I have a non symmetric matrix $AB$ where $A$ and $B$ are symmetric matrices. How can I find the eigenvectors and eigenvalues of $AB$?

In a paper( Fisher Linear Discriminant Analysis by M Welling), the author asks to find eigenvalues and eigenvectors of $B^{1/2} A B^{1/2}$ which is a symmetric matrix. But how can I get eigenvalues and eigenvectors of $AB$ from eigenvalues and eigenvectors of $B^{1/2} A B^{1/2}$?

Could someone please help me?

I tried the solution suggested below. But it doesn't work for me $A$=\begin{pmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{pmatrix} $B$=\begin{pmatrix}32&-12&8\\-12&34&-21\\8&-21&13\end{pmatrix} $B^{1/2}$=\begin{pmatrix}5.53308892146077&-0.950134037741956&0.694386274009086\\-0.950134037741957&4.93157708794602&-2.96256522898147\\0.694386274009086&-2.96256522898147&1.93417552628962\end{pmatrix} eigenvectors of $AB$=\begin{pmatrix}0.516537330395033&-0.781188319935242&-0.0177964973702446\\0.710088559129181&-0.185707982205180&0.521054279012559\\0.478501227273470&0.596034692062496&0.853337988726655\end{pmatrix}

eigenvectors of $B^{1/2} A B^{1/2}$= \begin{pmatrix}0.517933641073670&-0.855373946305353&-0.00895295628145857\\0.725005239437560&-0.444501257554748&0.526104585439372\\0.453995755742558&0.265996323309655&0.850372747536919\end{pmatrix}

$B^{-1/2}$ * (eigenvectors of $B^{1/2} A B^{1/2}$) =\begin{pmatrix}0.0692193234209673&-0.179045453961740&-0.160160577798552\\-0.0951564325356731&-0.0425635779871867&4.68925725410385\\0.0641222410443052&0.136608931923259&7.67966316565345\end{pmatrix}

eigenvectors of $AB$ is different from $B^{-1/2}$ * (eigenvectors of $B^{1/2} A B^{1/2}$). Is there any mistake in what I did?

  • $\begingroup$ Why do you think that this problem has a simple solution? $\endgroup$ – Fabian Aug 14 '14 at 9:26
  • $\begingroup$ @Fabian the new matrix $B^{1/2} A B^{1/2}$ is a symmetric matrix. I have already implemented a code to find eigenvalues and eigenvectors for a symmetric matrix. I think finding eigenvalues and eigenvectors for a nonsymmetric matrix is so complex $\endgroup$ – user3852441 Aug 14 '14 at 9:28

You simply multiply the eigenvectors with $B^{-1/2}$ (you need that $B$ is positive definite) and the eigenvalues stay the same.

Indeed, having $v$ and eigenvector of $B^{1/2} A B^{1/2}$ to the eigenvalue $\lambda$, i.e., $$ B^{1/2} A B^{1/2} v = \lambda v,$$ you can see easily that with $w= B^{1/2}v$ we have $$ B^{1/2} A B w = B^{1/2} A B^{1/2} v = \lambda v$$ and thus $ A B w = \lambda w$.

  • $\begingroup$ Thank u for your answer. But it didn't work for me. I have edited the question with what I have tried as per your suggestion. Is there anything wrong I am doing? $\endgroup$ – user3852441 Aug 14 '14 at 10:24
  • $\begingroup$ Whatever u have mentioned what right. Issue with what I did was, my new eigenvectors were a scalar mutiplied version of actual eigenvectors. Thanks a lot for your anser $\endgroup$ – user3852441 Aug 15 '14 at 7:58

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