# 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 $DA = P\Lambda P^{-1}$?

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–  Tobias Kienzler Aug 18 '11 at 9:06
So $D$ can have complex entries, or are you only considering $(1,-1)$ diagonals? –  Guess who it is. Aug 18 '11 at 9:17
@J.M.: yes, complex entries as well. The 2x2 case $D=$diag$(e^{-ix}a,e^{+ix}/a)$ would already be sufficient, though a more general solution might be interesting as well... –  Tobias Kienzler Aug 18 '11 at 9:23
Just curious, does this work all the time? It doesn't look like it would after I tried a few values. If $A=\begin{bmatrix} 1 & 2\\ 3 & 2 \end{bmatrix}$ and $D=\begin{bmatrix} 5 & 0\\ 0 & \frac{1}{5} \end{bmatrix}$, the eigenvalues of $A$ are $4$ and $-1$ but the eigenvalues for $DA$ are $6.06$ and $-0.66$. So how could the $\Lambda$ be the same for $A$ and $DA$? –  xenon Aug 20 '11 at 9:44
@xEnOn: It usually isn't; the question denotes them by $\Lambda_0$ and $\Lambda$, respectively. –  joriki Aug 23 '11 at 13:42

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, thanks –  Tobias Kienzler Sep 5 '11 at 6:01

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. –  Robert Israel Aug 18 '11 at 16:44