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?

  • $\begingroup$ 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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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