TITLE: What is an intuitive definition of the zero vector?
QUESTION [5 upvotes]: I'm learning now about vector spaces and subspaces, and one of three rules that determine if something is a subspace of a larger vector space is that it must contain the zero vector... but intuitively, I can't figure out why the zero vector wouldn't exist. When I first learned about vectors over the summer, the zero vector was basically described as "pick a point on the x-y plane. Move 0 units up and 0 units to the right. That is the zero vector... just a dot on the x-y plane". This is assuming of course that you're talking about a vector $\vec{v}\in\mathbb{R}^2$.
Now fast forward to today when I'm given the following definition and need to determine if it's a subspace or not based on the three rules:

Does it contain the zero vector?
Is the set closed under vector addition?
Is the set closed under scalar multiplication?

Take for example the question:

Let $a,b,c,d$ be constants, and let $U=\left\{\left[\begin{array}{r}x\\y\\z\end{array}\right]\in\mathbb{R}^{3}\;\middle|\; ax+by+cz=d\right\}$. Show that $U$ is a subspace of $V$ if and only if $d=0$.

By the first test above, $\left[\begin{array}{r}0\\0\\0\end{array}\right]$ is not in $U$ if $d\neq 0$ because $a(0)+b(0)+c(0)$ necessarily implies that $d=0$...
Consider also the line $x+y=1$ in $\mathbb{R}^2$. This also does not contain the zero vector.
It seems to me like "zero vector" is being used synonymously with "origin", but this doesn't fit the definition of a vector that I was given. Sure, an arrow can begin at the origin and extend outward, but not necessarily. Any arrow representing a vector is the same as any other as long as it has the same length and direction, no matter where the base of the arrow sits, the zero vector should still be the zero vector.
So I ask... How can the zero vector not be in any plane, if it is indeed properly understood as "pick a point, move 0 units on the x-axis, 0 units on the y-axis, and 0 units on the z-axis"?

REPLY [3 votes]: You are operating on the understanding of not just a vector space but an affine space. An affine space is basically a set on which a vector space (transitively) acts. The vectors are "directions" and the affine space is intuitively comprised of points in space. Any two points determine a displacement vector: the unique vector you use to go from one to the other. Thus a vector can be identified more or less with an equivalence class of pairs of points with the given vector as displacement vector.
You are thinking: "given any nonempty subset of the affine space, there are two points in this space with the zero vector as displacement." (Indeed, just pick the same point twice.) But this is not what we're talking about when we talk about subspaces of the vector space. When we talk about subsets of the vector space, we are not really thinking in terms of points but in terms of vectors.
You may think of it in terms of moving pieces on a board game (like chess or checkers). Let's work with the real numbers. Let $A=\Bbb R^2$ be the affine space. Your "piece" in the game is a point in this space (the space is the "board" in this metaphor). And let $V=\Bbb R^2$ be the vector space; it is your set of moves at any given position. If you pick the move $v\in V$ then you go from position $x\in \Bbb R^2$ to position $x+v\in\Bbb R^2$. It is important to distinguish positions (points) from moves (vectors).
This much is true: "given a nonempty set $S\subseteq A$ of positions, there is a position $s\in S$ for which we have the option to not move." (Not moving is equivalent to being displaced by $0\in V$.) That corresponds to what you're thinking. But the idea of a subspace is a set of moves rather than a set of positions, so a subset $M\subseteq V$. There are indeed move sets $M\subseteq V$ in general which do not contain the [go nowhere] option. For example let $M=\{(0,1)\}$; this is a set containing the [go up one unit] move. This is not the same as the [go nowhere] move. So even though all positions admit a [go nowhere] move, not all sets of moves contain the [go nowhere] move.