Why does the subspace need to go through the origin?

Question

I understand that the main difference between a subspace and a hyperplane is that the subspace must go through the origin. Why does need to happen? In other words, why does a subspace always have to go through the origin? What restricts it from doing otherwise?

I think what Levent wants to point at is that linear subspaces always inherit the parent's addition and multiplcation. If this were not the case, you could define new operations that "translate" the subspace. This is then called an affine subspace. — WorldSEnder, Mar 06 '18 at 15:35
Just wanted to add, if you use the word origin you are implicitly considering a geometric subspace, since the origin is a geometric point in a Euclidean space of dimension $n$. Furthermore euclidean space is usually identified with $\mathbb R^n$, though there is a subtle difference as euclidean space has no coordinates but $\mathbb R^n$ does. For more general vector spaces, it may not make sense to say that a subspace goes through origin e.g. a polynomial subspace. Unless your origin is the zero vector. Planes are geometric objects, polynomials are not. — john, Mar 11 '22 at 19:32

score 31 · Accepted Answer · answered Mar 06 '18 at 14:08

31

A subspace is a vector space, then it must satisfy all axioms for a vector space, including the existence of a zero vector.

answered Mar 06 '18 at 14:08

Joca Ramiro

1,589

3

... and, in particular, the inclusion must be a linear transformation, so its zero vector is the same as the zero of the total space. – Mar 06 '18 at 17:40
The zero vector is unique. How could the zero for a subspace be different from the zero of the entire space? Do you have any example? I would really appreciate to see one. – Joca Ramiro Mar 06 '18 at 17:52
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Sure. The line in the plane defined by $y=1$ is a vector space when given the addition operation is $(a,1) + (b,1) = (a+b,1)$, scalar multiplication $r \cdot (a,1) = (ra,1)$ and zero element $(0,1)$. – Mar 06 '18 at 18:05
But you are defining a different addition of vectors from that defined for vectors in the plane. I always thought that a subset of a vector space can be considered as a vector subspace if it is a vector space itself with the same vector addition defined for the entire space. – Joca Ramiro Mar 06 '18 at 18:21
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That’s right, but in your answer you’re tacitly assuming this inheritance. – amd Mar 06 '18 at 19:06
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Yes @amd, I am assuming that inheritance. In fact, I think that the inheritance must be assumed. You can think in terms of groups, recalling that a vector space is a abelian group: can we admit as a subgroup a subset with a different group operation from the entire group? – Joca Ramiro Mar 06 '18 at 20:05
Thanks @jobe, your comment made me look into the axioms. I feel ashamed that I had totally forgotten about these. From a quick wikipedia search, I think I may have gotten the answer to my question. The vector space must satisfy both Identity element of addition (There exists an element 0 ∈ V, called the zero vector, such that v + 0 = v for all v ∈ V.) and Inverse elements of addition (For every v ∈ V, there exists an element −v ∈ V, called the additive inverse of v, such that v + (−v) = 0.), which @Lazward also kindly pointed out – usr Mar 06 '18 at 21:39
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@jobe: That the addition (and other operations) must be the same is precisely what "the inclusion must be a linear transformation" implies. The point is that merely "satisfying all axioms for a vector space" isn't enough; you need that extra information. – Mar 06 '18 at 23:34
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I see @Hurkyl, I just thought that my answer was sufficient to help the asker and apparently it was. I was assuming that the vector addition is the same as the entire space because it seems to me absurd to admit a subset as a subspace if the addition is not the same. If such an idea is admitted for groups, for example, we could have an abelian group with a noncommutative subgroup, what I consider nonsense. Anyway, the "extra information" can be seen in these comments and in the answer by Irin Thirdwater, so anyone interested can absorb it. – Joca Ramiro Mar 07 '18 at 02:00

score 15 · Answer 2 · answered Mar 06 '18 at 14:11

15

You want to be able to operate vectors in the subspace without leaving it. If any vector $v$ is in there you want $(-1)v = -v$ also to be there, and also their sum $v+(-v) = 0$.

answered Mar 06 '18 at 14:11

Lazward

587

score 11 · Answer 3 · edited Mar 06 '18 at 21:43

You can indeed have two vector spaces over the same field $F$ such that the identities are different (i.e. does not go through the origin). However, these two vector spaces would simply be different since their additions and multiplications would necessarily have to be different as well. By thinking of the "origin" you are already implicitly referring to that one vector space of $\mathbb{R}^n$ with the usual $+$ and $\cdot$.
An example of this idea is already touched upon by WorldSEnder's comment above.

As for why you need some restrictions for one vector space to be a subspace of another, others have already provided very good answers.

Actually, this is a full-blown answer, not a comment. Good job. — Parcly Taxel, Mar 06 '18 at 17:03

score 6 · Answer 4 · answered Mar 06 '18 at 14:46

6

We need the zero vector in order to have $\forall \vec v\in V$

$\vec v+(-\vec v)=\vec 0 \in V$
$0\cdot \vec v=\vec 0 \in V$

answered Mar 06 '18 at 14:46

user

154,566

score 0 · Answer 5 · answered Mar 07 '18 at 17:50

To avoid confusion, let's start by restating the definition of a subspace:

A subspace of a vector space is a subset of its underlying set of points, equipped with the same addition and scalar multiplication operations, which itself satisfies the definition of a vector space.

Given this definition, it's not hard to see that a non-empty subset $A$ of a vector space $S$ that does not include the origin $\vec 0 \in S$ cannot be a vector space (and thus cannot be a subspace of $S$) because

it is not closed under scalar multiplication: for any $\vec v \in A$, $0 \cdot \vec v = \vec 0 \notin A$, and also because
it is not closed under vector addition and the taking of inverses: for any $\vec v \in A$, $\vec v + (- \vec v) = \vec 0 \notin A$.

(The empty subset $\emptyset \subset S$ does vacuously satisfy both of these closure properties, but trivially fails the requirement that a vector space must contain at least one point, namely its origin.)

In fact, we can show a bit more; namely that if $A \subset S$ contains an origin (i.e. an element $\vec \epsilon \in A$ such that $\vec \epsilon + \vec v = \vec v$ for all $\vec v \in A$), then this origin must be the same as the origin $\vec 0$ of $S$. Specifically, since $\vec \epsilon + \vec \epsilon = \vec \epsilon$, if follows from the vector space axioms on $S$ that $$\vec 0 = \vec \epsilon + (-\vec \epsilon) = (\vec \epsilon + \vec \epsilon) + (-\vec \epsilon) = \vec \epsilon + (\vec \epsilon + (-\vec \epsilon)) = \vec \epsilon + \vec 0 = \vec \epsilon.$$ Thus, conversely, if $\vec 0 \notin A$, then $A$ also directly fails the requirement that a vector space must have an origin.

Why does the subspace need to go through the origin?

5 Answers5