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Given a diagonalizable matrix $A = P_0\Lambda_0 P_0^{-1}$ and a diagonal matrix $D$ with $\det D=1$, is there any connection between $P_0$ and the matrix $P$ of the diagonalization of $DA = P\Lambda P^{-1}$?

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    @xEnOn: It usually isn't; the question denotes them by $\Lambda_0$ and $\Lambda$, respectively.2011-08-23

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I know that if A is diagonalizable, and D a diagonal matrix D it is not even true that DA need to be diagonalizable. An example is: $ D= \begin{bmatrix} 2 & 0 \\ 0 & 1/2 \end{bmatrix}, \,\, DA = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} $ and of course $A=D^{-1}DA$.

I leave it as an exercise for the reader to verify that indeed $A$ is diagonalizable and $DA$ not.

Although not a complete answer to the question (since this example doesn't say anything about the case where both A and DA are diagonalizable) it at least shows that the question should be posed more carefully.

Also the fact that you want $\det D = 1$ will ensure that the product of the elements in the diagonal of $\Lambda_0$ will be the same as those in $\Lambda$, but I guess you don't need that info since you care about $P$ and $P_0$

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    assumption failed... ok, thanks2011-09-05
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The columns of $P_0$ are the eigenvectors of $A$. The columns of $P$ are the eigenvectors of $DA$. So write down some $2\times2$ matrix $A$ and see whether there's any relation between its eigenvectors and those of $DA$.

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    You might also try finding an example where $DA$ is not diagonalizable but $A$ is.2011-08-18