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Finding X of (X^-1)AX = B

  1. Jan 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Let A =

    \begin{bmatrix}
    \lambda & a \\
    0 & \lambda \\
    \end{bmatrix}

    and B =

    \begin{bmatrix}
    \lambda & b \\
    0 & \lambda \\
    \end{bmatrix}

    Assuming that a ≠ 0, and b ≠ 0 ; find a matrix X such that X-1AX = B.




    2. Relevant equations

    (A- [itex]\lambda[/itex]I)v=0


    3. The attempt at a solution

    I tried using the following logic: Let B = {v1, v2,....vn}
    be the basis of Fn consisting of the columns of X. We know that column j of B is
    equal to [Avj ]B, that is, the coordinates of Avj with respect to the basis B.

    But because of the two matrices having the exact same eigenvalues, I just end up with a=0, and am unable to actually find an invertible matrix X. Am I misreading the question, the logic, etc..?
     
  2. jcsd
  3. Jan 24, 2012 #2
    Also in case the notation is different elsewhere, lambda = eigenvalue.
     
    Last edited: Jan 24, 2012
  4. Jan 24, 2012 #3
    Am I missing something because from what I can tell just set up a 2x2 matrix with 4 unknown elements setup 4 equations and find a matrix X which satisfies those four equations.
     
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