Definition: A vector space over a field $K$ consists of a set $V$ and two binary operations $+: V \times V \to V$ and $\cdot: K \times V \to V$ satisfying the following axioms:

  1. Commutativity of $+$.
  2. Associativity of $+$.
  3. Existence of an identity element $\mathbf{0}$ for $+$.
  4. Existence of inverses for $+$.
  5. Compatibility of $\cdot$ with multiplication in $K$.
  6. Distributivity of $\cdot$ over $+$.
  7. Distributivity of $\cdot$ over addition in $K$.
  8. $1_K$ is a left identity of $\cdot$.

Question: Are all seven of the previous axioms necessary (in the sense that weakening any one of them permits a structure which is not a vector space)? If not, which can be weakened (or removed)?

EDIT: user7530 has quite cleverly shown that the commutativity of $+$ can be derived from axioms 2-8. Supposing we throw this out, can the remaining axioms all be proven necessary?

EDIT 2: It was pointed out that axiom 3 cannot simply be thrown out, as the definition of an inverse in axiom 4 depends on the existence of $\mathbf{0}$. What if we tweak the statement of axiom 4 to axiom 4': "For every $x \in V$, there exists $y \in V$ such that $(x+y)+x = x$ and $(y+x)+y = y$"? Is this weakened version equivalent to the original, and if so, does it allow the removal of axiom 3?

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    $\begingroup$ I am also interested in this question. My guess is yes they are independent, otherwise all the textbooks would have adopted a weakened version as you said. (Just a guess) $\endgroup$ – Vim Aug 28 '15 at 17:59
  • $\begingroup$ In particular, I am having a hard time coming up with an example of a structure $(V,+,\cdot)$ which satisfies 1-4 and 6-8 but fails to satisfy 5. If anybody has such an example, it would be much appreciated. $\endgroup$ – David Zhang Aug 28 '15 at 18:01
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    $\begingroup$ 1-4 say $(V,+)$ is an abelian group. 6 means that $\cdot$ gives a mapping from $K$ to $End(V)$. 7 means that this mapping is a homomorphism of abelian groups. Having all of 5-8 mean that the mapping is a homomorphism of rings (End(V) is a ring with composition as the product). With this interpretation a structure satisfying 1-4 and 6-8 would be one where this mapping from $K$ to $End(V)$ fails to take a product to a product (at least sometimes). Shouldn't be too hard... $\endgroup$ – Jyrki Lahtonen Aug 28 '15 at 18:13
  • $\begingroup$ (cont'd) Like, take $K=\Bbb{R}$, and let $x_i, i\in I$ be a vector space basis of $K$ over $\Bbb{Q}$. Assume that $0\in I$, and $x_0=1$. If we redefine $\cdot$ to be $$(\sum_{i\in I}q_i x_i)r=q_0r,$$ don't we get a structure obeying everything except 5? $\endgroup$ – Jyrki Lahtonen Aug 28 '15 at 18:17
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    $\begingroup$ @Vim Actually, one axiom is enough: A vector space is an abelian group on which a field acts. :) $\endgroup$ – Hagen von Eitzen Aug 29 '15 at 9:22

I think they are redundant after all! Here's a proof that axiom 1 is redundant. Let $a,b\in V$, and consider $(1+1)\cdot (a+b)$. By axiom 7 and 8, this is equal to $(a+b)+(a+b)$; on the other hand by axiom 6 it is $(1+1)\cdot a + (1+1)\cdot b$, or $(a+a)+(b+b)$ by axiom 7 and 8. We can then use axioms 2, 3, 4 to show that \begin{align*} a^{-1} + (a+b) + (a+b) + b^{-1} &= a^{-1} + (a+a) + (b+b) + b^{-1}\\ b + a &= a + b \end{align*} and $V$ is Abelian.

Necessity of some of the other axioms:

4: Take $V=[0,\infty)$ under multiplication, and $K=\mathbb{R}$, with $z\cdot x \mapsto \begin{cases} x^z, & x\neq 0\\0, &x=0.\end{cases}$

5: Consider $K=\mathbb{C}$, $V=\mathbb{R}$ with $z\cdot x = \Re(z)x$.

6: Necessary once you toss out commutativity. Take $K=F_3$, and $V$ the Heisenberg group over $F_3$, with $z\cdot x = x^z$. Since all elements of $V$ have order dividing 3, axiom 7 is satisfied, but $$\left(\left[\begin{array}{ccc}1 & 0 &0\\0 & 1 & 1\\0 & 0& 1\end{array}\right]\left[\begin{array}{ccc}1 & 1 & 0\\0 & 1 & 0\\0 & 0 &1\end{array}\right]\right)^2 \neq \left[\begin{array}{ccc}1 & 0 &0\\0 & 1 & 1\\0 & 0& 1\end{array}\right]^2\left[\begin{array}{ccc}1 & 1 & 0\\0 & 1 & 0\\0 & 0 &1\end{array}\right]^2.$$

7: Take $K=\mathbb{C}$, $V=\mathbb{R}$, and $z\cdot x = |z|x$.

8: See comment by Jyrki below.

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    $\begingroup$ Good ones! However, it is possible to find structures that obey everything except 8. $K=\Bbb{R}$, $V=\Bbb{R}^2$ with the usual addition but $a\cdot(x,y)=(ax,0)$. $\endgroup$ – Jyrki Lahtonen Aug 28 '15 at 18:20
  • $\begingroup$ Very nice! Is it also possible to find a structure satisfying everything but 6 or everything but 1? $\endgroup$ – David Zhang Aug 28 '15 at 19:04
  • $\begingroup$ This is an interesting development! If we now throw out Axiom 1, can the remaining axioms all be proven necessary? We are sure about 5, 7, 8, and I believe $K = V = \mathbb{R}_{\ge 0}$ shows that 4 is necessary. 2, 3, and 6 remain to be shown. $\endgroup$ – David Zhang Aug 28 '15 at 20:35
  • $\begingroup$ $K$ cannot be $R_{\geq 0}$ but you can take all of $\mathbb{R}$. I will add this one. 2 is likely to be very difficult to analyze. 3 cannot be removed as without it 4 isn't well-defined. $\endgroup$ – user7530 Aug 28 '15 at 20:40
  • $\begingroup$ Oh yes, my mistake. I suppose that leaves 2 and 6 open. $\endgroup$ – David Zhang Aug 28 '15 at 20:44

You won't be able to entirely remove axiom 3, since otherwise $V=\emptyset$ would (vacuously) satisfy the other axioms. However, you can remove axiom 4 if you replace axiom 3 with this slightly stronger version (which I will call axiom 3*):

(Axiom 3*) There exists an element $\mathbb{0'} \in V$ such that for all $x \in V$, $0_K \cdot x = \mathbb{0'}$.

(Here, I use the notation $\mathbb{0'}$ to denote that this is a nonstandard definition of $\mathbb{0}$).

Axiom 3* implies axiom 3 and 4

That this element is an additive identity follows from axioms 6 and 8: we have $$\mathbb{0'} + x = 0_k \cdot x + 1_k \cdot x = (0_k + 1_k)\cdot x = 1_k \cdot x = x.$$

Also, for every $x \in V$, we have $$x + (-1_K)\cdot x = (1_K)\cdot x + (-1_k)\cdot x = (1_K + -1_K)\cdot x = 0_k \cdot x = \mathbb{0'}$$ so each $x \in V$ has as an inverse $(-1_K)\cdot x$.

Axioms 3 and 4 imply axiom 3*

We have that

$$(0_K)\cdot x + x = (0_k + 1_K) \cdot x = x$$

so, denoting the inverse of $x$ by $-x$,

$$(0_K)\cdot x + x + (-x) = x+(-x)$$ $$(0_K)\cdot x + \mathbb{0} = \mathbb{0}$$ $$(0_k) \cdot x = \mathbb{0}$$

  • $\begingroup$ Actually you can formulate Axiom 3* without naming $0'$: (Axiom 3*) $0_K\cdot x=0_K\cdot y$ for all $x,y\in V$. $\endgroup$ – celtschk Aug 29 '15 at 8:27
  • $\begingroup$ @celtschk Only if you assume that $V$ is nonempty. Otherwise, $V=\emptyset$ satisfies your axiom 3*. On the other hand, by asserting that $\exists \mathbb{0'} \in V$, we guarantee that $V$ is nonempty. $\endgroup$ – Strants Aug 30 '15 at 5:39
  • $\begingroup$ Ah, right, I didn't think of that. $\endgroup$ – celtschk Aug 30 '15 at 7:29

Selected progress on the definition of Vector Space:

At 1971, Bryant proved that the commutativity of $\oplus$ can be deduced by other axioms$^{(1)}$.

At 1973, Rigby and Wiegold proved that only 6 axioms are needed$^{(2)}$.

In fact, the set of axioms of a vector space may reduce to only 6 as described:

Definition. A vector space over a field $K$ consists of a set $V$ and two binary operations $\oplus: V\times V→V$ and $\odot: K×V→V$ satisfying the following axioms:

  1. $(a\oplus b)\oplus c=a\oplus (b\oplus c), \forall a,b,c\in V$,

  2. $\lambda\odot(a\oplus b)=(\lambda \odot a)\oplus(\lambda \odot b),\forall a,b\in V, \forall \lambda\in K,$

  3. $(\lambda +\mu)\odot a=(\lambda \odot a)\oplus(\mu \odot a),\forall a\in V,\forall\lambda,\mu\in K,$

  4. $(\lambda\times\mu)\odot a=\lambda\odot (\mu \odot a), \forall a\in V,\forall \lambda,\mu\in K,$

  5. $0\odot a=0\odot b,\forall a,b\in V,$

  6. $1\odot a=a, \forall a \in V.$

We have the follow theorems, in which, the Theorem 1. implies the existence of additive inverse and Theorem 2. implies the commutativity axiom.

Theorem 1. If $V$ satisfies axioms, $1, 3, 5, 6,$ then it have additive inverse.

Hints:$$ a =1\odot a=(1+0)\odot a=(1\odot a)\oplus(0\odot a)=a\oplus z$$ $$z=0\odot a=(1+(-1))\odot a=(1\odot a)\oplus(-1)a=a\oplus((-1)\odot a)$$

Denote the element $(-1)\odot a$ as the additive inverse of $a$.

Theorem 2. If $V$ satisfies axioms, $1,2,3,5,6,$ then it is a commutative additive under $\oplus$.

On counterexamples

The example corresponding to additive inverse axiom described in the answer (@user7530) doesn't satisfy the axioms $2, 3$, too. According to Theorem 1., it certainly failed for additive inverse axiom. Therefore, this example cannot prove the necessary of this axiom.

Last but not least

An important rule that intrinsically holds is VERY often to overlook! I call it Rule 0. as follows:

Rule 0. $\lambda\odot a\in V, \forall a\in V, \forall \lambda\in K. $ (Closure under scalar multiplication)

Consider the following counterexample for being a vector space:

$V=\{(a_1,...,a_n):a_i\in \mathbb{R}, i=1,...,n\}, K=\mathbb{C}$ with the operations of coordinatewise addition and multiplication.

It seems satisfy all the axioms listed above. But pay attention to it that $\exists \lambda\in K,$ s.t. $(\lambda\odot a)\notin V$. Now that so, how can we apply axioms $2,3,$ or $4$?


(1) Bryant, V. (1971). Reducing Classical Axioms. The Mathematical Gazette, 55(391), 38-40. doi:10.2307/3613304

(2) Rigby, J., & Wiegold, J. (1973). Independent Axioms for Vector Spaces. The Mathematical Gazette, 57(399), 56-62. doi:10.2307/3615171


I believe $6$ is indeed indispensible, and may be equivalent to $1$, here's my reasoning:

what $6$ actually says is we have an action of $(F,+)$ upon $(V,+)$, that is, the map $v \mapsto a\cdot v$ (let's call this map $\phi_a$) induces a group homomorphism (the operation being $+$):

$F \to V$ via $a \mapsto \phi_a(v)$ for any fixed $v \in V$.

Indeed, we can relax the vector space axioms to allow $R$ to be a commutative ring with unity, and obtain an $R$-bimodule. Now there is a unique homomorphism $\psi:\Bbb Z \to R$ sending $1 \mapsto 1_R$, and this allows us to define on any $R$-bimodule $M$, a $\Bbb Z$-action by:

$n\cdot m = \psi(n)\cdot m$.

Now the intuitive way to try to impose a $\Bbb Z$-action on a group $G$, is to try to set:

$n\cdot g = g^n$.

However, $g \mapsto g^n$ is an element of $\text{End}(G)$ for all $n \in \Bbb Z$ if and only if $G$ is abelian (the "if" part is obvious, the "only if" can be proved using $n = 2$, which is essentially user7530's argument).


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