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I need to find eigenvectors and eigenvalues of a matrix which is product of 2 symmetric positive definite matrix(SwInverseSbProd=SwInverse*Sb). Since SwInverseSbProd is non-symmetric and calculation of eigenvectors is so complex for non-symmetric matrices, I find eigenvectors and eigenvalues corresponding to SbSwSbProd=Squareroot(Sb)*SwInverse*Squareroot(Sb) which is a symmetric matrix (As explained in paper : Fisher Linear Discriminant Analysis by Max Welling).

But I don't know what is the relation between eigenvectors of SwInverseSbProd and eigenvectors of SbSwSbProd. Could anyone please tell me how can I find eigen vectors of SwInverseSbProd from eigenvectors of SbSwSbProd?

I tried the solution. 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?

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  • $\begingroup$ Please see the MathJax tutorial in order to improve readability of your question. $\endgroup$ – Jonas Dahlbæk Aug 13 '14 at 15:31
  • $\begingroup$ You copied some signs wrong, I think: $.71008...$ should be $-.71008...$ and $.725005...$ should be $-.725005...$. But your final result is correct. These are eigenvectors of $AB$. They don't have the same scaling as the ones you got directly, but a nonzero scalar multiple of an eigenvector is an eigenvector. $\endgroup$ – Robert Israel Aug 15 '14 at 0:52
  • $\begingroup$ Thanks a lot. Yes, u r right. I corrected the mistake. $\endgroup$ – user3852441 Aug 15 '14 at 7:54
  • $\begingroup$ I have one more question. Since different eigen vectors are scaled by different values, does that affect multiplication of a vector with this matrix of eigen vectors (I use the matrix with eigen vectors as a transformation matrix for linear discriminant analysis. So will that affect the output of linear discriminant analysis?) thank u... $\endgroup$ – user3852441 Aug 15 '14 at 13:31
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If $u$ is an eigenvector of $B^{1/2} A B^{1/2}$ for eigenvalue $\lambda$, i.e. $B^{1/2}AB^{1/2} u = \lambda u$, then $v = B^{-1/2} u$ is an eigenvector of $AB$ for the same $\lambda$, because $$ABv = B^{-1/2} (B^{1/2} A B^{1/2}) u = \lambda B^{-1/2} u = \lambda v$$

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  • $\begingroup$ Thank u for your answer. But can I get all the eigen vectors for AB from u by multiplying with B^−1/2? Because when I try that, I get different values $\endgroup$ – user3852441 Aug 14 '14 at 8:27
  • $\begingroup$ As long as $B$ is positive definite (and therefore nonsingular) you should get all of them. If $v$ is an eigenvector of $AB$, then $v = B^{-1/2} u$ where $u = B^{1/2} v$ is an eigenvector of $B^{1/2} A B^{1/2}$. $\endgroup$ – Robert Israel Aug 14 '14 at 15:11
  • $\begingroup$ thank u for your reply. I have modified the question with what I have tried. Could u please tell me where I have gone wrong? $\endgroup$ – user3852441 Aug 14 '14 at 15:29
  • $\begingroup$ sorry for unreadable formatting of my question. I have modified it. It will be of great help for me if you can help me to find me find what has gone wrong. I really stuck at this point. Thank you. $\endgroup$ – user3852441 Aug 14 '14 at 18:35

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