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Similarity Transformation Doesn't seem to work

  1. Dec 6, 2014 #1
    if A is a square matrix, and A' = B-1AB is its similarity transform (with a non-singular similarity transformation matrix B), then the eigenvalues of A and A' are supposed to be the same. I can verify this for all most all cases of A. But, it doesn't seem to work, when the eigen values of A are all zero. For example, it doesn't work when A = [0 1 0;0 0 0; 1 0 0].
    Yet I can't see why, it shouldn't work.
    from http://mathworld.wolfram.com/SimilarityTransformation.html
    the eigen values of A' is given as a solution of
    Inline13.gif =0
    Inline13.gif = 0
    Inline15.gif = 0
    Inline18.gif = 0
    Inline21.gif = 0
    Inline24.gif =0
    which is the solution for eigen values of A.
    Last edited by a moderator: Apr 29, 2017
  2. jcsd
  3. Dec 6, 2014 #2


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    Homework Helper

    for what matrix B did you find that the eigenvalues of B-1AB to be nonzero?
  4. Dec 6, 2014 #3
    I created a random matrix T,
    T = rand(3)
    T =
    0.4509 0.7447 0.1835
    0.5470 0.1890 0.3685
    0.2963 0.6868 0.6256
    and, the eigen values I got are:
    ans =
    1.0e-05 *
    0.4418 + 0.0000i
    -0.2209 + 0.3826i
    -0.2209 - 0.3826i
    Oh! now that I post it, I realize that, it may have to do with numerical errors, the values are very small anyway.
  5. Dec 7, 2014 #4


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    yeah, since the eigenvalues are of order 10^-5, it's likely they are nonzero just due to numerical error.
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