Question

Let (v, +, *) be a vector space. Show that if \[v \in V\] satisfies v'+v=v' for all \[v' \in V\] then v=0, the additive identity.

Is it enough to say that
a) because (v,+,*) is a vector space, then there exists a 0 in V s.t. for all v in V, v+0=0,
b) subtract v' from both sides showing v=0, therefore ... Im not sure.

This is for a linear algebra class, not real analysis so I'm pretty sure this should be done with the properties of a vector space, and not more formal methods.. any suggestions on what is rigorous enough?

Answer 1

whoops, a) should be "exists a 0 in V s.t. for all v in V, v+0=v"

Answer 2

The condition they give is the definition of an additive identify. So v is an additive identity. Now you have to show it is *the* additive identity 0

Answer 3

Ok, so the definition of THE additive identity that I have is: There exists 0 in V s.t. for all v in V, v+0 = v.

Since the problem states that if v satisfies the equation for all v' in V, shouldn't that be enough? I guess Im having trouble wrapping my brain around whatever step I am missing in this.

Answer 4

In "theory" there could be 2 additive identities: 0 and v (separate and distinct)

but you could say
v+0 = (by the property of v) 0

but 0+v = v (by the property of 0)
we have
v=0
showing that v is in fact 0

Answer 5

Okay, that makes some sense. I didnt see the potential for 2 additive identities. Thanks for the help!

Answer 6

I didnt see the potential for 2
that's because there really isn't any... as we just showed.

Answer 7

Ug, I will have to spend more time wrapping my brain around this, but I do appreciate your time. I see what you mean though, not a potential for a 2nd, but a need to demonstrate the one.