Eigenvalue/vector Induction Proof

[RickStrut95]

Homework Statement

Let A be n x n, λ ∈ ℂ, let v be n x 1, and suppose that A ⋅ v = λ ⋅ v. Show that A^j ⋅ v =

^j ⋅ v for each positive integer j.

Homework Equations

The Attempt at a Solution

I haven't been able to get very far but,

^j ⋅ v - A^j ⋅ v = 0_{n x 1}
v(

^j - A^j) = 0_{n x 1}

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

 

[RUber]

What if you tried induction?

 

[RickStrut95]

Where j = 0

v(

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

v(

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

v(I_n-1) = 0n x 1

v ⋅ I_n - 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?

 

[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...

 

[RickStrut95]

I see what you're saying about

$A^2v=A \lambda v$.
And for any integer...

and vice-versa for

but I don't see how that helps?
My skills with induction are sub par to say the least.

 

[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.