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A Eigenstates of "summed" matrix

  1. Apr 1, 2016 #1
    Hi to all.

    Say that you have an eigenvalue problem of a Hermitian matrix ##A## and want (for many reasons) to calculate the eigenvalues and eigenstates for many cases where only the diagonal elements are changed in each case.

    Say the common eigenvalue problem is ##Ax=λx##. The ##A## matrix is a sum of two matrices ##A=B+C## and the ##C## matrix contains only the diagonal elements non-zero. For example

    \begin{equation}
    \begin{pmatrix}
    E_{1}& a & b \\
    c & E_{2}& d \\
    e & f & E_{3}
    \end{pmatrix}
    =
    \begin{pmatrix}
    0& a & b \\
    c& 0 & d \\
    e& f & 0
    \end{pmatrix}
    +
    \begin{pmatrix}
    E_{1}& 0 & 0 \\
    0& E_{2} & 0 \\
    0& 0 & E_{3}
    \end{pmatrix}
    \end{equation}


    Say you want to calculate the eigenstates of the ##A## matrix but in a different way.

    I was thinking is there is a method which is used to split the problem in two other sub-problems. The first step to be something like to calculate something for the ##B## matrix which remains unchanged in each case so the calculation is performed only one time, in the second step you only perform calculations for the ##C## matrix which in any case onlyn the diagonal elements are changed so you have to work only with the ##C## matrix problem and finally you combine the data of these two matrices in order to obtain the final result for the ##A## matrix.

    Whats your opinion?
     
  2. jcsd
  3. Apr 1, 2016 #2

    andrewkirk

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    Science Advisor
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    Gold Member

    If the diagonal elements are only changed by adding the same amount to all three then there is a simple solution. First solve the eigenvalue problem for ##A##. Then the eigenvalues for the matrix that is ##A## with ##h## added to all diagonal elements are simply the eigenvalues of ##A## with ##h## added to them. The eigenvectors are unchanged by the addition of ##h## along the diagonal.

    For more general changes to the diagonal elements, I see no reason to expect there to be a solution that is any easier than just re-solving the eigenvalue problem in each case.
     
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