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Determinant of exponential matrix

  1. Dec 22, 2017 #1
    1. The problem statement, all variables and given/known data

    upload_2017-12-23_1-26-42.png
    2. Relevant equations


    3. The attempt at a solution



    Det( ## e^A ## ) = ## e^{(trace A)} ##

    ## trace(A) = trace( SAS^{-1}) = 0 ## as trace is similiarity invariant.

    Det( ## e^A ## ) = 1

    The answer is option (a).

    Is this correct?

    But in the question, it is not given that S AS-1 is a similarity matrix of A. I assumed it without checking.
    So, should I check it by calculating eigen values of both S and ## SAS^{-1}## and their eigen vectors?
     
  2. jcsd
  3. Dec 22, 2017 #2

    Orodruin

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  4. Dec 22, 2017 #3
     
  5. Dec 22, 2017 #4

    Orodruin

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    ##S## is clearly a unitary matrix. Anyway, all you really need to know is that ##S## is invertible (otherwise ##S^{-1}## does not exist) and the cyclic property of the trace.
     
  6. Dec 22, 2017 #5
    If there exists a non – singular matrix P, then ## P^{-1} A P ## is known as a similarity matrix of A.

    Similarity matrix is the name of ## P^{-1} A P ##.

    Why is ## P^{-1} A P ## called a "similiarity matrix" of A? Is there any intuitive reason behind the word " similiarity"?

    Now, A and its similiarity matrix have the following properties:

    1) They have the same set of eigen values.

    2) If x is eigen vector of A, then ## P^{-1} x ## is the eigen vector of corresponding similiarity matrix ## P^{-1} A P ##.

    3) Trace( A ) = Trace (## P^{-1} A P ## )
    ##SAS^{-1}## is a similarity matrix of A by definition. There is no need to check it.

    Is this correct?
     
  7. Dec 22, 2017 #6

    Orodruin

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    Yes, but the point is that you do not need all those properties to solve this problem. You just need the cyclicity of the trace.
     
  8. Dec 22, 2017 #7
    Thanks, I got it.
     
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