Is work a vector quantity in physics?

[Orodruin]

Strictly speaking, $\mathbb{R}^n$ is just the set of n-tuples {(x₁,..,x_n)}. To make it a vector space you have to specify two sets, the set of vectors and the field of scalars, then define how the vector elements add and what the product of a scalar and a vector is.
That we commonly think of $\mathbb{R}^n$ as the natural vector space $\{\mathbb{R}^n, \mathbb{R}\}$ is just a convenient shorthand.
Is there only one way to do that with $\mathbb{R}^n$ as a set of vectors over the field $\mathbb{R}$?
What if I take the n-tuples $\mathbb{C}^n$ over the field $\mathbb{R}$? Doesn't that make a vector space in the obvious way?

The original discussion was regarding $\mathbb R$ itself, not $\mathbb R^n$. Any field is a vector space over itself without any additional structure as it already has addition and multiplication by scalars. I therefore do not think it makes sense to claim that $\mathbb R$ is not a vector space - it has all of the required properties. That does not mean it cannot also be made into a vector space over a different field, but in itself it has all of the required properties to be called a vector space.

 

[PeroK]

Strictly speaking, $\mathbb{R}^n$ is just the set of n-tuples {(x₁,..,x_n)}.

What is $\mathbb{R}$ for that matter? Strictly speaking it has no properties until you define it by some sort of construction. It's a set of equivalence classes of Cauchy sequences of rationals if you want to start at the beginning.

There's no reason arbitrarily to decide that $\mathbb{R}$ needs "construction" specifically to be a vector space, but no construction to be a field of scalars.

 

[haruspex]

What is $\mathbb{R}$ for that matter? Strictly speaking it has no properties until you define it by some sort of construction. It's a set of equivalence classes of Cauchy sequences of rationals if you want to start at the beginning.

There's no reason arbitrarily to decide that $\mathbb{R}$ needs "construction" specifically to be a vector space, but no construction to be a field of scalars.

$\mathbb{R}$ is defined (as I understand it) not just as a set but all the operations and axioms necessary to be a field. I accept that one could also define something else with the same name (a pun) which is a vector space constructed with the elements of the set underlying $\mathbb{R}$ as its vectors, over the field $\mathbb{R}$, with all the addition and multiplication rules defined in the obvious way.
But it is an interesting question as to whether there is a nonobvious construction that makes a different vector space.

 

[jbriggs444]

$\mathbb{R}$ is defined (as I understand it) not just as a set but all the operations and axioms necessary to be a field. I accept that one could also define something else with the same name (a pun) which is a vector space constructed with the elements of the set underlying $\mathbb{R}$ as its vectors, over the field $\mathbb{R}$, with all the addition and multiplication rules defined in the obvious way.
But it is an interesting question as to whether there is a nonobvious construction that makes a different vector space.

So the idea is that one is using R as the set of elements of the vector space and R as the set of scalars.
With ordinary real addition for the addition of two vectors and ordinary real multiplication for the product of a vector and a scalar.

Given that, I do not see that there is any freedom to define anything non-obvious. Everything is already nailed down.

 

[haruspex]

With ordinary real addition for the addition of two vectors and ordinary real multiplication for the product of a vector and a scalar.

Those are choices that nail it down.
My question is whether you could define the vector addition and scalar x vector multiplication differently and still arrive at a structure satisfying vector space axioms. Even if you can, is the resulting vector space isomorphic to the usual one?

 

[PeroK]

So the idea is that one is using R as the set of elements of the vector space and R as the set of scalars.
With ordinary real addition for the addition of two vectors and ordinary real multiplication for the product of a vector and a scalar.

Given that, I do not see that there is any freedom to define anything non-obvious. Everything is already nailed down.

Every n-dimensional real vector space is isomorphic to $\mathbb{R}^n$. You are right, there are no other options.

There is flexibility when you define an inner product or norm.

The starting point for physics, I would say, is that $\mathbb{R}^3$ is a real vector space equipped with the usual inner product, leading to the usual Euclidean 2-Norm: $|v| = \sqrt{v_1^2 + v_2^2 + v_3^2}$.

When we talk about the magnitude of a vector, there's not any doubt we are talking about the usual 2-Norm.

It's not entirely clear, I guess, when elementary kinematics is taught in one-dimension exactly what is assumed, as an underlying mathematical framework.