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Linear Algebra: Find a formula for a matrix M to any power

  1. Sep 6, 2014 #1
    1. The problem statement, all variables and given/known data
    sCWkyXV.png


    2. Relevant equations
    A = SDS-1

    Under some specific conditions,

    An=SDnS-1


    3. The attempt at a solution
    det(A-λI) = 0
    (16-λ)(-λ) - (-64)(1) = 0
    λ2 - 16λ + 64 = 0

    λ = 8 Multiplicity 2.

    This is as far as I got because you need 2 eigenvalues to get 2 eigenvectors to create the S/D matrices. Is there another way to solve this, because this is the only way I learned. Also there has to be an answer for this because it is on webwork with fill in the blank answers.
     
  2. jcsd
  3. Sep 6, 2014 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    For an ##n \times n## matrix ##A##, if ##r_1, r_2, \ldots, r_p## are the distinct eigenvalues of multiplicities ##m_1, m_2, \ldots, m_p##, it follows from the Jordan canonical form that there are matrices ##E_{i,k_i}, i = 1, \ldots, p, k_i = 1, \ldots, m_i## such that
    [tex] P(A) = \sum_{i=1}^p [E_{i1} P(r_i) + E_{i2} P'(r_i) + \cdots + E_{i,m_i} P^{(m_i-1)}(r_i) ] [/tex]
    The matrices ##E_{ik}## are fixed, and are the same for all functions ##P##.

    In your case, ##P(M) = E_1 P(8) + E_2 P'(8)##, where the matrices ##E_1, E_2## are the same for any polynomial ##P## (and, in fact, for any analytic function ##f(M)##). You can get them from two known values ##P(M)##, such as for ##P(x) = 1 \Longrightarrow P(M) = I## (the identity matrix) and ##P(x) = x \Longrightarrow P(M) = M##. For ##P(x) = 1, P'(x) = 0## we get ##I = E_1 + 0 E_2## and for ##P(x) = x, P'(x) = 1## we get ##M = 8 E_1 + 1 E_2##. You can solve for ##E_1, E_2##, then get ##M^n = E_1 8^n + E_2\, n 8^{n-1}##.
     
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