Matrices

**Showcase_22** · October 23rd 2008, 10:49 AM

Suppose A and P are n x n matrices, and that P is invertible, and n is a positive integer. Show that $(P^{-1}AP)^n=P^{-1}A^n P$

I think i've sort of got this. I just want someone to check what i've done.

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

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

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

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

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

Is this right?

**Moo** · October 23rd 2008, 10:56 AM

Hello !!

Originally Posted by Showcase_22

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

And how do you know this ??

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

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

Otherwise, you can do it roughly from the formula $(P^{-1}AP)^n=P^{-1}APP^{-1}AP\dots P^{-1}AP$ , because you're not told $P^{-1}AP$ will yield the matrix with the eigenvalues.

**Showcase_22** · October 24th 2008, 01:42 PM

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!

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