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Math Help - Computing A^k

  1. #1
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    Computing A^k

    So it gives me this matrix
    A=\left(\begin{matrix} -1 &0 &1\\5 &3 &2\\6 &0 &4\end{matrix}\right)
    And asks me to compute A^k, where k is any positive integer.
    I know I have to use to eigenvectors which I found to be this matrix:
    P=\left(\begin{matrix} -5 &0 &2\\3 &1 &17\\5 &0 &12\end{matrix}\right)
    and the diagonal matrix made out of the eigenvalues:
    D=\left(\begin{matrix} -2 &0 &0\\0 &3 &0\\0 &0 &5\end{matrix}\right)
    Now I know A^k=P*D^k*P^-1 but when I do this I just get something really messy,
    Any help will be appreciated.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Computing A^k

    Quote Originally Posted by farmeruser1 View Post
    Now I know A^k=P*D^k*P^-1 but when I do this I just get something really messy. Any help will be appreciated.
    Firstly check if P^{-1}AP=D or equivalently if AP=PD .
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  3. #3
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    Re: Computing A^k

    I have checked and yes they give the same answer. It is from there I do not know what to do.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: Computing A^k

    Quote Originally Posted by farmeruser1 View Post
    I have checked and yes they give the same answer. It is from there I do not know what to do.
    What is the problem? A^k=P\textrm{diag}((-2)^k,3^k,5^k)P^{-1}=\ldots (only computations).
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  5. #5
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    Re: Computing A^k

    Quote Originally Posted by farmeruser1 View Post
    So it gives me this matrix
    A=\left(\begin{matrix} -1 &0 &1\\5 &3 &2\\6 &0 &4\end{matrix}\right)
    And asks me to compute A^k, where k is any positive integer.
    I know I have to use to eigenvectors which I found to be this matrix:
    P=\left(\begin{matrix} -5 &0 &2\\3 &1 &17\\5 &0 &12\end{matrix}\right)
    and the diagonal matrix made out of the eigenvalues:
    D=\left(\begin{matrix} -2 &0 &0\\0 &3 &0\\0 &0 &5\end{matrix}\right)
    Now I know A^k=P*D^k*P^-1 but when I do this I just get something really messy,
    Any help will be appreciated.
    Not that messy really. For D^k, just write it as \begin{bmatrix} (-2)^k &0 &0\\0 &3^k &0\\0 &0 &5^k\end{bmatrix}. Then you have to find the inverse of P. Again, that should not be too hard, except that the entries will be fractions (with denominator 70). Finally, it's straightforward to multiply the three matrices together.
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  6. #6
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    Re: Computing A^k

    So I just computed D^k(P^-1), and as you can see its getting pretty messy, am I doing it a long way?
    A^k= P* D^k(P^-1)=\left(\begin{matrix} (-2^k)(-6/35) &0 &(-2^k)(1/35)\\(3^k)(-7/10) &3^k &(3^k)(-13/10)\\(5^k)(1/14) &0 &(5^k)(1/14)\end{matrix}\right)
    *So that matrix is just D^k*P^-1, I still have to multiply it by P.
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  7. #7
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    Re: Computing A^k

    Quote Originally Posted by farmeruser1 View Post
    So I just computed D^k(P^-1), and as you can see its getting pretty messy, am I doing it a long way?
    A^k= P* D^k(P^-1)=\left(\begin{matrix} (-2^k)(-6/35) &0 &(-2^k)(1/35)\\(3^k)(-7/10) &3^k &(3^k)(-13/10)\\(5^k)(1/14) &0 &(5^k)(1/14)\end{matrix}\right)
    *So that matrix is just D^k*P^-1, I still have to multiply it by P.
    Well Mr Farmeruser, sometimes a farmer just has to put his boots on and get muddy. Keep going, you're doing it correctly.
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Re: Computing A^k

    Another way: using the euclidean division, x^k=c(x)(x+2)(x-3)(x-5)+ax^2+bx +c\;\;[1] . Using the Cayley-Hamilton theorem, A^k=aA^2+bA+cI\;\;[2] . Substituting x by -2,3,5 in [1] we get the system

    \begin{Bmatrix}4a-2b+c=(-2)^k\\ 9a+3b+c=3^k\\25a+5b+c=5^k\end{matrix}\;\;[3]

    Solving [3] and substituting in [2] we get A^k . This is not necessarily a faster method but has the advantage that is also valid for non diagonalizable matrices (differentiating [1] conveniently).
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