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Period of a matrix

  1. Dec 2, 2011 #1
    can someone please tell me that "how period of a matrix can be determined?"
     
  2. jcsd
  3. Dec 3, 2011 #2

    chiro

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    Hey helen01 and welcome to the forums.

    Can you provide some context to your question? Are you talking about periodicity with respect to Markov chains? Is it some other area? How do you define periodicity?
     
  4. Dec 3, 2011 #3
    A square matrix A for which A power k+1=A (where k being a positive integer) is called periodic. I did not understand this definition of periodicity. I want to know that if i have a matrix and i have to determine a period then what i have to do?
     
  5. Dec 3, 2011 #4

    HallsofIvy

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    Looks like Markov chains or, more generally, "discrete dynamics". We are thinking of the sequence A, A2, A3, ...

    If, for some k, Ak+1= A, then set of matrices A, A2, ..., Ak repeats over and over again. That sequence is "periodic" with period k. In order that A be invertible (otherwise, each Ak(V) has lower dimension than the previous and we can never get A again) and we must have [itex]A^k= I[/itex]. That, in turn, means that we must have [itex]\lambda_i^{n_i}= 1[/itex] for every eigenvalue [itex]\lambda_i[/itex] and corresponding [itex]n_i[/itex]. The period of A is the least common multiple of all the [itex]n_i[/itex].
     
  6. Dec 3, 2011 #5

    I like Serena

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    I believe A does not have to be invertible.
    An eigenvalue of zero is also perfectly acceptable.
    Consider for instance [ 0 0; 0 1].

    I'd say the period of A is the lowest k such that for each eigenvalue λi of A holds that (λi)k+1i.
    Btw, an extra condition is that A is diagonalizable (I'll explain if you're interested).
    This includes the possibility for eigenvalues to be zero, in which case A is not invertible, but it is periodic.
     
  7. Dec 3, 2011 #6
    I can't understand what is k? when solving the matrix what should be the value of k?
    2 3 0
    4 9 3
    1 2 6
     
  8. Dec 3, 2011 #7

    I like Serena

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    Your matrix A needs to be square, so you can multiply it by itself.
    When you repeatedly multiply A by itself, and it comes out back as A, you have found that it is periodic.

    The matrix you show is not square, so periodicity for it is not defined.

    Edit: Ah, I see you made it square now.
     
    Last edited: Dec 3, 2011
  9. Dec 3, 2011 #8
    How many times i multiply? is there any limit?
     
  10. Dec 3, 2011 #9

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    There's no upper limit.

    Consider for instance the matrix for rotation over an angle of 1 degrees.
    You need to apply it 361 times before it comes back to itself.
    Its period is 360.

    What you can do, is determine the eigenvalues of the matrix, and deduce from them whether the matrix is periodic and what the period will be.
     
  11. Dec 3, 2011 #10

    micromass

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    Are you familiar with characteristic and minimal polynomials.
     
  12. Dec 3, 2011 #11

    I like Serena

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    The determinant can give away whether it could be periodic or not.
    Since det(Ak)=det(A)k, you can see that if |det(A)| is different from 0 and 1, then A can not be periodic.
     
  13. Dec 3, 2011 #12
    Ok! as you said that my matrix is not square but i think that it is a square matrix because it has number of rows and columns equal, which is the definition of square matrix.
     
  14. Dec 3, 2011 #13

    Evo

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    Homework must go in the homework section.
     
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