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?
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 $$ABv = B^{-1/2} (B^{1/2} A B^{1/2}) u = \lambda B^{-1/2} u = \lambda v$$