How to interpret 'vector space is closed under scalar multiplication' for physics vector

Question

First of all, I am not familiar with mathmatical language, so please forgive me if I use incorrect or inaccurate expressions.

I have seen lots of descriptions about vector space:

vector space is closed under addition and scalar multipliciation

So, theses mean that with vectors $\vec v, \vec u \in V$, and a scalar $s$,

1. $\vec v + \vec u \in V$
2. $s\vec v \in V$

Now, suppose $V$ is a real 3D position vector space. That is, any vector in $V$ can be written as,

$\vec v = x\vec e_x + y\vec e_y +z\vec e_z $

where $(x,y,z)$ denotes 3d coordinate such as (1m, 0.5mm, 3km) and $\vec e_i (i=x,y,z)$ denotes unit vector of each $i$-th axis.

Now, I assume that the scalar of this vector space is length value because it is the component(element) of the vector although I am not sure this is right assumption.

If I multiply any scalar to this vector, the result won't be any more 3d coordinate because for instance, $1\text m\times 1\text m=1\text m^2$ which is no more scalar of 3d position vector space.

This conflicts with the second property of vector space: closed under scalar multiplication.

How should I interpret the vector property for vectors from physics?

Is it just difference kind of vector with that of linear algebra? Or, have I made wrong assumption?

Answer 1

Think of it like this. Your vectors have units (say m in your example), but the scalars over which your vector space is formed is unitless. When you have a unitvector $\hat{v}=\frac{\vec{v}}{\|\vec{v}\|}$, you have divided by a dimensionful quantity, not a scalar of your vector space, which means you are now in a new vector space. And this is good, because if we were still in the same vector space, we would need to make sense of expressions like $\hat{v}+\vec{v}$ which are nonsensical if $\vec{v}$ has units. So think of your vector space itself having units, but the scalars do not have units. Then also be careful when you multiply your vector by a dimensionful quantity (which we would in physics calla scalar too, but it isn't the scalar of our vector space, since those are always dimensionless) that you are actually moving it into a new vector space with the new appropriate units. When the vector space is unitless, then we are in what mathematicians would usually consider to be a vector space.

If this seems way too complicated, it might be useful to think about it as being a mathematical vector space (vectors themselves don't have units) being implicitly multiplied by some dimensionful quantity. This is a it less elegant, and I feel can lead to future confusion.