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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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    @GerryMyerson: Awesome. /headdesk/2012-05-15

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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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    @GerryMyerson: My default is brute force...2012-05-15
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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!2012-05-15