# The zero vector

I am having trouble grasping an intuitive understanding of why a subspace needs to have the zero vector. I understand to satisfy the axioms the zero vector is needed, like: "a vector $v$ is an element of a subspace then $-v$ must be also and $v+(-v) = 0$" but I am not having trouble understanding the properties of a subspace, instead I am having trouble grasping an intuitive understanding of the zero vector. For example why would a line that does not pass through the origin not be considered a subspace of $\Bbb R^n$? I know that line would not satisfy the zero vector but wouldn't that line still satisfy addition and scalar axioms? Can anyone provide me an explicit example to why the zero vector must be contained?

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Have you actually checked if a line which does not go through the origin satisfies the addition axioms? Take $n=2$, pick your favorite line not going through the origin and try it! –  Mariano Suárez-Alvarez Sep 22 '12 at 2:19
You may want to compare the Wikipedia articles on linear subspaces and affine subspaces. –  joriki Sep 22 '12 at 2:21
If you keep the requirement that addition and scalar multiplication are closed and ask for the subspace to be non-empty, then as a consequence zero must be in it. We just include that axiom to get rid of the empty set. –  Javier Sep 22 '12 at 2:27
It is a good idea to have a neutral element to your operation. –  Sigur Sep 22 '12 at 2:52
...and the masses of experts in semi-group theory sighed... –  Mariano Suárez-Alvarez Sep 22 '12 at 2:59

If a subspace is non-empty and satisfies the scalar multiplication axiom, then the zero vector comes "for free", since it is just any nonzero vector multiplied by the zero scalar. (In particular, any non-empty set that doesn't contain zero cannot possibly be closed under scalar multiplication. It's harder, but not all that hard, to show that lines/planes that don't go through the origin aren't closed under addition, either).

Furthermore, if you do not insist on the zero vector axiom, you will find that all the other axioms are conditions on the vectors in the space – e.g. that for every pair of vectors, they have a sum in the space – and hence are trivially satisfied by a set with no vectors. So the zero vector axiom is actually the only axiom that forces a vector (sub)space to be non-empty.

Hence, a vector (sub)space containing a zero vector is equivalent (in the presence of the other axioms) to it containing any vector at all. Since the zero vector is usually the easiest to find anyway, it's convenient to use its existence as an axiom.

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Thank you very much! I see my mistake in my reasoning. I naively assumed a vector to be a point, I don't know what I was thinking. But thank you very much! –  diimension Sep 22 '12 at 4:01