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Eigenvalues of AX - XA

  1. Aug 18, 2010 #1
    Hi all,

    Here is this problem that I have been at for some time now: find eigenvalues and corresponding eigenvectors of the following linear mapping on a vector space of real 2 by 2 matrices:

    [tex]L(X) = AX - XA[/tex], where A is 2 by 2 symmetric matrix that is not a scalar multiple of identity.

    It is clear that 0 is one eigenvalue of the above with an eigenspace consisting of all matrices X that commute with A, I do not know however what its algebraic multiplicity would be, except that it has to be less than 4. To find the other eigenvalues I looked at

    [tex]AX - XA = \lambda X [/tex]

    a column at a time to obtain a 4 by 4 system.

    [tex]Ax_{i} - \sum^{2}_{j = 1}a_{ji}x_{i} = \lambda x_{i} [/tex]

    for i = 1, 2, with j indexing the rows of A.

    This approach requires me to find eigenvalues and eigenvectors of a 4 by 4 matrix, which may get messy if you have to do it by hand. Do you guys know any other, more efficient way of approaching this problem? I have attempted to use the fact that A can be orthogonally diagonalized in hopes of simplifying the above but with no success.

  2. jcsd
  3. Aug 18, 2010 #2
    Diagonalize A: let A = P-1DP, where D = diag(k1, k2). Given any matrix X, write
    [tex]X = P^{-1} \begin{pmatrix}a&b\\c&d\end{pmatrix}P.[/tex]
    You can compute that
    [tex]L(X) = P^{-1} \begin{pmatrix}0&(k_1-k_2)b\\(k_2-k_1)c&0\end{pmatrix}P.[/tex]
    You can now directly read off the eigenvalues of L: 0, 0, k1 - k2, and k2 - k1. The eigenvectors are also obvious.

    The same process applies to arbitrary diagonalizable matrices. If A is a n × n diagonalizable matrix and L(X) = AX - XA, and A has eigenvalues k1, ..., kn, then the eigenvalues of L are ki - kj, for each pair of i, j from 1, ..., n.
  4. Aug 19, 2010 #3
    This is indeed a lot more tractable, thanks a lot adriank!
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