Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
2  
    
So $D$ can have complex entries, or are you only considering $(1,-1)$ diagonals? –  J. M. 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
add comment

2 Answers 2

up vote 3 down vote accepted

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 $P_0$

share|improve this answer
    
assumption failed... ok, thanks –  Tobias Kienzler Sep 5 '11 at 6:01
add comment

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 $DA$.

share|improve this answer
1  
You might also try finding an example where $DA$ is not diagonalizable but $A$ is. –  Robert Israel Aug 18 '11 at 16:44
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.