1. ## Vector space proof

I was doing some homework problems and was proving several facts about vector spaces (and was quite proud at how easily I was doing it) and suddenly noticed my proofs were circular. So I picked one to prove from scratch. I did it, but I want someone to verify that it is, in fact, a proof. Here it goes.

In the following, V is an Abelian group and F is a field. We have a defined function: $\cdot :F x V \rightarrow V$ defined as scalar multiplication. Under these assumptions, we may call V an F-vector space.

Proposition: Given $1 \epsilon F$ and v any vector in V, show that $1 \cdot v = v$.

Proof: I will assume that there exists an $f \epsilon F$ such that $f \cdot v = v$ for all $v \epsilon V$. Existence of such an f will be demonstrated.

Now, there exists an element 1 of F such that 1 is the multiplicative identity (F is a field). We have that 1f = f (for all f) for multiplication in F. Thus:
$f \cdot v = (1f) \cdot v = 1 \cdot (f \cdot v)$
(The last step is a part of the definition of the scalar product.)

We defined above that $f \cdot v = v$. Inserting this in the first and last steps of the previous line we find:
$v = 1 \cdot v$

By comparing this result to $v = f \cdot v$ we see that the required f = 1. Therefore such an f exists and the theorem is proved.
(End of proof.)

It just seems too...convenient I guess...to assume an f, then basically cancel the darned thing out, then state what it is. I can't find an argument against the above being a valid proof, but I'm uncomfortable with it.

Any thoughts? Thanks!

-Dan

2. The proposition you're trying to prove is an axiom of vector spaces, so it really need not be proven. What you can do is prove the UNIQUENESS of the multiplicative identity, if you want to.

3. (Sigh) I looked at the wrong line in my book notes. (I hate it when I do that! ) You're right, that's one of the axioms.

Still, I am unable to find a logical reason to abandon the "proof" in my earlier post, except for the fact that I am uncomfortable with it for some unconscious reason. IS there a problem with it being a proof or not?

Thanks!

-Dan

(And it appears we have a Robert Ludlum fan among us! )

4. Given 1 in F, then 1v=v by definition of scalar multiplication. There's really nothing to prove.

The problem with your proof is that you want to show the existence of the scalar identity, but it already exists because F is a field.

Ludlum rocks!