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Thread: matrix exponents

  1. #1
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    matrix exponents

    Calculate limit of A^n for large n where A=[1/2 0 0; 1/2 1 3/4; 0 0 1/4]

    i have found eigenvalues but i cant find the eigenvectors
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    You should try to compute $\displaystyle A^2,\,A^3,\,A^4$ to see what happens and then you'll be able to show by induction what is $\displaystyle A^n$ in function of $\displaystyle n$.
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    I expect that the eigenvalues that you got were 1, 0.5 and 0.25?

    The eigenvectors are found by solving

    [0.5-y 0 0;0.5 1-y 0.75;0 0 0.25-y]x=0 (where y is an eigenvalue)

    or in matlab by typing

    [V,D]=eig(A)

    To give the three eigenvectors:

    [0 1 0], [0.707 -0.707 0], [0 -0.707 0.707]

    Now A = V*D*inv(V) and it is easy to calculate A^n
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  4. #4
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    Quote Originally Posted by Jojo123 View Post
    Calculate limit of A^n for large n where A=[1/2 0 0; 1/2 1 3/4; 0 0 1/4]

    i have found eigenvalues but i cant find the eigenvectors
    Where's your trouble here?

    Let the eigenvector be $\displaystyle \left( \begin{array}{c}
    a \\
    b \\
    c \end{array} \right)$.

    $\displaystyle {\color{red}\lambda = 1}$: $\displaystyle \frac{a}{2} = a \Rightarrow a = 0$, $\displaystyle b = b$, $\displaystyle \frac{c}{4} = c \Rightarrow c = 0$.

    Therefore the eigenvector has the form $\displaystyle \left( \begin{array}{c}
    0 \\
    b \\
    0 \end{array} \right)$ and a normalised eigenvector is $\displaystyle \left( \begin{array}{c}
    0 \\
    1 \\
    0 \end{array} \right)$.



    $\displaystyle {\color{red}\lambda = \frac{1}{2}}$: $\displaystyle \frac{a}{2} = \frac{a}{2} \Rightarrow a = a$, $\displaystyle \frac{c}{4} = \frac{c}{2} \Rightarrow c = 0$, $\displaystyle \frac{a}{2} + b = \frac{b}{2} \Rightarrow b = -a$ (since c = 0).

    Therefore the eigenvector has the form $\displaystyle \left( \begin{array}{c}
    a \\
    -a \\
    0 \end{array} \right)$ and a normalised eigenvector is $\displaystyle \frac{1}{\sqrt{2}} \left( \begin{array}{c}
    1 \\
    -1\\
    0 \end{array} \right)$.


    You should try finding the eigenvector for $\displaystyle \lambda = \frac{1}{4}$. A normalised eigenvector is $\displaystyle \frac{1}{\sqrt{2}} \left( \begin{array}{c}
    0 \\
    -1\\
    1 \end{array} \right)$.


    So now you can easily diagonalise your matrix (well, you still have to get the inverse of a 3 x 3 matrix but that's just a mechanical process) and then easily raise to any power you want.
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  5. #5
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    Quote Originally Posted by Kiwi_Dave View Post
    I expect that the eigenvalues that you got were 1, 0.5 and 0.25?

    The eigenvectors are found by solving

    [0.5-y 0 0;0.5 1-y 0.75;0 0 0.25-y]x=0 (where y is an eigenvalue)

    or in matlab by typing

    [V,D]=eig(A)

    To give the three eigenvectors:

    [0 1 0], [0.707 -0.707 0], [0 -0.707 0.707]

    Now A = V*D*inv(V) and it is easy to calculate A^n Mr F adds: to two (maybe three) decimal places of accuracy.
    ..
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