# Eigenvalue/vector Proof

1. Nov 11, 2014

### RickStrut95

1. The problem statement, all variables and given/known data
Let A be n x n, λ ∈ ℂ, let v be n x 1, and suppose that A ⋅ v = λ ⋅ v. Show that A^jv = ^j ⋅ v for each positive integer j.

2. Relevant equations

3. The attempt at a solution
I haven't been able to get very far but,

^j ⋅ v - A^jv = 0n x 1
v( ^j - A^j) = 0n x 1

Not sure how to prove that for every positive j that this is true. Any thought would be appreciated. Thanks.

2. Nov 11, 2014

### RUber

What if you tried induction?

3. Nov 11, 2014

### RickStrut95

Where j = 0

v( ^j - A^j) = 0n x 1

v( ^0 -
A^0) = 0n x 1

v(In-1) = 0n x 1

v ⋅ In
- v= 0n x 1

v - v = 0n x 1

Since v is n x 1

0n x 1
= 0n x 1

Because this is true I can assume this for all j?

4. Nov 11, 2014

### RUber

I don't think so, since it would be true for just about anything raised to the zero.
Start with the original form $Av=\lambda v$. This is your base case.
$A^2v=A \lambda v$.
And for any integer...

5. Nov 11, 2014

### RickStrut95

I see what you're saying about
and vice-versa for but I don't see how that helps?
My skills with induction are sub par to say the least.

6. Nov 12, 2014

### RUber

Assume $A^n v = \lambda^n v$ show that $A^{n+1}v=A A^n v = \lambda^{n+1}v$.
The argument for induction is: It works for j=1, assume it works for some j and show it is true for the next j.