# Tough Eigenvalue problem

## Homework Statement

[PLAIN]http://img28.imageshack.us/img28/5227/79425145.jpg [Broken]

## The Attempt at a Solution

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Dick
Homework Helper

Is it lambda^2(x)

Dick
Homework Helper

Is it lambda^2(x)
Sure. So that means x is an eigenvector of A^2 with eigenvalue lambda^2, right? The statement for general N>0 follows in the same way.

So the solution is simply:
Ax = lambda x
Therefore A^n x = lambda^n x ?

Is there something I'm missing?

Dick
Homework Helper

So the solution is simply:
Ax = lambda x
Therefore A^n x = lambda^n x ?

Is there something I'm missing?
That's what the problem is asking you to show, isn't it?

That's what the problem is asking you to show, isn't it?
So what you showed me with A^2 is all I need to answer the question?

Dick
Homework Helper

Yes it is, but is that all there is to it?
If you understand why it's true, then yes, that's all there is to it. If you want to be formal about proving it you might want to present it as an induction proof.

Here is my solution:

A^2(x) = A(A(x))
A^2(x) = lambda^2(x)

Therefore x is an eigenvector of A^2 with eigenvalue lambda^2. The general statement for n>0 follows in the same way.

Is this complete?

Mark44
Mentor

You should probably use induction rather than say "The general statement for n>0 follows in the same way." That's very vague.

You should probably use induction rather than say "The general statement for n>0 follows in the same way." That's very vague.
What do you mean by induction? I've never learned it.

Dick