# Two different definitions of vector space

I have two different linear algebra books and realized that the definitions of vector space on them are slightly different.

One of the definition has the following statement for the condition of scalar multiplication and the other does not:

"For all $a,b \in \mathbb F, u \in V$ implies $(ab)u=a(bu)$."

I cannot derive it from the other conditions in the definition of vector space nor I can give the example that satisfies other definitions of vector spaces but not this particular one.

http://en.wikipedia.org/wiki/Vector_space#Definition
One of my textbooks has the same definition as Wikipedia. and the definition on the other book is the same as this except that it does not have "For all a,b∈F,u∈V implies (ab)u=a(bu)."

Would you help me figure out if the two definitions are the same or not?

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If you want us to say something about it, you should include the other axioms and/or give the text books you are citing. –  Julian Kuelshammer Mar 11 '13 at 16:02
The axiom appears ineludible, so it must be there in disguise in the other set of axioms. –  Andreas Caranti Mar 11 '13 at 16:04
en.wikipedia.org/wiki/Vector_space#Definition one of my textbooks has the same definition as wikipedia. and the definition on the other book is the same as this except that it does not have "For all a,b∈F,u∈V implies (ab)u=a(bu)." –  Tengu Mar 11 '13 at 16:08
Please tell us which books. –  lhf Mar 11 '13 at 16:38

The approach below contains an error! See if you can find it.

It appears that your book may have omitted the axiom. However, I have often heard the quip that the final axiom of vector spaces is that Linear Algebra textbook authors turn four axioms into eight. So really, you can get away with defining fewer axioms more carefully, but you still need to basically accept the field axioms as acting on the vector space in the expected way.

Assuming you have the scalar multiplication axiom in both definitions (that is, $a\mathbf{u} \in V$ for $\mathbf{u} \in V, a \in \Bbb F$):

For nonzero $a,b \in \Bbb F$, $$a^{-1}a(b\mathbf{u}) = b\mathbf{u}, \\ b^{-1}b\mathbf{u} = \mathbf{u}.$$ This is justified without associativity. Assume $b^{-1}(b\mathbf{u}) = \mathbf{w} \neq \mathbf{u}.$ Then, $b\mathbf{u}-b\mathbf{u} = \mathbf{0} = b\mathbf{w}-b\mathbf{u}$, so by uniqueness of the additive inverse, $\mathbf{w} = \mathbf{u}$.

Hence $$b^{-1}a^{-1}a(b\mathbf{u}) = \mathbf{u}.$$

But, $b^{-1}a^{-1} = (ab)^{-1}$, so $$b^{-1}a^{-1}(ab)\mathbf{u} = (ab)^{-1}(ab)\mathbf{u} = \mathbf{u},$$

so the two are evidently equal.

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Thank you for your answer! I am wondering what you meant by "uniqueness of the additive identity". –  Tengu Mar 11 '13 at 16:17
@Tengu I meant to write "additive inverse". Basically, one of the axioms states that for every $\mathbf{v} \in V$, there is an additive inverse $-\mathbf{v}\in V$ such that $\mathbf{v}+(-\mathbf{v}) = \mathbf{0}$, and it follows that there is only one such element -- hence it is unique. –  Arkamis Mar 11 '13 at 16:18
However, proving that it is unique would require accepting either associativity or distributivity of scalars, so I am not entirely convinced that this argument isn't wrong. –  Arkamis Mar 11 '13 at 16:19
I think with the uniqueness of additive inverse we can only say bu=bw. –  Tengu Mar 11 '13 at 16:20
Look, it might be me, but... you have $b^{-1} (b u) = w$. You infer $b u = b w$. Ho do you do that? One could multiply $b^{-1} (b u) = w$ by $b$, to get $b(b^{-1}(bu)) = b w$, but then? Your first two formulas are ok, but they really mean $(a^{-1} a) (b u) = b u$. It might be me, though, please clarify. –  Andreas Caranti Mar 11 '13 at 16:21

I believe the associativity axiom $(ab)v = a(bv)$ (more precisely, compatibility of scalar multiplication with field multiplication) to be independent of the others, but please do check my argument below.

Consider the additive group $V$ of the real numbers $\mathbf{R}$.

Now consider a bijective function $\varphi: \mathbf{R} \to \mathbf{R}$, different from the identity $1_{\mathbf{R}}$, that is an automorphism of $\mathbf{R}$ as a $\mathbf{Q}$-vector space, and such that $\varphi(1) = 1$. (There are plenty of choices for such a $\varphi$, just consider a Hamel basis containing $1$.)

What we really need of $\varphi$ is that $\varphi(a+b) = \varphi(a)+\varphi(b)$ for all $a, b \in \mathbf{R}$.

Now define a structure of $\mathbf{R}$-vector-space-without-associativity on $V$ by declaring the product of the scalar $a$ by the vector $v$ (all elements of $\mathbf{R}$, of course) as $$a \cdot v = \varphi(a) v,$$ where RHS is just the usual multiplication in $\mathbf{R}$.

All axioms but associativity are satisfied. The subtlest is probably $$(a + b) \cdot v = \varphi(a+b)v = (\varphi(a) + \varphi(b)) v = \varphi(a) v + \varphi(b) v = a \cdot v + b \cdot v.$$

But associativity does not hold. In fact if $(a b) \cdot 1 = a \cdot (b \cdot 1)$ for all $a, b \in \mathbf{R}$, then $$\varphi(ab) = \varphi(ab) 1 = \varphi(a) ( \varphi(b) 1) = \varphi(a) \varphi(b)$$ for all $a, b \in \mathbf{R}$.

So associativity holds if and only if $\varphi$ is a ring automorphisms of $\mathbf{R}$. However the only such map is the identity, and we have chosen $\varphi \ne 1_{\mathbf{R}}$.

PS A simpler example can be obtained by replacing $\mathbf{R}$ with $\mathbf{C}$, and taking $\varphi(a+bi)=a+2bi$, say, for $a, b \in \mathbf{R}$.

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i checked this just now. nicely done. –  user29743 Mar 11 '13 at 17:10
@countinghaus, thanks. Perhaps I should add that my way of remembering the axioms of an $F$-vector space $V$ is that they are defining a morphism $\varphi$ of rings with identity from $F$ to the ring of abelian group endomorphisms $\operatorname{End}(V)$. So I just constructed a morphism $\varphi$ of abelian groups that sends $1$ to $1$ but it is not a ring morphism. –  Andreas Caranti Mar 11 '13 at 17:14
Without the compatibility of multiplication with the field multiplication, scalar multiplication just gives you a morphism of (Abelian) groups from the additive group of the field to the endomorphisms of the additive group of $V$. So for an example showing that this is not enough, you need such a map where multiplication in the field does not correspond to composition of scalar multiplications. The easiest example I can think of right now is where $V$ is already an honest $\mathbf R$ vector space, and one equips is with a (dishonest) "$\mathbf C$-vector space structure" by having $z\in\mathbf C$ act as scalar multiplication by the real number $\operatorname{Re}z$. This clearly satisfies all the other axioms, but not compatibility of multiplicative structures.
If you want a smaller example, you can play the same trick with the finite field $\mathbf F_4$ "acting" on a $\mathbf F_2$-vector space, multiplication by one element $\alpha\in \mathbf F_4\setminus\mathbf F_2$ being multiplication by $0$, and multiplication by $\alpha+1$ then being the identity.