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Matrix to the Power of Something

  1. Mar 6, 2009 #1
    1.

    Hi could someone kindly give me a clue as to how I would go about calculating a matrix to the power of something?

    eg A=[1 3^(1/2);
    3^(1/2) 1]

    what would A^12 be?

    ...
    3.

    Do I use P,P^(-1) and a diagonal matix somehow? I am really a beginner at matrices sorry!
     
  2. jcsd
  3. Mar 6, 2009 #2

    lurflurf

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    A^2=2A+2I
    so
    A^12=49920A+36544I
     
  4. Mar 7, 2009 #3
    Sorry but that's not what I get when I check this with Matlab. I get :

    A=
    [1, 3^(1/2);
    3^(1/2), 1]

    B=A^12

    B=
    [14.74, 28.15;
    28.15, 53.78]

    But no idea how this is done with pencil and paper. Please enlighten me someone!
     
  5. Mar 7, 2009 #4

    gabbagabbahey

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    Gold Member

    Matlab's answer is incorrect...are you sure you typed in in correctly?

    If you don't like lurflurf's method (or don't follow it), you can of course start by diagonalizing A...If D=PAP-1 is diagonal, then A^12=(P-1DP)^12=(P-1DP)(P-1DP)(P-1DP)......(P-1DP)=P-1D12P since PP-1=I
     
  6. Mar 8, 2009 #5
    I typed it in correctly I'm pretty sure of that. There was a scalar value in front which I multiplied with to give the above matrix. Oh well looks like it's wrong anyway. :(



    Okay must admit I don't understand this. This is complex notation?
     
  7. Mar 8, 2009 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, there's nothing "complex" about it!

    As lurflurf told you, A^2= 2A+ I. (He got that by finding its characteristic equation. Every matrix satisifies its own characteristic equation.)

    So A^4= (A^2)^2= (2A+ 2I)^2= 4A^2+ 8A+ 4I= 4(2A+ 2I)+ 8A+ I= 16A+ 9I and, then, A^8= (A^4)^2= (16A+ 9I)^2= 256A^2+ 288A+ 81I= 256(2A+ 2I)+ 288A+ 81I= 800A+ 593I. Finally, A^12= (A^8)(A^4)= (800A+ 593I)(16A+ 9I)= 12800A^2+16688A+ 5337I= 12800(2A+ 2I)+ 16688A+ 5337I. See what that gives. Unless I have made an arithmetic error, that should give 49920A+ 36544I.
     
  8. Mar 8, 2009 #7

    lurflurf

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    I=a^0={{1,0},{0,1}}
     
  9. Mar 8, 2009 #8
    Aaah alright starting to make a bit more sense now.

    Thanks for your patience !
     
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