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A vector space is a set V along with an addition on V and a scalar multiplication on V such that the following properties hold, according to S. Axler's Linear Algebra Done Right:

  1. Commutativity
  2. Associativity
  3. Existence of additive identity
  4. Existence of additive inverse
  5. Multiplicative identity
  6. Distributive Properties

Now, by definition, an addition of two elements in the set V should also be a member of V. Using that fact, and (4) from the list above, shouldn't (3) be provable?

Note: A few more moments of thought made me question my statement. What is $0$? Unless (3) is taken, does (4) make any sense? Am I on the right track?

Also, S. Axler says, about subspaces, that they have to satisfy three conditions for them to be considered a subspace (excluding the condition that they have to be a subset of a vector space, which would ensure that the other properties (distributive, commutative) are satisfied $\forall v \space \epsilon\space Subspace$).

  1. $0 \space\epsilon\space Subspace$
  2. Scalar multiplication is closed.
  3. Addition is closed.

In addition to this, he also says we can replace (1) here with a similar condition:

  1. The subspace is non-empty.

He says that since scalar multiplication is closed within the subspace, and that $0v = 0$ (the proof of this involves the additive inverse axiom and the fact that the "Zero" produced by the inverse CAN be added to a vector from $V$; Does this zero have to exist within $V$ for us to define "addition" between this zero and a vector from $V$?) for any $v\space\epsilon\space Subspace$, this would imply that $0\space \epsilon \space Subspace$.

So here, is this replacement possible only because (3) from the vector space axioms defines "zero"?

Note: I tried my best to organize my thoughts and questions, but something seems amiss. I know my questions aren't sequential and coherent but I am struggling to understand which part of this is the head and which one's the tail.

Vishal Subramanyam
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    Yes, 4) makes sense only relative to 3) – Simon Jul 13 '20 at 20:07
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    You could, however, have a weaker version of the inverses axiom without an identity, viz. e.g. $\forall a\exists b\forall c(c+a+b=b+a+c=c)$. – J.G. Jul 13 '20 at 20:10
  • Also yes, Axler's "trick" for formally weakening the axioms for a subspace depends on the fact that 0 exists in the larger space. – Simon Jul 13 '20 at 20:11
  • @Simon but is it necessary that zero should exist in the vector space for any of its properties? – Vishal Subramanyam Jul 13 '20 at 20:12
  • @J.G. But since addition is closed, wouldn't your axiom automatically imply that this "inverse" (a+b) belongs to the vector space? – Vishal Subramanyam Jul 13 '20 at 20:15
  • @Vishal Subramanyam, do you mean do the other axioms imply the existence of a zero, or do you mean is the property of vector spaces that they have a zero ever used ? The answers are (I think) no and (I'm sure) absolutely yes, respectively. – Simon Jul 13 '20 at 20:16
  • @J.G. I think you just condensed two axioms into one. – Vishal Subramanyam Jul 13 '20 at 20:18
  • @VishalSubramanyam But it doesn't imply $a+b$ is the same vector each time, or that it's the same as $b+a$. Well, it depends on your other axioms. But all these examples show the moral of the story is, an axiom's meaning (well, relevance at least) depends on its context, e.g. what other axioms are used. – J.G. Jul 13 '20 at 20:28
  • For the first part, another problem is that you can't say "let $a \in V$, and then define $0 = a + (-a)$", because none of the axioms other than the identity axiom state that the vector space is not empty! This is very much related to the non-empty condition in the second part. Generally, requiring an identity is the existentially quantified assumption for a lot of algebraic objects, which stops the empty set from qualifying. If you removed the identity axiom here, the empty set would indeed become a vector space. – Izaak van Dongen Jul 13 '20 at 20:38
  • See this post for a way to replace 4 with something that is intelligible in the absence of 3, and an explicit proof that each axiom is independent. – Arturo Magidin Jul 13 '20 at 20:55
  • If I mark this as a duplicate of that post, it will automatically close this one because I have a gold linear-algebra badge. Please check that post and see if it answers your question. – Arturo Magidin Jul 13 '20 at 20:57
  • See https://math.stackexchange.com/a/479005/589 – lhf Jul 14 '20 at 00:56
  • @ArturoMagidin, I read your answer to the question you linked. My thoughts were exactly similar to your observations. Thanks. Close this. – Vishal Subramanyam Jul 14 '20 at 19:13

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(1) "Does ${4}$ and the addition of two vectors being in ${V}$ imply ${(3)}$?" No. The definitions are intertwined. In order to define an additive inverse, you need a definition of the identity element, since its definition is based on the identity element. By definition, if you take a vector ${v}$, its additive inverse, ${(-v)}$ is the vector ${\in V}$ such that

$${v + (-v) = 0_V}$$

where ${0_V}$ is the additive identity vector in the vector space. If we have no concept of what ${0_V}$ is, the definition for this additive inverse doesn't really make sense. It's like asking "what does blue mean?" without knowing the concept of a colour.

(2) Exactly. Given any subset of ${V}$, (call it ${U}$. That is ${U\subseteq V}$) it's either empty or non-empty. If it's empty, it's not a subspace (since if it's empty it does not contain ${0_V}$, the zero vector). If it's non-empty, then there must exist at least one vector in the subspace. Take any vector ${u \in U}$. Then we have

$${0_F\times u = 0_V}$$

And since the subspace is closed under scalar multiplication we have that for any non-empty subset of ${V}$ that is closed under scalar multiplication and vector addition that ${0_V}$ exists within this subset. Hence the first condition that ${0_V}$ exists within the subspace can be replaced with it being non-empty. In this context, both statements are equivalent.

Ultimately - you cannot get away without defining ${0}$. If you don't define it - how do you then define additive invertibility?

Edit: As @ArturoMagidin has pointed out in the comments - it is possible to come up with a different set of axioms (by different I mean that you replace $4$) that don't include $0$, but that satisfies all other conditions. This is different from the standard vector space axioms, but is pretty cool - so you should check it out! :D

Riemann'sPointyNose
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    Actually, the comment points to a question in which we replace the inverses axiom with a cancellation axiom, while leaving the $0$ axiom in place, so that you can remove the $0$ axiom and see if you can construct a structure that satisfies all “other” axioms. – Arturo Magidin Jul 13 '20 at 21:39
  • @ArturoMagidin indeed you are right - I misread it. The question poster asks a different question, but as part of the answer another user explained also that the $0$ axiom cannot just be removed from the standard set of axioms. Apologies – Riemann'sPointyNose Jul 13 '20 at 21:42
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    No problem; “another user” being me, as it happens. ;-). – Arturo Magidin Jul 13 '20 at 21:44
  • @ArturoMagidin So it is! Holy moly this is from 2011 I only just noticed! :O that's awesome – Riemann'sPointyNose Jul 13 '20 at 21:46