Is there a connection between the diagonalization of a matrix $A$ and that of the product $DA$ with a diagonal matrix $D$? 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 $DA = P\Lambda P^{-1}$?
 A: 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 $P_0$
A: 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 $DA$. 
