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I am solving one generalized Eigenvalue problem as follows:

$$BAX=\lambda X$$

where $B$ is diagonal, and $A$ is tridiagonal. I have obtained the eigenvalues/eigenvectors of $A$:

$$A Y=\alpha Y$$

How can I obtain the eigenvalues/eigenvectors of $BA$? Considering that here $B$ is diagonal, I think there may be a way to obtain the eigenvalues/eigenvectors of $BA$ just by combing $Y, \alpha, B$. Can anyone give some help on this? Thank you.

One more questions, usually for such generalized Eigenvalue problem $BAX=\lambda X$, the calculated eigenvectors $Z$ (through lapack) satisfies: $Z^{T}B^{-1}Z=I$. However, when I directly solve the eigenvalues/eigenvectors of matrix $BA$, its eigenvectors should satisfy $Z^{T}Z=1$. So, to my understanding, the solution of eigenvectors for $BA$ matrix and solution under generalized Eigenvalue problem $BAX=\lambda X$ is not the same. Is it true? Thank you.

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I think finding the eigenvalues of $A$ and $BA$ are two different problems. Solving one won't help you the other. However, most of your comments are mysterious to me. Finding the eigenvalues of $BA$ is the same as finding the generalized eigenvalues $AX = \lambda B^{-1} X$, assuming $B$ is invertible. Isn't it? –  Stephen Montgomery-Smith Dec 15 '13 at 3:35
    
Since B is diagonal it commutes with A and life is good: math.stackexchange.com/questions/6258/…; if A is not diagonalizable things are more complicated. –  Betty Mock Dec 15 '13 at 3:57
    
Yes, Finding the eigenvalues of $BA$ is the the same as finding the generalized eigenvalues $AX=\lambda B^{-1}X$, so, are they the same as each other? Since I find the eigenvectors (for each eigenvalue) satisfies different normalization, for the solution of $BA$, it satisfies $X^{T}X=1$, while for the eigenvectors for generalized problem,it satisfies $X^{T}BX=1$. Suppose I now have the eigenvalues/eigenvectors for generalized problem, how can get the solution for $BA$? Thanks a lot. –  Hui Zhang Dec 15 '13 at 6:55

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