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Powers of a permutation matrix.

  1. Aug 24, 2009 #1
    1. The problem statement, all variables and given/known data

    If you take powers of a permutation matrix,
    why is some [tex] P^k [/tex] eventually equal to [tex] I [/tex]?

    2. Relevant equations

    -

    3. The attempt at a solution

    From the solutions manual of the book:

    There are n! permutation matrices of order n.

    Eventually, two powers of [tex] P [/tex] must be the same:
    [tex] P^r = P^s [/tex]

    Miltiply [tex] (P^-1)^s [/tex] to find [tex] P^{r-s} [/tex].

    Certainly [tex] r-s \leq n! [/tex]

    I do not quite see how this answers the question.
    I understand that there are n! permutation matrices.

    That two powers of P must be the same is also understandable, since taking powers of P just rearranges the rows. Since there are a finite number of ways to rearrange the rows, two powers will eventually be the same.

    The two other points, I do not understand.

    Would someone be so kind to explain this to me in some detail?

    Thanks
     
  2. jcsd
  3. Aug 24, 2009 #2

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    You are missing two key things here. Eventually two different powers of [itex]P[/itex] must be the same:

    [tex]P^r = P^s\quad, r \ne s[/tex]

    You can assume that [itex]r>s[/itex]. The other thing you are missing is that you need to multiply both sides of the equality by the inverse of [itex]P^s[/itex].

    See if you can take it from here.
     
  4. Aug 24, 2009 #3
    Hi D H,

    Is this what you are leading me towards:
    [tex] P^r(P^s)^{-1} = P^s(P^s)^{-1} [/tex]

    [tex] P^{(r-s)} = P^0 [/tex]

    [tex] P^{(r-s)} = I [/tex]

    Thank you.
     
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