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Using diagonalization to find A^k

  1. Nov 17, 2012 #1
    1. The problem statement, all variables and given/known data

    [tex]A = \begin{pmatrix}
    1 & 4\\
    2 & -1
    \end{pmatrix}[/tex]

    Find [itex]A^n[/itex] and [itex]A^{-n}[/itex] where n is a positive integer.


    2. Relevant equations



    3. The attempt at a solution

    [tex](xI - A) = \begin{pmatrix}
    x-1 & -4\\
    -2 & x+1
    \end{pmatrix}[/tex]
    [tex]det(xI - A) = (x-3)(x+3)
    [/tex]
    [tex]λ_1 = 3\quad λ_2 = -3[/tex]

    [tex]\begin{pmatrix}
    2 & -4\\
    -2 & 4\end{pmatrix}
    \begin{pmatrix}
    a\\
    b
    \end{pmatrix}
    = \begin{pmatrix}
    0\\
    0
    \end{pmatrix}\quad
    \begin{pmatrix}
    a\\
    b
    \end{pmatrix}
    = \begin{pmatrix}
    2\\
    1
    \end{pmatrix}[/tex]
    [tex]\begin{pmatrix}
    -4 & -4\\
    -2 & -2\end{pmatrix}
    \begin{pmatrix}
    c\\
    d
    \end{pmatrix}
    = \begin{pmatrix}
    0\\
    0
    \end{pmatrix}\quad
    \begin{pmatrix}
    c\\
    d
    \end{pmatrix}
    = \begin{pmatrix}
    -1\\
    1
    \end{pmatrix}[/tex]
    [tex]P = \begin{pmatrix}
    λ_1 & λ_2
    \end{pmatrix}
    = \begin{pmatrix}
    2 & -1\\
    1 & 1
    \end{pmatrix}[/tex]
    [tex]P^{-1} = \begin{pmatrix}
    \frac{1}{3} & \frac{1}{3}\\
    \frac{-1}{3} & \frac{2}{3}
    \end{pmatrix}[/tex]
    [tex]D = \begin{pmatrix}
    λ_1 & 0\\
    0 & λ_2
    \end{pmatrix}
    = \begin{pmatrix}
    3 & 0\\
    0 & -3
    \end{pmatrix}[/tex]
    [tex]PD^nP^{-1} = A^n
    = \begin{pmatrix}
    2 & -1\\
    1 & 1
    \end{pmatrix}
    \begin{pmatrix}
    3^n & 0\\
    0 & -3^n
    \end{pmatrix}
    \begin{pmatrix}
    \frac{1}{3} & \frac{1}{3}\\
    \frac{-1}{3} & \frac{2}{3}
    \end{pmatrix}[/tex]
    [tex]= \begin{pmatrix}
    2(3^n) & -(-3)^n\\
    3^n & (-3)^n
    \end{pmatrix}
    \begin{pmatrix}
    \frac{1}{3} & \frac{1}{3}\\
    \frac{-1}{3} & \frac{2}{3}
    \end{pmatrix}[/tex]
    [tex]= \begin{pmatrix}
    \frac{2}{3}(3^n) + \frac{1}{3}((-3)^n) & \frac{2}{3}(3^n) - \frac{2}{3}((-3)^n)\\
    \frac{1}{3}(3^n) - \frac{1}{3}((-3)^n) & \frac{1}{3}(3^n) + \frac{2}{3}((-3)^n)
    \end{pmatrix}[/tex]

    I think that everything I've done so far is correctly, but I can't find any way to simplify this equation any further, and I don't think that I could find [itex]A^{-n}[/itex] with an equation this complicated, so I must be missing something.
     
  2. jcsd
  3. Nov 17, 2012 #2

    lurflurf

    User Avatar
    Homework Helper

    There are different ways of writing that out, you might not have pick the most simple. Another way is something like

    A^n=(3^n)(1/2)((1+(-1)^n)I+(1/3)(1-(-1)^n)A)

    which you can see is lot like yours, note all that (-1)^n stuff is just to unify the even and odd terms

    even A^n=(3^n)I

    odd A^n=(3^n)(1/3)A

    I=A^0 the 2x2 identity matrix
     
  4. Nov 18, 2012 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Frankly, the first thing I would have done would be to note that
    [tex]A^2= \begin{pmatrix}1 & 4 \\ 2 & -1\end{pmatrix}^2= \begin{pmatrix}9 & 0 \\ 0 & 9\end{pmatrix}= 9\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}[/tex]

    From that it follows immediately that if n is even, [itex]A^n= 3^nI[/itex] and if n is odd, [itex]A^n= 3^{n-1}A[/itex]
     
  5. Nov 18, 2012 #4
    Thanks for your responses. Are those equations self-evident or is there a proof that they apply for all values of [itex]n[/itex]?
     
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