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 $DA = P\Lambda P^{-1}$?
Asked
Active
Viewed 405 times
1
-
2related: Eigenvalues of product of a matrix and a diagonal matrix – Tobias Kienzler Aug 18 '11 at 09:06
-
So $D$ can have complex entries, or are you only considering $(1,-1)$ diagonals? – J. M. ain't a mathematician Aug 18 '11 at 09:17
-
@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 09:23
-
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 09:44
-
@xEnOn: It usually isn't; the question denotes them by $\Lambda_0$ and $\Lambda$, respectively. – joriki Aug 23 '11 at 13:42
2 Answers
3
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 $P_0$
J. M. ain't a mathematician
- 76,540
Maarten Derickx
- 165
1
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 $DA$.
Gerry Myerson
- 185,413
-
1You might also try finding an example where $DA$ is not diagonalizable but $A$ is. – Robert Israel Aug 18 '11 at 16:44