Strange question regarding eigenvectors / eigenvalues

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
6 replies · 2K views
lubricarret
Messages
34
Reaction score
0

Homework Statement



Suppose that the 2x2 matrix A has eigenvalues lambda = 1,3 with corresponding eigenvectors [2,-1]^T and [3,2]^T. Find a formula for the entries of A^n for any integer n. And then, find A and A^-1 from your formula.

Homework Equations



Ax = lambda X
(P^-1)AP = D
A = PDP^-1

The Attempt at a Solution



I've been trying to figure this out for a long time. I'm really not sure where to start... If anyone could provide some sort of guidance on how to begin this problem... that would be really helpful. I've tried using the above formulas, but I'm just not sure how to get a generalized formula. I could find A and A^-1 individually without the generalized formula, but the problem asks for it; thanks in advance for even the slightest hint.
 
Physics news on Phys.org
D=P^(-1)AP is [[1,0],[0,3]], right? And you have P and P^(-1), I hope. Then isn't A^n=P(D^n)P^(-1)? Since A^n=PDP^(-1)*PDP^(-1)*... n times. D^n is pretty easy to figure out. It's not a strange question at all. It's easy.
 
Hi Dick.

Thanks for your explanation! So, my formula would be:
A^(n) = P(D^n)P^(-1)

P is obviously just
[2 3
-1 2]

and P^(-1) would then just be its inverse, or
[(2/7) (-3/7)
(-1/7) (2/7)]

Then you just use this P, P^(-1) and the given D^n to get the answer of A^n.

So, then with A^(-1), I would just get the inverse of D, and plug it into the formula...?



Thanks!
 
Hi

Thanks. Just for clarification though, why would I not say

A^(n) = (PDP^(-1))^n

Why is it just D that I raise to the n, and not the whole right of the equation?
 
Because they are both the same thing and it's much easier to raise D^n than A^n. Take A^3. A^3=(PDP^(-1))*(PDP^(-1))*(PDP^(-1)). Don't you see how the P^(-1)*P parts cancel in the middle?
 
Ahhh, ok. Thanks for clearing that up for me; understand it now.

Thanks!