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Matrix exp. question

  1. Nov 14, 2006 #1
    I need to find e^A of this 2x2 matrix....A= 1 1
    -1 -1

    When I do det(A-(lambda)I) I get 0 for the eigenvalues, which makes no sense. Am I doing something wrong?
  2. jcsd
  3. Nov 14, 2006 #2


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    Probably not. If the matrix is singular then at least one of the eigenvalues will be zero
    [tex]det(A-I \lambda )=(-\lambda+1)(-\lambda-1)+1=\lambda^2[/tex]
    So [itex]0[/itex] is a double root, and both of the eigenvalues are zero.

    Why don't you see what happens if you apply this matrix to a vector?
    Last edited: Nov 14, 2006
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