# Clarification on subspace definition.

In the 2nd edition of Hoffman and Kunze's Linear Algebra, four requirements must be met for an arbitrary subset W of a vector space V to be a subspace of V.

1. The 0 vector must be in W
2. a + (-a) = 0 where -a exists for all a
4. Closed under scalar multiplication

Every other resource I've looked at only mentions three requirements (the second requirement is omitted in other resources). My thought is that the 0 vector being in W implies that there is an additive inverse for every vector in W. This comes from the definition of the vector addition operation for a vector space: a + (-a) = 0. Thus, if 0 is in W, 0 = a + (-a), implying the existence of -a for every a. Is this right?

Additionally, it seems that if W is assumed to be a non-empty set, both requirements 1 and 2 no longer need to be checked. Now, I can see requirements 1 and 2 being treated as trivial for a particular case, but I don't see why it would be true arbitrarily, i.e. for any arbitrary set.

Any help would be much appreciated.

The second requirement follows from the fourth, because $$(-1) \mathbf{a} + \mathbf{a} = (-1 + 1)\mathbf{a} = \mathbf{0}$$, so $$(-1) \mathbf{a} = -\mathbf{a}$$.
You are correct that if $$W$$ is non-empty and the final two requirements are satisfied, then $$\mathbf{0} \in W$$. This follows because, for arbitrary $$\mathbf{a} \in W$$, $$-\mathbf{a} \in W$$ as above, and then the third requirement gives $$\mathbf{a} + -\mathbf{a} = \mathbf{0} \in W$$. However, we do need $$W$$ to be non-empty, and the empty set is not a vector space (no additive identity), so it's important to stop it counting as a subspace.
There's an alternative, concise definition of a subspace, which is $$\emptyset \subset W \subseteq V$$, and $$W$$ is closed under arbitrary linear combinations.
• @SillyGoose Yes, that also works. I probably didn't need to show $-\mathbf{v} = (-1) \mathbf{v}$ (or maybe should have shown $0\mathbf{v} = \mathbf{0}$ as well). It's not an axiom, but it is one of the first properties usually proved. Commented May 25, 2022 at 9:12