Vector Spaces: Redundant Axiom?

Question

Question

Why are the axioms for vector space independent?

More precisely $1x=x$ seems redundant...

(I take the axioms from: Wikipedia)

Explanation

One has for zero vector: $$\lambda0+\lambda0=\lambda(0+0)=\lambda0\implies\lambda0=0$$

And for zero scalar: $$0x+0x=(0+0)x=0x\implies0x=0$$

In familiar form: $$\lambda x=0\implies\lambda=0\lor x=0$$

Threrefore one calculates: $$1(1x+x^{-1})=1(1x)+1(x^{-1})=(11)x+1(x^{-1})=1x+1(x^{-1})=1(x+x^{-1})=10=0$$

Hence for nontrivial field: $$1\neq0\implies1x+x^{-1}=0\implies1x=x$$

But where is the flaw in that check??

What exactly do you mean by saying two axioms are "independent"? — Gregory Grant, May 15 '15 at 20:42
@GregoryGrant I'm assuming what he means to ask is if we get $1v = v$ for free from the other axioms. — Cameron Williams, May 15 '15 at 20:43
@GregoryGrant: Oh that may be wrong phrasing. I mean it follows from the others. — C-star-W-star, May 15 '15 at 20:43
What is $x^{-1}$? In a vector space we do not necessarily have multiplicative inverses. Do you mean $-x$? — Gregory Grant, May 15 '15 at 20:44
While there are many ways to axiomatize the notion of vector spaces, I have never seen $\lambda x=0\implies \lambda =0\lor x=0$ listed as an axiom of vectorr spaces. - By the way, you need only a single axiom: A vector pace $V$ over a field $F$ is an abelian group on which $F$ acts as uintary ring. — Hagen von Eitzen, May 15 '15 at 20:46
@CameronWilliams: I especially chose this notation to be careful not assuming unproven identities like $-x=(-1)x$. — C-star-W-star, May 15 '15 at 20:47
@Freeze_S $-x$ is the notation used for the inverse in an abelaian group where the group operation is symbolized by $+$. — Hagen von Eitzen, May 15 '15 at 20:48
I suspect that your first property ($0\lambda = 0$) is actually using the fact that $1\lambda = \lambda$. — Cameron Williams, May 15 '15 at 20:49
@HagenvonEitzen: Yes I know this is common notation but it may tempt to abuse $(-1)x=-x$. — C-star-W-star, May 15 '15 at 20:50
@CameronWilliams I don't think he did, it's true in an abelian group that $a+b=a+c$ $\Rightarrow$ $b=c$. Where did he use $1\lambda=\lambda$? — Gregory Grant, May 15 '15 at 20:52
The listed axiom are not independent: indeed, commutativity of addition is a consequence of the other axioms. — egreg, May 16 '15 at 12:37

score 29 · Accepted Answer · answered May 15 '15 at 20:52

29

The axiom system you quote does not have $$\lambda x=0\implies \lambda=0 \lor x=0 $$ as an axiom.

If we drop the axiom $1v=v$, the following becomes an example of a "vector space" over $\mathbb R$:

$V=(\mathbb R,+)$, $F=(\mathbb R,+,\cdot)$
for $\lambda \in F$ and $v\in V$ let $\lambda v=0$.

We do not want this to happen.

answered May 15 '15 at 20:52

Hagen von Eitzen

374,180

I think you may have nailed it. – Gregory Grant May 15 '15 at 20:53
But I didn't assume this, I prove it! – C-star-W-star May 15 '15 at 20:53
@Freeze_S As your "proof" would be applicable to my counterexample, it must be wrong. Do you see where? – Hagen von Eitzen May 15 '15 at 20:54
Hmm yes that is what I actually need your help to find a flaw... – C-star-W-star May 15 '15 at 20:55
4

@GregoryGrant Sometimes a simpe counterexample is simpler than a seventeen comment discussion :) – Hagen von Eitzen May 15 '15 at 20:55
1

@Freeze_S Simple. To conclude from $\lambda0=0$ and $0x=0$ that $\lambda x=0\implies \lambda=0\lor x=0$ is a non-sequitur. For example in the ring $\mathbb Z/4\mathbb Z$ we have $\lambda 0=0$ and $0x=0$ for all $\lambda, x$, but $\lambda x=0\not\implies \lambda=0\lor x=0$ because $\lambda=x=2+4\mathbb Z$ is a counterexample. – Hagen von Eitzen May 15 '15 at 20:56
2

@Freeze_S You did not prove $\lambda x=0\implies \lambda=0$ or $x=0$. You proved the other implication. – Mike Earnest May 15 '15 at 20:57
@Freeze_S You showed $\lambda0=0$ and $0x=0$, you did not show $\lambda x=0$ $\Rightarrow$ $\lambda=0$ or $x=0$. – Gregory Grant May 15 '15 at 20:57
Aaaaahh thanks yes that was the flaw. :) – C-star-W-star May 15 '15 at 21:01
@HagenvonEitzen Why don't we "want" such a thing to happen as in your answer? – Aniruddha Deshmukh Mar 13 '19 at 15:18

score 24 · Answer 2 · answered May 15 '15 at 20:57

A mistake: You only showed that $\lambda=0$ or $x=0$ $\implies$ $\lambda x=0$. You did not show the reverse implication: $\lambda x=0 \implies \lambda=0$ or $x=0$. A proof of that implication uses the axiom $1x=x$.

A standard counterexample of a structure that satisfies all the other axioms save $1x=x$ is the following:

$V=\Bbb{F}^2$
$(x_1,x_2)+(y_1,y_2)=(x_1+y_1,x_2+y_2)$, i.e. the usual componentwise vector addition
$a*(x_1,x_2)=(ax_1,0)$.

The system $(V,+,*)$ satisfies all the other axioms but $1x=x$. Note that that reverse implication does not hold in this system: $$1*(0,1)=(0,0)=0_V.$$

Mike Earnest · Answer 3 · 2015-05-16T22:00:37.070

In order to prove $$ \lambda x=0\implies \lambda=0\vee x=0\tag1 $$ you need to use the axiom $1x=x$.

Here is how you prove (1): if $\lambda\neq0$, then $\lambda x=0$ implies $$ \lambda^{-1}(\lambda x)=\lambda^{-1}(0) $$

$$ (\lambda^{-1}\lambda)x= 0 $$ $$ 1x =0 $$ $$x=0$$ so $\lambda\neq0$ implies $x=0$, or equivalently, $\lambda=0\vee x=0$.

Thus, your proof of the axiom $1x=x$ being redundant goes in circles.

egreg · Answer 4 · 2015-05-16T15:38:16.847

The axioms of the list you cite are not independent, but $1x=x$ is not the problem.

Commutativity of addition follows from the other axioms; let $x,y\in V$ and set $$ z=(1+1)(x+y) $$ Then $$ z=1(x+y)+1(x+y)=x+y+x+y $$ (parentheses can be omitted because of associativity). On the other hand $$ z=(1+1)x+(1+1)y=1x+1x+1y+1y=x+x+y+y $$ Therefore, being $V,+$ a group, we can do $$ (-x)+x+y+x+y+(-y)=(-x)+x+x+y+y+(-y) $$ which gives $$ y+x=x+y $$

An correct objection would be that, removing the commutativity of addition axiom, only right zero and right opposites are assumed. However a general result about monoids applies.

Let $M$ be a (multiplicative) semigroup, with right identity $e$. If every element has a right $e$-inverse, then every element has a left $e$-inverse and the right identity $e$ is also a left identity.

The assumption is that $ae=a$, for all $a\in M$, and that, for all $a\in M$, there exists $b\in M$ such that $ab=e$.

Let $a\in M$ and $b\in M$ such that $ab=e$. Then $bab=be=b$, so, if $c\in M$ and $bc=e$, we have $babc=bc$, hence $ba=e$. Moreover, $ea=aba=a$. Thus $e$ is also a left identity and $b$ a left inverse of $a$.

The commutativity axiom is indeed redundant, but your proof is incomplete. The list only postulates the existence of right zero and right inverses, not left zero or left inverses, so you need to prove those. — Emil Jeřábek, May 16 '15 at 15:21

Vector Spaces: Redundant Axiom?

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