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Matrix Powers

  1. Nov 13, 2011 #1
    1. The problem statement, all variables and given/known data
    Let B be a matrix with characteristic polynomial λ2-λ√6+3. Evaluate B4.

    2. Relevant equations
    Bn=PDnP-1

    3. The attempt at a solution
    I can find the eigenvalues from the characteristic equation and those would form the diagonal entries of D. But how would I find P, which contains the eigenvectors, if I don't have a matrix?

    Side Note: How can I quickly find an inverse for a 2 by 2 matrix. Is it just dividing the 2x2 matrix by it's determinant, then negating the diagonal entries going from a11 to a22 and swapping a12 and a21?
     
  2. jcsd
  3. Nov 13, 2011 #2

    I like Serena

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    Hi shaon0! :smile:

    Let's first see how far we can get before concluding you have too little information shall we?

    What did you find for the eigenvalues?
    What do you get if you calculate D4?


    On your side note: almost, but you need to swap the diagonal entries a11 and a22, and negate a12 and a21.
    Check by multiplying your matrix with the supposed inverse. It should yield the identity matrix.
     
  4. Nov 13, 2011 #3
    Hi I like Serena :);

    I've found the answer. Thanks for the help, needed another to look at the problem.
     
    Last edited: Nov 13, 2011
  5. Nov 13, 2011 #4

    micromass

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    You don't really need to calculate the eigenvalues or [itex]D^4[/itex] for this.

    Cayley-Hamilton says that

    [tex]B^2=\sqrt{6}B-3I[/tex]

    Thus

    [tex]B^4=B^2*B^2=B^2(\sqrt{6}B-3I)=\sqrt{6}B^3-3B^2=\sqrt{6}B(\sqrt{6}B-3I)-3(\sqrt{6}B-3I)=...[/tex]
     
  6. Nov 13, 2011 #5

    I like Serena

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    Neat! :smile:

    Btw, it's also neat to see how Bn=PDnP-1 cancels out. :cool:
     
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