Does zero vector in the context of vector space imply origin?

Question

Let's consider for example $\mathbb R^n$. Here, we know of vectors whose components (its elements) consist of $\mathbb R$ but we do not interprets it as points in linear algebra (vectors as what we consider). Now keeping that in mind how can we say that zero vector is same as origin given that it implies no incremental change?

Answer 1

There is a subtle difference between a coordinate system (parameterization) and a vector space.

A coordinate system is a mapping $\phi:\mathbb{R}^n\to M$ between tuples of real numbers and some geometric object $M$. The origin is the image of the zero vector under $\phi$. You can think of the parameterization domain $\mathbb{R}^n$ as a vector space, but $M$ does not have to be a vector space, and $\phi$ does not have to be linear; if you try to map a vector $v\in\mathbb{R}^n$ to an ``image vector'' $\phi(v)-\phi(0)$ you will see that these image vectors do not add "tip to tail" in general. For a simple example, consider the polar coordinate parameterization $\phi(r,\theta)$ of the plane. The vector $(r=1,\theta=0)$ maps to $(x=1,y=0)$ in the plane, and $(r=0,\theta=1)$ maps to $(0,0)$, but $(1,1)$ maps to $(\cos 1, \sin 1)$, and not $(1+0,0+0)$.

The best you can do is to associate tangent vectors of points $p$ in $\mathbb{R}^n$ with tangent vectors at the image $\phi(p)$; both of these tangent spaces are vector spaces and the mapping between them is given by the derivative $d\phi$ of $\phi$. But again, it is the tangent spaces here that are vector spaces, not $M$.

I know it's confusing; the key point is that we are used to treating the parameterization domain $\mathbb{R}^n$ as a vector space, but this structure is a red herring as it is incidental and does not tell us anything about the structure of $M$.

Some keywords for learning more include "affine connection" and "parallel transport."

Now if you just have a vector space, you can of course imagine giving it trivial geometric structure with origin given by the zero vector.

Answer 2

At each point $P(x_1,...,x_n)$ in $\mathbb R^n$ corresponds a vector from the origin $v=(x_1,...,x_n)$ and viceversa then zero vector $v_0=(0,...,0)$ corresponds to $O(0,...,0)$.