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??

Answer 1

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.

Answer 2

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: $$1*(0,1)=(0,0)=0_V.$$

Answer 3

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.

Answer 4

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 $$ y+x=x+y $$

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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 $a$.