# Matrices

• Oct 23rd 2008, 10:49 AM
Showcase_22
Matrices
Quote:

Suppose A and P are n x n matrices, and that P is invertible, and n is a positive integer. Show that $\displaystyle (P^{-1}AP)^n=P^{-1}A^n P$
I think i've sort of got this. I just want someone to check what i've done.

$\displaystyle P^{-1}AP=\begin{pmatrix} {\lambda_1}&{0}\\ {0}&{\lambda_2} \end{pmatrix}$

Where $\displaystyle \lambda_1$ and $\displaystyle \lambda_2$ are eigenvalues.

Therefore $\displaystyle (P^{-1}AP)^n=\begin{pmatrix} {\lambda_1}&{0}\\ {0}&{\lambda_2} \end{pmatrix}^n=\begin{pmatrix} {\lambda_1^n}&{0}\\ {0}&{\lambda_2^n} \end{pmatrix}$

$\displaystyle P^{-1}A^nP=\begin{pmatrix} {\lambda_1^n}&{0}\\ {0}&{\lambda_2^n} \end{pmatrix}$

and since they are the same then the proof is complete.

Is this right?
• Oct 23rd 2008, 10:56 AM
Moo
Hello !!
Quote:

Originally Posted by Showcase_22
$\displaystyle P^{-1}A^nP=\begin{pmatrix} {\lambda_1^n}&{0}\\ {0}&{\lambda_2^n} \end{pmatrix}$

And how do you know this ??

If it's not a formula from your notes, then you can prove it by induction :)

$\displaystyle P^{-1}A^{n+1}P=P^{-1}A^n AP=P^{-1}A^n \underbrace{P P^{-1}}_{=\text{Id}}A P=(P^{-1}A^nP)(P^{-1}AP)$

Otherwise, you can do it roughly from the formula $\displaystyle (P^{-1}AP)^n=P^{-1}APP^{-1}AP\dots P^{-1}AP$, because you're not told $\displaystyle P^{-1}AP$ will yield the matrix with the eigenvalues.
• Oct 24th 2008, 01:42 PM
Showcase_22
Great response as ever Moo!

I was reading ahead in my notes and couldn't remember how this bit worked. I just got shown today and thought it was pretty good.

Thanks!