Linear algebra -- vector space definiton

[johnk]

Hello everyone,

Knowing the usual definition of a vector space, like the one on wikipedia: http://en.wikipedia.org/wiki/Vector_spa ... definition
I need to prove that a vector space can also be defined by substituting axioms 3 (identity element) and 4 (inverse element)
3: \(\displaystyle \Large \exists 0 \in V, ~\forall v \in V: v+0=v\)
4: \(\displaystyle \Large \forall v \in V~ \exists w, v + w = 0\)

by this one axiom:
5: \(\displaystyle \Large \exists \Theta \in V,~ 0\cdot x = \Theta~~ \forall x \in V\)
(the definitions should be equivalent)

I think I need to prove that axiom 5 (combined with all the other axioms (except 3 and 4 obviously)) implies both 3 and 4, and that 3 and 4 imply 5.
I have a textbook which shows how 3 and 4 imply 5:
\(\displaystyle \large (1+0)\cdot u = 1\cdot u\)
\(\displaystyle \large u + u\cdot 0 = u\)
\(\displaystyle \large u+ u \cdot 0 - u = u - u\)
\(\displaystyle \large u\cdot 0 = \Theta\)

I was also able to prove that 5 implies 3 (which seems easy):
\(\displaystyle \large u\cdot 0 = \Theta\)
\(\displaystyle \large u\cdot 0 +u = \Theta + u\)
\(\displaystyle \large u\cdot (1+0)=\Theta+u\)
\(\displaystyle \large u = \Theta + u\)

But I don't know how to show that 5 implies 4 (existence of an inverse element). Can someone help me?

 

[pka]

Not knowing the depth of rigor required in your course, I will nonetheless make some recommendations and hints.

Do you know or have proven that \(\displaystyle ( - 1)u = - u\)?
Do you know or have proven that \(\displaystyle - u\) is unique?
Have you considered this: \(\displaystyle (0)u = \left( {1 + ( - 1)} \right)u = (1)u + ( - 1)u\)

 

[johnk]

Not knowing the depth of rigor required in your course, I will nonetheless make some recommendations and hints.

Do you know or have proven that \(\displaystyle ( - 1)u = - u\)?
Do you know or have proven that \(\displaystyle - u\) is unique?

We've been shown proofs for both of these for scalars, I suppose the proofs will be similar for vectors. (Actually, now I see that the first one is the key to my problem).

Have you considered this: \(\displaystyle (0)u = \left( {1 + ( - 1)} \right)u = (1)u + ( - 1)u\)

That must be it . I had a problem understanding why -1 must exist without axiom 4, but now realise that 1 is a scalar, so it has an inverse by definition of a field.

Thank you very much, I think I understand it now.

 

[pka]

. I had a problem understanding why -1 must exist without axiom 4, but now realise that 1 is a scalar, so it has an inverse by definition of a field.

In defining a vector space we start with a set and a field. Therefore, we have all the properties of the underlying field that we call the scalar field. Hence you know that –1 exist as a scalar and so (-1)u must exist.