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Math Help - Matrices

  1. #1
    Super Member Showcase_22's Avatar
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    Matrices

    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}<br />
{\lambda_1}&{0}\\ <br />
{0}&{\lambda_2}<br />
\end{pmatrix}

    Where \lambda_1 and \lambda_2 are eigenvalues.

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

    P^{-1}A^nP=\begin{pmatrix}<br />
{\lambda_1^n}&{0}\\ <br />
{0}&{\lambda_2^n}<br />
\end{pmatrix}

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

    Is this right?
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  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
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    Hello !!
    Quote Originally Posted by Showcase_22 View Post
    P^{-1}A^nP=\begin{pmatrix}<br />
{\lambda_1^n}&{0}\\ <br />
{0}&{\lambda_2^n}<br />
\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.
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  3. #3
    Super Member Showcase_22's Avatar
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    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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