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Value of cos(x) where x is multiple of a matrix

  1. Feb 15, 2017 #1
    1. The problem statement, all variables and given/known data
    Given a matrix M={{2,1},{1,2}} then value of cos( (π*M)/6 )


    2. Relevant equations


    3. The attempt at a solution
    Eigen values are π/6 and π/2 and eigen vectors are (π/6,{-1,1}) and (π/2,{1,1}).
    Diagonalize matrix is {{π/6,0},{0,π/2}}
    I got same value (√3/2)M
     
  2. jcsd
  3. Feb 15, 2017 #2

    Ray Vickson

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    I get something different: ##\cos(M \pi /6)## is NOT a scalar multiple of ##M##.
     
  4. Feb 16, 2017 #3
    Of course it isn't !
    I found similar problem like this in which we have to find cos(A) where A=(π/2) \begin{pmatrix}
    1 & 1 \\
    1 & 1 \\
    \end{pmatrix}. The process is they first find eigen vectors and then used diagonalization formula.
    cos(A)= PDP-1
    I'm not sure it is D or cos(D).
     
    Last edited: Feb 16, 2017
  5. Feb 16, 2017 #4

    Mark44

    Staff: Mentor

    Neither, really.
    If ##A = PDP^{-1}##, then ##\cos(A) = \cos(PDP^{-1})##. Now, write the right side as the Maclaurin series for cosine.
     
  6. Feb 16, 2017 #5

    Ray Vickson

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    You say "of course it isn't"---but in post #1 you said the opposite.

    If ##A = P^{-1} D P## with diagonal ##D##, then we have ##f(A) = P^{-1} f(D) P##, and
    $$D = \pmatrix{d_1 & 0 & \ldots \;\; 0\\ 0 & d_2 & \ldots\;\; 0 \\
    \vdots & \vdots&\; \vdots\\
    0 & 0 & \ldots\;\; d_n} \: \Longrightarrow
    f(D) =
    \pmatrix{f(d_1) & 0 & \ldots\;\; 0\\ 0 & f(d_2) & \ldots\;\; 0 \\
    \vdots & \vdots&\; \vdots\\
    0 & 0 & \ldots \;\; f(d_n) }
    $$
     
  7. Feb 17, 2017 #6
    It gives me expansion of cosine series. I think for final answer may be I have to put value of π.

    Thanks for help.
     
  8. Feb 17, 2017 #7
    By M is not scalar multiple of cos(π M/6) I meant cos(π M/6) ≠ M cos(π/6).

    I was calculating cos(A)=cos(PDP-1) but instead of ##f(A) = P f(D) P^{-1}## I was doing ##f(A) = f(PDP^{-1})##.Thanks for pointing that out.

    Thank you for help.
     
  9. Feb 17, 2017 #8

    Ray Vickson

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    They are the same: ##f(P D P^{-1}) = P f(D) P^{-1}## for any analytic function ##f(\cdot)##.
     
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