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Linear Algebra Problem

  1. Oct 6, 2013 #1
    Hey guys, I'm having problems with a question.

    Let P be an invertible matrix and assume that A = PMP[itex]^{-1}[/itex]. Where M is

    M = [{3,1,0}{0,3,0}{0,0,2}]

    Find a matrix B(t) such that e[itex]^{tA}[/itex] = PB(t)P[itex]^{-1}[/itex].

    Now this might be an easy problem, but I really have no idea what to do because my lecturer is so bad and the book for the course doesn't cover this material.

    I have seen something about A= PBP[itex]^{-1}[/itex] implying e[itex]^{tA}[/itex] = Pe[itex]^{tB}[/itex]P[itex]^{-1}[/itex] so I have tried computing the exponential of M, but to no avail. Any advice is much appreciated.
  2. jcsd
  3. Oct 6, 2013 #2


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    Yes, matrix exponential. This problem is fairly easy because you can split Mt into the sum of a diagonal matrix D and an offdiagonal matrix N which is nilpotent. And they commute with each other. So you can use exp(D+N)=exp(D)exp(N). Finding the exponential of each matrix is pretty easy.
    Last edited: Oct 6, 2013
  4. Oct 6, 2013 #3
    Thank you so much, you hero! Just to check I haven't gone completely wrong, should I have

    B(t) = e[itex]^{2t}[/itex]*[{e[itex]^{t}[/itex],e[itex]^{t}[/itex],0}{0,e[itex]^{t}[/itex],0}{0,0,1}]?
  5. Oct 6, 2013 #4


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    Close, but no cigar. That's only correct for t=1. Your nilpotent matrix is N={[0,t,0],[0,0,0],[0,0,0]}. What's exp(N)?
  6. Oct 6, 2013 #5
    Aha, of course! I think I see where I went wrong. Is it

    B(t) = e[itex]^{2t}[/itex]*[{e[itex]^{t}[/itex],t*e[itex]^{t}[/itex],0}{0,e[itex]^{t}[/itex],0}{0,0,1}]?
  7. Oct 6, 2013 #6


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  8. Oct 6, 2013 #7
    Thank you so much, I really appreciate it!
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