# Finding Transition Matrix

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? – Cameron Buie May 15 '12 at 2:37
Yea we studied them for awhile a couple weeks ago – Mathstudent May 15 '12 at 2:39
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. – Cameron Buie May 15 '12 at 2:41
Ok Thanks a lot! – Mathstudent May 15 '12 at 2:42
Something's odd here: $\,A\,,\,B\,$ don't have the same trace so there can't exist a matrix $\,P\,$ a required... – DonAntonio May 15 '12 at 3:40

This is impossible. $A$ has a real eigenvalue greater than 3. $B$ has a real eigenvalue less than 2. $B$'s other two eigenvalues are not real.

(I just evaluated the characteristic polynomial of $A$ at 1,2,3. The form of $B$ shows one real eigenvalue and a complex pair.)

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More simply, as Don Antonio notes in the comments, $A$ and $B$ don't have the same trace. – Gerry Myerson May 15 '12 at 3:57
@GerryMyerson: My default is brute force... – copper.hat May 15 '12 at 3:58

Hints:

1. $A, B$ have the same eigenvalues... why?

2. Recall what eigendecomposition is..

3. What does the eigendecomposition on both $A, B$ in $P^{-1} A P = B$ would look like?

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$A,B$ should have the same eigenvalues...but apparently they don't! – Gerry Myerson May 15 '12 at 3:58