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Linear Algebra: 2x2 matrix raised to the power of n

  1. Jun 24, 2016 #1
    1. The problem statement, all variables and given/known data
    If n is a positive integer, then 2x2 matrix [-32,252] [-4,32] raised to the power of n is....

    2. Relevant equations
    I know that first I should diagonalize the given matrix, something I also seem to have a hard time with.

    3. The attempt at a solution
    I determined the eigenvalues to be 32 and -32 and determined eigenvectors to be 16,1 and 252,1 , respectively. I am not sure if this is right or where to even go from there....
     
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  3. Jun 24, 2016 #2

    blue_leaf77

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    I think you are supposed to calculate
    $$
    \left( \begin{array}{cc}
    -32 & 252 \\
    -4 & 32 \\
    \end{array} \right)^n
    $$
    with the help of the diagonal form of the original matrix. By the way, the eigenvalues you got are not right.
     
  4. Jun 25, 2016 #3

    ehild

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    You need not diagonalize the matrix. What is the square of it? What is its cube? Try, it is surprising:smile:.
    To make it easier, pull out the factor 4 from the matrix.
     
    Last edited: Jun 25, 2016
  5. Jun 25, 2016 #4

    Ray Vickson

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    If you have a 2x2 matrix ##A## with two distinct eigenvalues ##r_1, r_2##, then there are two constant 2x2 matrices ##E_1, E_2## such that ##p(A) = E_1 p(r_1) + E_2 P(r_2) ## for any polynomial ##p##, so ##A^n = E_1 r_1^n + E_2 r_2^n##. You can figure out ##E_1## and ##E_2## by applying that to two simple examples of ##p##, such as ##p(x) = x^0 =1 \Rightarrow p(A) = E_1 r_1^0 + E_2 r_2^0 = I## (the identity matrix) and ##p(x) = x \Rightarrow p(A) = A = E_1 r_1 + E_2 r_2##. That gives you two equations in the two "unknowns" ##E_1## and ##E_2##.

    Your eigenvalues are incorrect; start again.
     
  6. Jun 25, 2016 #5
    So, I redid my calculations for the eigenvalues and after taking out the factor of 4 i found -1 and 1, do i need to multiply theses values by the factor of 4 that I took out? There is also a hint that says my answer will be a formula that involves n
     
    Last edited: Jun 25, 2016
  7. Jun 25, 2016 #6

    ehild

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    Yes, the eigenvalues of the original matrix are 4 and -4.
    BUT: try my previous hint. The original matrix is A=4B. A2=42B2. Determine B2. What matrix do you get?
     
  8. Jun 25, 2016 #7
    [1,0],[0,1]
     
  9. Jun 25, 2016 #8
    I was able to solve the problem guys, thanks for your help
     

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  10. Jun 25, 2016 #9

    ehild

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    If B=(1/4)A

    B2=E

    and B3=B.

    So the resulting matrix is 4n E if n is even and 4n B if n is odd.
     
  11. Jun 26, 2016 #10

    Ray Vickson

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    Alternatively, using the method in #4 we have
    [tex] A^n = 4^n E_1 + (-4)^n E_2, \; n= 0,1,2, \ldots , [/tex]
    giving
    [tex] I = E_1+E_2\\
    A = 4E_1 - 4E_2. [/tex]
    Thus
    [tex] E_1 = \frac{1}{8}A + \frac{1}{2}I = \pmatrix{-7/2 & 63/2 \\ -1/2 & 9/2} [/tex]
    and
    [tex] E_2 = -\frac{1}{8} A + \frac{1}{2} I = \pmatrix{9/2 & -63/2 \\ 1/2 & -7/2} [/tex]
     
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