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I know that every vector space needs to contain a zero vector. But all the vector spaces I've seen have the zero vector actually being zero (e.g. $\mathbf{0}=\langle0,0,\ldots,0\rangle$). Can't the "zero vector" not involve zero, as long as it acts as the additive identity? If that's the case then are there any graphical representations of a vector space that does not contain the origin?

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    $\begingroup$ Considering that all real vector spaces of finite dimension are isomorphic to $\mathbb R^n$, you'll not have much luck finding a graphical representation of a vector space that looks any different that $\mathbb R^n$. The zero vector might not obviously "involve" zero, but it's still the origin. $\endgroup$ – Milo Brandt Mar 26 '16 at 17:13
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    $\begingroup$ You'll normally label a vector $(0,\ldots,0)$ because it is the zero vector. In actual applications, you're most often not interested in $\mathbb{R}^n$ itself, but just use it as a convenient representation for some other vector space. $\endgroup$ – leftaroundabout Mar 27 '16 at 13:01
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Here's an example. Let $V$ be the set of all $n$-tuples of strictly positive numbers $x_1,\ldots,x_n$ satisfying $x_1+\cdots+x_n=1$. Define "addition" of such vectors by

$$ (x_1,\ldots,x_n) \mathbin{\text{“}{+}\text{''}} (y_1,\ldots,y_n) = \frac{(x_1 y_1,\ldots,x_n y_n)}{x_1 y_1 + \cdots + x_n y_n }. $$

This is a vector space whose zero element is $$ \left( \frac 1 n , \ldots, \frac 1 n \right). $$ The additive inverse of $(x_1,\ldots,x_n)$ is $$ \frac{\left( \dfrac 1 {x_1}, \ldots, \dfrac 1 {x_n} \right)}{\dfrac 1 {x_a} + \cdots + \dfrac 1 {x_n}}. $$ This operation is involved in a basic identity on conditional probabilities: $$ (\Pr(A_1),\ldots,\Pr(A_n)) \mathbin{\text{“}{+}\text{''}} k\cdot(\Pr(D\mid A_1),\ldots,\Pr(D\mid A_n)) = (\Pr(A_1\mid D),\ldots,\Pr(A_n\mid D)) $$ where $k$ is whatever it takes to make the sum of the entries $1$. However, in practice, one wouldn't bother with $k$; just multiply term by term and then normalize.

Here's a more down-to-earth example. Look at $\mathbb R^3$ and say you want to put the zero point at $\vec p = (2,3,7)$. Then define "addition" as follows: $$ \vec a \mathbin{\text{“}{+}\text{''}} \vec b = \underbrace{\vec p + (\vec a - \vec p) + (\vec b - \vec p)}_{\begin{smallmatrix} \text{These are the usual} \\ \text{addition and subtraction.} \end{smallmatrix}}. $$

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    $\begingroup$ I assume you mean "satisfying $x_1 + \cdots + x_n = 1$". I'm curious: how is scalar multiplication defined in this space? $\endgroup$ – Bungo Mar 26 '16 at 17:13
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    $\begingroup$ @Bungo I assume scalar multiplication is $c(x_1,\ldots,x_n)=\frac{(x_1^c,\ldots,x_n^c)}{x_1^c+\ldots + x_n^c}$ - this is at least correct for integer $c$ due to the definition of addition, and certainly satisfies what it's supposed to. $\endgroup$ – Milo Brandt Mar 26 '16 at 17:15
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    $\begingroup$ @MiloBrandt: Thanks, that makes sense. Nice example, I haven't seen this space before. $\endgroup$ – Bungo Mar 26 '16 at 17:16
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    $\begingroup$ If you take the logarithms of the coordinates, you can see this space as simply the quotient of $\mathbb{R}^n$ by the span of $(1,\ldots,1)$. $\endgroup$ – filipos Mar 27 '16 at 13:54
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Michael Hardy provides a very good answer. I want to explain what's so exceptional about it.

If you have a vector space (let's say finite dimensional), once you choose a basis for that vector space, and once you represent vectors in that basis, the zero vector will always be $(0,0,\ldots,0)$. Of course, the coordinates here are with respect to that basis.

We usually describe elements of $\mathbb R^n$ using coordinates that are of course the coordinates of the most obvious basis of $\mathbb R^n$. And the same for any subspace. So this question doesn't come up there.

The exotic examples only happen when you use coordinates that are not really indigenous to the vector space. The coordinates may have some interesting mathematical structure, but one structure they will not have is the structure of the vector space they are representing. Calling them "coordinates" is almost a lie, since they don't act like vector space coordinates at all, for instance, $(0,0,\ldots,0)$ is not the zero vector.

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  • $\begingroup$ This is a good answer and it's a shame it's so far removed (spatially) from Hardy's. $\endgroup$ – Eric Stucky Mar 27 '16 at 20:33
  • $\begingroup$ Theoretically, the underlying field of a vector space could have an additive identity that is not denoted by the glyph "$0$", say "$\zeta$", and then the zero vector would have components from that field of $(\zeta ,\zeta ,\ldots ,\zeta)$, but that would be arbitrarily less-clear notation for that field and merely a difference in glyphs used, not actual meaning. $\endgroup$ – Todd Wilcox Mar 28 '16 at 13:42
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In linear algebra textbooks one sometimes encounters the example $V = (0, \infty)$, the set of positive reals, with "addition" defined by $$ u \oplus v = uv $$ and "scalar multiplication" defined by $$ c \odot u = u^{c}. $$ It's straightforward to show $(V, \oplus, \odot)$ is a vector space, but the zero vector (i.e., the identity element for $\oplus$) is $1$.

(The pleasure "relabeling" this example to look like a more familiar space is left as an exercise.)

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The zero vector in a vector space depends on how you define the binary operation "Addition" in your space.

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For an example that can be easily visualized, consider the tangent space at any point $(a,b)$ of the plane $\mathbb{R}^2$. This is the space of "bound vectors" that start at the point $(a,b)$. Any such vector can be written as $(a,b) + t(c,d)$ for some $t \geq 0$ and $(c,d) \in \mathbb{R}^2$. The zero element in this vector space is the base point $(a,b)$. Pictures from Lee's "Introduction to Smooth Manifolds":

enter image description here

Or consider the tangent plane to a sphere at any point, which is also naturally a vector space:

enter image description here

Note however that it is generally meaningless to speak of "zero" without having a vector space structure (or some other algebraic structure with the notion of zero) in mind already.

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In the vector plane of the oriented segments with addition defined with the parallelogram law the zero vector is just a segment consisting of a single point, no zeros involved. Only after you introduce a vector base and represent the vector with coordinates you have that the zero vector is represented by $(0,0)$.

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That's because once you establish an isomorphism of a vector space with $\bigoplus_{i\in I} F$ (in which I am of course talking about the standard addition and scaling ), the zero vector is the only candidate that the zero of the 'abstract' vector space could map to. There is nothing else such that $v+v=v$.

Since it is common practice to represent vector spaces this way, that is why you've seen it so often.

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  • $\begingroup$ In standard usage $\oplus$ appears in things like $F\oplus G$ and $F_1\oplus\cdots\oplus F_n$ and $\bigoplus$ appears in things like $\bigoplus_{i\in I} F$. I edited accordingly. $\qquad$ $\endgroup$ – Michael Hardy Mar 26 '16 at 16:57
  • $\begingroup$ @MichaelHardy whatever, dude. $\endgroup$ – rschwieb Mar 27 '16 at 0:05
  • $\begingroup$ @djechlin I think you're thinking of $v+v=0$ $\endgroup$ – rschwieb Mar 28 '16 at 11:57
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One more comment- in the specific vector space, $\mathbb R^n$, the zero vector is the $n$-tuple $(0, 0, \ldots, 0)$. Any $n$-dimensional vector space is "isomorphic" to $R^n$, with, given a basis $e_1, e_2, \ldots , e_n$, the isomorphism that maps $e_n$ to $(0, 0, \ldots, 1, \ldots, 0)$ where the "$1$" is in the $n$th place so we can always express vectors in the $(a, b, \ldots, z)$ notation.

However there exist infinite dimensional vector spaces for which we cannot use that notation.

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    $\begingroup$ As long as you believe in the Axiom of Choice, and are willing to use infinitely long tuples, then you can use this notation even in infinite-dimensional vector spaces. $\endgroup$ – LSpice Mar 27 '16 at 20:27

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