1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Invertible Matrix

  1. Feb 14, 2006 #1
    "Let A be an invertible matrix with entries in Z_p. Show that A is diagonalizable if and only if its order (the least t such that A^t=1 in GL_n(Z_p)) divides p-1."

    I got the => direction, but I'm having trouble with the backwards direction. Any hints?
     
  2. jcsd
  3. Feb 14, 2006 #2

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    EDIT: I originally made three posts, but I'll put them all in one:

    ----------------------------------------------------------------

    POST 1:

    Suppose A is diagonalizable but it's order does not divide p-1. Let D = (dij) be the corresponding diagonal matrix. Then use:
    - Fermat's Little Theorem
    - the fact that D is diagonal
    - the fact that A and D are similar
    - then use the division algorithm together with assumption that the order of A does not divide p-1 to derive a contradiction which essentially says "if t is the order of A, i.e. if t is the least positive natural such that At = 1, then there exists a t' such that 0 < t' < t but such that At' = 1"

    EDIT TO POST 1: Oops, I guess that's the direction you already proved. I'll have to think some more.

    ----------------------------------------------------------------

    POST 2:

    Just throwing out some ideas:

    1) the characteristic polynomial of a diagonal matrix splits (it's irreducible factors are all linear)
    2) matrices with degree dividing p-1 form a normal subgroup of GLn(Zp) - maybe the orbit-stabilizer theorem or the class equation can be used here (you want to show that every matrix whose order divides p-1 contains a diagonal matrix in its conjugacy class).

    ----------------------------------------------------------------

    POST 3:

    I'm rusty on the linear algebra, but how about this:

    The order of A divides p-1
    implies
    The minimal polynomial of A is xt - 1, where t is the order of A
    implies
    The minimal polynomial of A splits (since xt - 1 = 0 has solutions in Zp iff t | p-1)
    implies
    The char poly of A splits
    implies
    A is diagonalizable (I think there's a theorem showing that the char poly splits iff A is diagonalizable).

    EDIT TO POST 3: Actually, it wouldn't surprise me if the "implies"s can be changed to "iff"s, but at the same time, it wouldn't surprise me if some of the "implies"s were wrong altogether. It's been well over a year since I did any linear algebra, especially anything to do with diagonalization. And I've never really done any linear algebra over finite fields. So check your theorems in your book, and see if the above proof a) is correct, and b) can be strengthened so the "implies"s can become "iff"s, which would then prove both directions of the theorem simultaneously, and then get back to me about it.
     
    Last edited: Feb 14, 2006
  4. Feb 14, 2006 #3
    The order of A divides p-1 does not necessarily imply that the minimal polynoial of A is x^t -1, but it does imply that the minimal polynomial of A DIVIDES x^t -1. I'm not sure if your implications still follow; thinking.
     
  5. Feb 15, 2006 #4

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    We know the minimal poly divides x^{p-1}-1

    This now tells us everything about the minimal poly we know. Think what the multiplicities of the eigenvalues can be (think Fermat's Little Theorem). Think back to the other question you posted too about multiplicity one eigenvalues.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Invertible Matrix
  1. Invertible matrix (Replies: 5)

  2. Invertible Matrix (Replies: 1)

  3. Invertible matrix (Replies: 0)

  4. Invertible matrix (Replies: 3)

  5. Invertible matrix (Replies: 11)

Loading...