1. Proving vector space properties

The question:

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

My attempt:

a) Let v be an element of vector space $V^n$.

2v = $2\begin{pmatrix} x_1 \\ x_2 \\ ...\\ x_n \end{pmatrix}$ = $\begin{pmatrix} 2x_1 \\ 2x_2 \\ ...\\ 2x_n \end{pmatrix}$

v + v = $\begin{pmatrix} x_1 \\ x_2 \\ ...\\ x_n \end{pmatrix}$ + $\begin{pmatrix} x_1 \\ x_2 \\ ...\\ x_n \end{pmatrix}$ = $\begin{pmatrix} 2x_1 \\ 2x_2 \\ ...\\ 2x_n \end{pmatrix}$

Therefore 2v = v + v

Is this correct, or am I approaching this the wrong way? Thanks!

Can you continue with the use of induction?

3. 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).

4. Is the above correct? Thanks.

5. In your initial problem nothing is said about the vector space being of finite dimension yet when you introduce " $V^n$" 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).