Why do you have doubts about your proof?
Can you continue with the use of induction?
Prove that the following properties are true for every vector space.
a) 2v = v + v
b) nv = v + ... + v, where there are n terms on the right
a) Let v be an element of vector space .
2v = =
v + v = + =
Therefore 2v = v + v
Is this correct, or am I approaching this the wrong way? Thanks!
Not sure, I guess it's different to the answer in my text which is:
2v = (1 + 1)v = 1v + 1v = v + v
As for part b, there's no answer in the text, but I got this:
P(n) = nv = v + ... + v
P(1) = 1v = v (by axiom 8)
P(n+1) = (n+1)v = nv + v (by axiom 9)
Not sure if that's a proper proof (I've never really done induction proofs before).
In your initial problem nothing is said about the vector space being of finite dimension yet when you introduce " " you are assuming that. Typically "Linear Algebra" classes assume finite dimension but if that is not explicetely said, you should not assume it- and for this proof you don't need to.
The proof your book gives is directly from the "axioms" for a vector space:
For all scalars, a, b, and vectors, u, v, a(u+ v)= au+ av and (a+ b)v= av+ bv.
Also, 1v= v. That is why you can write 2v= (1+ 1)v= 1v+ 1v= v+ v.
Your induction looks good. More formally stated:
For n= 1, we just have v= v which is true.
Assume that, for some k, kv= v+ v+ v+ ...+ v (k times).
Then (k+ 1)v= kv+ v= (v+ v+ ...+ v)+ v= (v+ v+ ...+ v) (k+1 times).