to do mentally, and allows you to get a good sense of how subtraction and addition interact.

I'm not a fan of charter schools, but if I had money, I'd start an elementary charter school where the subjects were taught by people who understood that and not generalists. And students would get like multiple hours of recess a day, especially those first few years. Just pure, unstructured play time. But that's a rant for another day.

Going back around the millennium or before when I last taught A level maths at college, we had them in over the summer before term started for a two week intensive algebra and basics course.

Seemed to help.

The original author (Tim Gowers, a Fields medallist and professor of mathematics at Cambridge) has a totally hilarious blog post about being asked to coach a teenager doing A level maths...

https://gowers.wordpress.com/2012/11/20/what-maths-a-level-d...

As philosopher Daniel Dennett puts it: competence comes before comprehension. It's totally possible to do something well, as animals do, without understanding what one is doing. But for comprehension one needs to have something in place to reason about and make connections.

And this isn't the full picture. Motivation comes before competence. One needs reasons to acquire skills: they have to address problems in one's mind if the mind is to fully engage. Which is why coercive education with its curricula, exams, etc, largely fails.