Why are the axioms for vector space independent?

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

(I take the axioms from: Wikipedia)


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??

  • 1
    $\begingroup$ I'm guessing you're using these axioms? $\endgroup$ – Cameron Williams May 15 '15 at 20:42
  • 4
    $\begingroup$ What is $x^{-1}$? In a vector space we do not necessarily have multiplicative inverses. Do you mean $-x$? $\endgroup$ – Gregory Grant May 15 '15 at 20:44
  • 9
    $\begingroup$ @GregoryGrant $x^{-1}$ is (bad) notation for $-x$. $\endgroup$ – Cameron Williams May 15 '15 at 20:45
  • 1
    $\begingroup$ @CameronWilliams Ah got it, yes that is bad. $\endgroup$ – Gregory Grant May 15 '15 at 20:45
  • 1
    $\begingroup$ We're actually using $1a=a$ to deduce $0a=0$. @Freeze_S $\endgroup$ – Alexey Burdin May 15 '15 at 20:46

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.

| cite | improve this answer | |
  • $\begingroup$ I think you may have nailed it. $\endgroup$ – Gregory Grant May 15 '15 at 20:53
  • $\begingroup$ But I didn't assume this, I prove it! $\endgroup$ – C-Star-W-Star May 15 '15 at 20:53
  • 4
    $\begingroup$ @GregoryGrant Sometimes a simpe counterexample is simpler than a seventeen comment discussion :) $\endgroup$ – Hagen von Eitzen May 15 '15 at 20:55
  • 1
    $\begingroup$ @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. $\endgroup$ – Hagen von Eitzen May 15 '15 at 20:56
  • 2
    $\begingroup$ @Freeze_S You did not prove $\lambda x=0\implies \lambda=0$ or $x=0$. You proved the other implication. $\endgroup$ – Mike Earnest 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: $$1*(0,1)=(0,0)=0_V.$$

| cite | improve this answer | |

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.

| cite | improve this answer | |

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 $a$.

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Emil Jeřábek May 16 '15 at 15:21
  • $\begingroup$ @EmilJeřábek Done. $\endgroup$ – egreg May 16 '15 at 15:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.