# Proving vector space properties

• Nov 25th 2010, 03:48 PM
Glitch
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!
• Nov 25th 2010, 04:46 PM
Plato

Can you continue with the use of induction?
• Nov 25th 2010, 06:41 PM
Glitch
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).
• Nov 26th 2010, 06:05 PM
Glitch
Is the above correct? Thanks.
• Nov 27th 2010, 04:53 AM
HallsofIvy
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).