What examples of the vector space axiom $u + 0 = u$ exist where the so-called "zero vector" is not an actual zero vector in $\mathbb{R}^n$? Sometimes the axiom holds because there is some "zero vector" s.t. $u + 0 = u$ but the "zero vector" is not what one would normally consider a $0$ vector. Could someone help me conceptualize this through an example?
If that isn't clear, feel free to ask for clarification.
 A: One of the main ideas in elementary linear algebra is that every finite-dimensional vector space is isomorphic to $\mathbb{R}^n$ or $\mathbb{C}^n$, so the usual idea of a "zero" vector will not be particularly enlightening beyond your normal intuition of "zero" in arithmetic.
A slightly non-trivial version might be considering the vector space of linear functions $T\colon\mathbb{R}^n\rightarrow\mathbb{R}^m$ (a nice exercise would be verifying this is a vector space), and in this case the "zero vector" would be the linear function $T$ with the property that $Tx=0$ for all $x\in\mathbb{R}^n$.
A: Here is a family of vector spaces over $\mathbb R$. Choose your favourite $k \in \mathbb R$. Now define the following operations for vectors $\mathbf u, \mathbf v \in \mathbb R$ and constant $a \in \mathbb R$:
$$\mathbf u \oplus \mathbf v = \mathbf u + \mathbf v+k$$
$$a \otimes \mathbf u = a \mathbf u + k(a-1)$$
Addition and multiplication on the right-hand sides of the above definitions are the usual ones in $\mathbb R$. Then the "zero vector" is $-k$. 
A: *

*Take your vector space to be some function space, for example, continuous function on $[0,1]$. If the vector space operation is addition, then the zero element is the zero function.

*Consider the following finite vector space. It consists of two symbols, which we will write as $0$ and $1$. Addition is defined as per usual, except that $1+1 = 0$. As for multiplication by scalars, we take the scalars to be the same elements $0$ and $1$, and perform multiplication as usual. As you can see, the scalars in a vector space do not always have to be $\mathbb{R}$, but just some algebraic structure called a field, which our $0$ and $1$ are. Now we have a vector space with two elements, and the $0$ symbol is our zero.

*Here's a more technical example. Picture in your mind a shape, like a sphere. A vector field on the sphere is assignment to each point on the sphere a direction/vector. 

The collection of all possible vector fields forms a vector space because you know how to add to vector fields: simply add up the vectors at each point. You can multiply by scalars, which just mean you scale the direction vectors to make them bigger of smaller. The zero vector is just the zero vector field that assigns the zero direction (don't go anywhere!) at every point on the sphere.
A: Consider $\mathbb R_{>0}$ as a vector space over $\mathbb R$, with the following operations on vectors $\mathbf u,\mathbf v\in\mathbb R_{>0}$ and scalar $a\in\mathbb R$:
$$\begin{align}
\mathbf u \oplus \mathbf v &= \mathbf u\mathbf v, \\
a \otimes \mathbf u &= \mathbf u^a.
\end{align}$$
The zero vector is $\mathbf 1$.
This is nothing but the usual vector space on $\mathbb R$ transformed by $\exp$. That is, if we denote $\mathbf u = \exp u'$, then
$$\begin{align}
\mathbf u \oplus \mathbf v &= \mathbf w \quad \iff & u' + v' &= w', \\
a \otimes \mathbf u &= \mathbf v \quad \iff & a u' &= v'. \\
\end{align}
$$
Similarly, Théophile's example is the usual vector space on $\mathbb R^n$ transformed by $\mathbf u = \mathbf u' - \mathbf k$.
In general any bijection $f$ from a vector space to an arbitrary set $S$ induces a vector space on $S$ whose zero vector is $f(\mathbf 0)$.
