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: $\displaystyle \cdot :F x V \rightarrow V$ defined as scalar multiplication. Under these assumptions, we may call V an F-vector space.

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

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

$\displaystyle f \cdot v = (1f) \cdot v = 1 \cdot (f \cdot v)$

(The last step is a part of the definition of the scalar product.)

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

$\displaystyle v = 1 \cdot v$

By comparing this result to $\displaystyle v = f \cdot v$ 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