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Problem: Find the transition matrix P such that $P^{-1}AP=B$ where: $$A=\begin{bmatrix} 3 & -1 & 0 \\ -1 & 0 & -1 \\ 0 & 1 & 1 \end{bmatrix} \quad\text{and}\quad B=\begin{bmatrix} \frac{2}{\sqrt{3}} & 0 & 0 \\ \frac{4}{\sqrt{2}} & \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{2}{\sqrt{6}} & \frac{-3}{\sqrt{6}} & \frac{-3}{\sqrt{6}} \end{bmatrix}$$

So I am unsure how to do this because we have not discussed it. I was just wondering If somebody could give me some ideas about how to find transition matrices and what exactly they are. Any help is appreciated. Thanks

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    It's quite all right. I can tell you off the top of my head that transition matrices have all non-negative real entries, and that the entries in each column (or row, or both) sum to 1. You need such a matrix $P$, which will be $3\times 3$, and satisfy $AP=PB$. That probably isn't the most elegant way to go about it, though. Have you discussed eigenvalues and eigenvectors, yet?2012-05-15
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    Yea we studied them for awhile a couple weeks ago2012-05-15
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    Actually, now that I think about it, I'm not sure that will help you, since $B$ isn't diagonal. I'm blanking, here. Long day. Go ahead with the one approach I gave, if you like (it isn't too laborious). Eventually someone else will come along and give you better advice.2012-05-15
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    Ok Thanks a lot!2012-05-15
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    Oh yes ok I remember this. Yes I can thanks a lot I understand now.2012-05-15
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    Something's odd here: $\,A\,,\,B\,$ don't have the same trace so there can't exist a matrix $\,P\,$ a required...2012-05-15
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    @Cameron, you are assuming this is a Markov chain problem, but the phrase "transition matrix" can be used in other contexts, where your assertions about the entries need not hold.2012-05-15
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    @GerryMyerson: Awesome. /headdesk/2012-05-15

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