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Diagonalising Matrices / Recurrence Relations Watch

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    For parts (a) and (b) I've found the eigenvalues to be -\frac{1}{3} and -1 with corresponding eigenvectors \begin{bmatrix} -1 \\ 3 \end{bmatrix} and \begin{bmatrix} -1 \\ 1 \end{bmatrix} respectively.

    Now for part (c) I know there is a way of solving this by diagonalising matrices but I can't remember the method.

    The recurrence relation can be written as \begin{bmatrix} a_n \\ a_{n-1} \end{bmatrix} = \begin{bmatrix} -\frac{4}{3} & -\frac{1}{3} \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{n-1} \\ a_{n-2} \end{bmatrix}

    We can diagonalise A = \begin{bmatrix} -\frac{4}{3} & -\frac{1}{3} \\ 1 & 0 \end{bmatrix} by:

    letting D = \begin{bmatrix} -\frac{1}{3} & 0 \\ 0 & -1 \end{bmatrix} and P = \begin{bmatrix} -1 & -1 \\ 3 & 1 \end{bmatrix} so that we have A= PDP^{-1}

    Now how do I find a_n from here?

    EDIT: I see that \begin{bmatrix} a_n \\ a_{n-1} \end{bmatrix} = A^{n-1} \begin{bmatrix} a_1 \\ a_0 \end{bmatrix}
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    Assuming you're aiming to get a matrix A to the power n:

    Find P such that:
    P^-1 A P = D, where D is diagonal. Then:
    A = (P D P^-1) and:
    A^n = (P D P^-1)(P D P^-1)(P D P^-1)....

    The P^-1 P terms cancel and you're left with:
    A^n = P D^n P^-1

    ... and this is much easier to calculate as D^n is trivial
 
 
 
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