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.
Mind clarifying this part? As someone else already pointed out, the integers are multiplicatively closed, but I suspect you're using "multiplicatively closed" to also mean "closed under multiplicative inverses". But I don't see how linear combinations stop being possible, e.g., "3x + 2y" is still a linear combination in a Z-module, or what it means for a linear combination to be invertible.
(Also not sure what exactly you mean by linear relations not existing if you have a module and not a vector space...)
The integers violate the field axiom that every element in the field (other than 0, the additive identity) must have a multiplicative inverse in the field. But that has nothing to do with the requirement of being closed under multiplication.
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.
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.
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.
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.
Typically, students don't start on an algebra track until after a fair amount of analysis, but there is no real reason for that to be the case. Its a shame too since, as someone who prefers algebra myself, I (totally unfairly) blame analysis for giving math a bad image.
Isn't this a 1 dimensional space. Eg. Consider the vector space R2 over R.
If you have the vector (1,0), there is no way to arrive at the vector (1,1) through just scalar multiplication.
I'm afraid you have badly misremembered this stuff.
One of us is very confused. It seems to me that I also can't get from (2,3,4) to (5,6,7) by pure scalar multiplication even if fractions are allowed. If I pick a scalar factor of 2.5 to make 2 -> 5 work, then I get (5, 7.5, 10). If I pick anything else, the result won't start with 5.
>>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.
Really no. You can only access parallel vectors by scalar multiplication. E.g. if your vector space is R2, given a starting vector and scalar multiplication, you can anything in a line with the direction of that vector, but nothing pointing in a different direction. That's more or less why it's called "scalar" multiplication - it scales the original vector, but doesn't change its direction.
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)?
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.
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.
Because it's a definitional thing. A "scalar" is routinely defined as a real number, not an integer.
And you're absolutely right that it makes no sense, which is the whole point of the multiple-choice question. Four of those answers are plausible, the other requires you to make assumptions (like a redefinition of scalar) not in the question as posed.
"Or perhaps they wouldn’t like A because the scalar field [the complex numbers] is the same as the set of vectors (unless, that is, they thought that the obvious scalars were the real numbers)."
In this case, while there is an an acknowledgement that you could take the reals as your scalars, it is regarded as the secondary of the "natural" choices.
Or, in my example, example, there is no way to view Q as a vector space over R, but it is clearly a vector space. There is an entire field of algebra (field theory), that relies on the fact that, for example, Q(sqrt(2)) is a 2 dimensional vectorspace over Q.
If you've seen someone define a scalar as a real number, that's really only because they're informally stating their underlying field is R.
The whole purpose of this exercise is to see if there was a way to come up with a straightforward, reasonably informal, multiple choice question that would expose a fundamental understanding in basic university math concepts like "vector space" in the same way we see in primary math.
And instead all you people want to do is natter over the ways in which someone could cleverly make the "wrong" answer right. It's... beyond missing the point, it's actively working against the whole goal of the exercise.
By itself, this is a minor complaint (you cannot include every example in you choices, although I do think that an example which could not be viewed as an R-vector space would be good to include). However, when you explain why Z3 is not a vector space, your explanation must be correct. An explanation which also excludes Q3 is incorrect.
Well, this is totally untrue. A scalar is defined as a non-vector quantity, a single element as opposed to a multidimensional list of them.
I give up on this thread. It's a bunch of people not just willfully misunderstanding the linked article, but actively campaigning against the whole idea of math education in an attempt to prove how much smarter than each other they are. This is... awful, folks.
> Could one devise a university-level question that would catch a significant proportion of people out in a similar way? I’m not sure, but here’s an attempt.
> Which of the following is not a vector space with the obvious notions of addition and scalar multiplication?
> ...
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.
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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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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Two twice differentiable functions can be added together to yield a twice differentiable function
(f(x)+g(x))''=f''(x)+g''(x)
(cf(x))''=c(f''(x))
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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)
P(x)+Q(x) = (f(x)+g(x))(x^2+x+1)
c*P(x) = cf(x)(x^2+x+1)
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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);
}
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Adding two sequences
(x_1,...x_n) + (y_1...,y_n)
(x_1+y_1...,x_n+y_n)
(x1+y1) + (x2+y2)... + (xn+yn) = (x1+x2+..xn) + (y1+...+yn) = (0) + (0)
Also, scalar scaling works fine.
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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)