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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:

  1. Does it contain the zero vector?
  2. Is the set closed under vector addition?
  3. 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"?

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    $\begingroup$ If a nonempty subset is closed under scalar multiplication, it necessarily contains the zero vector ($0 {\bf x} = \bf 0$) $\endgroup$ – Robert Israel Sep 26 '13 at 1:23
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    $\begingroup$ It's not the zero vector you misunderstand: the problem is you're confusing two different vector spaces. You're asked a question about a subset of one vector space, but you're thinking about your answer in terms of some entirely different vector space. You may be interested in reading up on the notion of affine space. $\endgroup$ – user14972 Sep 26 '13 at 2:51
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From your question it appears you are confused with the meaning of the word vector. You describe a vector in, say, $\mathbb R^2$ is a displacement operation with an arbitrary starting point. However, a it's not quite that. We consider two such displacements to be essentially the same if they have the same direction and the same magnitude. This essential sameness defines an equivalence relation on the set of all such displacement. In that context a vector is an equivalence class, not just a representative of it.

The modern approach is to abandon these inconveniences all together and adopt an axiomatic approach. A vector space is a set with extra structure satisfying a certain list of axioms. Then, a vector is, by definition, an element of a vector space. The zero vector is then the (provably) unique vector in a given vector space which behaves neutrally with respect to addition of vectors. That is what the zero vector is. It should be emphasized that the precise name of the zero vector is highly sensitive to the vector space structure. For instance, $\mathbb R$ with its usual vector space structure admits $0$ as the zero vector. However, for every $a\in \mathbb R$ it is possible to endow $\mathbb R$ with a vector space structure such that $a$ is the zero vector. This is what happens when we shift from a definition of what vectors are (upon which your understanding of vectors currently relies) to not caring about what they are and only caring about what you can do with them (this is the axiomatic approach). There is good reason why the latter is the prevalent choice in modern mathematics. Nobody cares, nor should we care about what something is. All that matters is what we can do with it. It saves a lot of headaches and endless philosophical quarrels if you don't even attempt to define what something is and instead simply resort to listing the things you can do with these things.

With all that said, other than the cases you mention where the zero vector does not belong to a given set, consider the empty subset of any vector space. It is never a vector subspace since it does not contain the zero vector (nor any other vector).

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  • $\begingroup$ It can be noted that the empty set is the only case where it matters whether we explicitly require the zero vector to be in a subspace. For every non-empty set the two other conditions are sufficient, because if the subspace contains some $a$ it must also contain $0\cdot a$ which we already know is the zero vector. $\endgroup$ – hmakholm left over Monica Sep 28 '13 at 10:36
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The "spirit" of the $0$-vector is this: it is the identity element in the "addition" operation defined on the vector space.

So, whether or not your set contains "the" $0$-vector is really a question about how you've defined addition: is there an element $\vec{0}$ in your set such that $\vec{x}+\vec{0}=\vec{0}+\vec{x}=\vec{x}$ for all $\vec{x}$ in the set?

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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.

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The reason the $0$ vector cannot be contained in a subset such as $U = \{(x,y) \in \mathbb{R^2}: x + y = 1; x,y \in \mathbb{R}\}$, is because you essentially need $0$ as a hypothetical "starting" point to get anywhere. When we graph vectors, we draw them as a point in space connected to $0$ by a line segment. The line $x + y = 1$ is not geometrically connected to $0$, so we can't draw a single vector $v$ such that $\text{span}(v) = U$.

Consider another case:

$W = \{(x,0) \in \mathbb{R^2}: x,y \in \mathbb{R}\}$. This subspace of $\mathbb{R^2}$ can be thought of as the x axis, which contains the origin (and it turns out that addition and scalar multiplication are "well-behaved" along the axis). Therefore, we can find a single vector $v$ such that $\text{span}(v) = W$.

In both cases, we had lines in $\mathbb{R^2}$. A line that is a subspace will always be spanned by a single vector. If we have a line in space, that cannot be spanned by a single vector, then is it a vector space (How can a line that does not contain the origin, be thought of as some arbitrary $(x,y)$)? For subspace $U$, there is a quite a substantial amount of $x$'s and $y$'s where $x,y \in \mathbb{F}$ that will not satisfy $x + y = 1$.

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If the zero vector doesn't exist, $\mathbf{R^n}$ doesn't have a solution $\mathbf{a}: \mathbf{x} + \mathbf{a} = \mathbf{x}$.

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    $\begingroup$ I think you're confusing the number $0$ with the zero vector. We usually wouldn't regard $\Bbb{Z}$ as a vector space, so talking about the integer being $0$ as a vector is confusing. $\endgroup$ – user61527 Sep 26 '13 at 2:03
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    $\begingroup$ @T.Bongers I had no idea I was writing this, I guess I was on auto-pilot. fixed $\endgroup$ – Don Larynx Sep 26 '13 at 13:05

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