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Matrix Powers

  1. Apr 18, 2016 #1
    1. The problem statement, all variables and given/known data
    If A100 is some 3x3 matrix, find the base matrix A.

    2. Relevant information
    Eigenvalues, diagonalization, etc.

    3. The attempt at a solution
    So far, I've been finding the eigenvalues and diagonalizing the matrix via A = P-1DP where D is the diagonal matrix and P is a matrix with A's eigenvalues. hoping to find the base matrix with powers of n. Any suggestions to guide me?

    There was a similar discussion to this earlier, trying to reference that as well.
  2. jcsd
  3. Apr 18, 2016 #2


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    I think your method sounds good. What problem are you having with it?
    If A = P-1DP, then A2 = P-1DPP-1DP = P-1D2P, right?
  4. Apr 18, 2016 #3

    Ray Vickson

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    Some matrices are not diagonalizable. What would you do in those cases?
  5. Apr 18, 2016 #4
    After finding the diagonal, the P matrix, and the inverse of P, I'm setting them equal to An (with the numeric digits in the diagonal also set to the nth power). The cross produce of these three matrices supposedly give me the power formula of the base matrix A.

    Example: A 3x3 matrix of {{1,1,1},{2,2,2},{3,3,3}} turns out to be all the same digits to the nth power. I can see how this works moving forward, but it's not giving me the root matrix.
  6. Apr 18, 2016 #5
    Hmm, I'm not quite sure. I did a check to make sure the one I'm dealing with is, but I would be interested in knowing what to do in other cases as well. The book I'm using gives examples on how to go forward from A to A100 via use of patterns, but those matricies are diagonalizable as well.
  7. Apr 18, 2016 #6


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    I think your example may hint at Ray's question.
    With a diagonalizable matrix A=P-1DP, you have found that A100=P-1D100P. And a diagonal matrix raised to a power is the same as each term in that matrix raised to the same power.
    I question whether your example is diagonalizable, since your rows are all colinear.
  8. Apr 19, 2016 #7

    Ray Vickson

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    Here is an example:
    [tex] A = \pmatrix{11/10&19/10&21/10\\
    1/10& 19/10& 11/10\\
    -1/10&-9/10&-1/10} [/tex]
    The eigenvalues of ##A## are ##1,1,9/10##, and there are only two eigenvectors; thus, ##A## is not diagonalizable. In fact, if
    [tex] P = \pmatrix{1&0&1\\0&1&1\\-1&1&-1}, \; P^{-1} = \pmatrix{2&-1&1\\1&0&1\\-1&1&-1} [/tex]
    we have
    [tex] A = P^{-1} J P \;\; \text{where} \;\; J = \pmatrix{1&1&0\\0&1&0\\0&0&9/10} [/tex]
    Here, ##J## is the Jordan canonical form of ##A##.

    We have ##A^n = P^{-1} J^n P##, and ##J^n## is easily computed:
    [tex] J^n = \pmatrix{1 & n & 0\\0 & 1 & 0\\0 & 0 & (9/10^n} [/tex]

    In general, if the Jordan canonical form of a matrix is
    [tex] J = \pmatrix{a & 1 & 0\\0 & a & 0 \\ 0 & 0 & b} [/tex]
    [tex] J^n = \pmatrix{a^n & n a^{n-1} & 0 \\ 0 & a^n & 0 \\ 0 & 0 & b^n} [/tex]
  9. Apr 19, 2016 #8
    Ah, that actually makes sense! Thank you for going through the details.
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