Tough Eigenvalue problem

Re: Eigenvalue

Start with A^2. A^2(x)=A(A(x)). What's that in terms of lambda?

 

Re: Eigenvalue

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.

 

Re: Eigenvalue

That's what the problem is asking you to show, isn't it?

 

Re: Eigenvalue

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.

 

Re: Eigenvalue

Induction is the formal way to show it. If you've never heard of it and aren't expected to use it the alternative is 'hand waving'. It's easy enough to prove A^3(x)=lambda^3 x using A^2(x)=lambda^2 x. That makes it easy to show A^4(x)=lambda^4 x, etc, etc. So we see it's true for all n. Induction is just the formal way to state 'etc etc'. It's up to you to decide whether the course requires it.