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
  3. Closed under vector addition
  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.


1 Answer 1


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.

  • $\begingroup$ For your first comment on the second requirement following from the fourth, is this another accurate way of showing so: Suppose W is closed under scalar multiplication. Let c be a scalar and v a vector. Then, cv is in W. Thus, 0v = (1 - 1)v = 1v + (-1)v = v + (-1)v = 0. Thus, -v = (-1)v. Also, to clarify why does the fact that -v = (-1)v need to be verified in such a proof? Is it just not stated in any of the axioms for vector spaces? $\endgroup$ Commented May 25, 2022 at 8:51
  • 1
    $\begingroup$ @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. $\endgroup$ Commented May 25, 2022 at 9:12

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