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?

Answer 1

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

Answer 2

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 $v+(-v) = 0$.

Answer 3

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

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

Answer 4

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.

Answer 5

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

1. it is not closed under scalar multiplication: for any $\vec v \in A$, $0 \cdot \vec v = \vec 0 \notin A$, and also because
2. 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.