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Math Help - Calculating a matrix to a high power using Cayley-Hamilton theorem

  1. #1
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    Calculating a matrix to a high power using Cayley-Hamilton theorem

    Hi,

    Does anyone know how I can use the Cayley-Hamilton theorem to compute a high power of a matrix, for instance

    A = [0, 2, 0; 1, 1, -1; -1, 1, 1]

    A^1000 = ?

    I know the theorem states that plugging the matrix A into its own characteristic polynomial will result in the 0 matrix, but I don't yet see how this can be used to solve this problem.
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  2. #2
    MHF Contributor

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    Did you actually calculate the characteristic equation for this matrix? It is -\lambda^3- 2\lambda^2= 0. That is the same as saying that \lambda^3= -2\lambda^2 and so A^3= -2A^2. There is not that much difference between a cube and a square so this is tedious but doable.

    1000/3= 333 and 1/3 so A^{1000}= (A^3)^333(A)= -2^{333}(A^2)^{333}(A)= -2^{333} A^{334}. Now 334/3= 111 and 1/3 so A^{334}= (A^3)^{111}(A)= (-2A^2)^{111}(A)= -2^{111} A^{223}.

    So far we have that [tex]A^{1000}= 2^{444}A^{223}. 223/3= 74 and 1/3 so A^{223}= (A^3)^{74}(A)= (-2A^2)^{74}(A)= 2^{74}A^{159}: A^{1000}= 2^{518}A^{159}.

    159/3= 53 so A^{159}= (A^3)^{53}= (-2A^2)^{53}= -2^{53}A^{106}. A^{1000}= -2^{571}A^{106}.

    106/3= 35 and 1/3 so A^{106}= (A^3)^{35}(A)= (-2A^2)^{35}(A)= -2^{35}A^{71}. A^{1000}= 2^{605}A^{71}.

    Keep doing that until you have reduced this to a number times A^2.
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