# Eigenvalue problem

• Apr 4th 2010, 03:20 PM
temaire
Eigenvalue problem
http://img28.imageshack.us/img28/5227/79425145.jpg

This is my attempt at a 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 statement for general N>0 follows in the same way.

Am I missing anything?
• Apr 4th 2010, 04:46 PM
tonio
Quote:

Originally Posted by temaire
http://img28.imageshack.us/img28/5227/79425145.jpg

This is my attempt at a 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 statement for general N>0 follows in the same way.

Am I missing anything?

Not really: now just round up the proof doing some induction.

Tonio
• Apr 4th 2010, 04:51 PM
temaire
Quote:

Originally Posted by tonio
Not really: now just round up the proof doing some induction.

Tonio

I haven't learned what induction is. Or maybe its a word for something I already know. What do you mean by it?
• Apr 4th 2010, 07:52 PM
vincent
A proof by induction works like this: first show your statement holds for $\displaystyle n=1$ or $\displaystyle 0$ or the first $\displaystyle n$ for which it is true, then assume it to be true for $\displaystyle n=m$ (this is your induction hypothesis) and then show it to be true for $\displaystyle n=m+1$ using your hypothesis. if $\displaystyle \lambda$ is an eigenvalue of$\displaystyle A$, $\displaystyle Ax=\lambda x$ and $\displaystyle A^2x=A\lambda x =\lambda Ax=\lambda^2 x$ so $\displaystyle \lambda^n$ is an eigenvalue of $\displaystyle A^n$ for $\displaystyle n=2$. suppose it is true for $\displaystyle n=m$,then $\displaystyle A^{m+1}x=AA^mx=A\lambda^mx=\lambda^mAx=\lambda^{m+ 1}$
it is essentially what you said but a a bit more rigorous
induction is a really useful tool, you should practice using it