Answer to Question #182317 in Algebra for Ziyanda Nyele

Question

Show that in the definition of a vector space V the condition about

existence of additive inverse can be replaced with the condition:

0v = v, for all v is the element of V .

Here the 0 on the left side of the equation is the scalar 0, and on right side of

the equation is the additive identity of V .

Answer 1

1. For all v
2. v, there exists a w
3. w such that v+w=0
4. v+w=0, and
5. For all v
6. v, we have 0v=0
7. 0v=0

are equivalent, given the other axioms: If you have all the other axioms of a vector space, then it doesn't matter which of these two conditions you choose; they'll both deem exactly the same objects to be (or not to be) vector spaces.

You're definitely heading in the right direction, as you've shown one direction: That 1. ⟹

⟹ 2., given the other axioms. This is because, given the existence of such a w

w, you know

0=w+v=0v, hence 0=0v for all v, as desired.

0=w+v=0v, hence 0=0v for all v, as desired.

Now all that's left is showing 2. ⟹

⟹ 1., given the other axioms. So, you'll need to show that, if 0v=0

0v=0 for all v

v, then for all v

v, there exists some w

w such that w+v=0

w+v=0.