Thanks for your suggestion, Chuck. The equation arises in the substraction of two harmonic potentials V and V': V' = 1/2 x^t * A^(-1) * x V= 1/2 x^t * B^(-1) * x V'-V = 1/2 x^t * ( A^(-1) - B^(-1) ) * x = 1/2 x^t * Z^(-1) * x
A is the covariance matrix of the coordinates x in a molecular dynamics simulation, A = <x * x^t >. Same goes for B. -Jose On Fri, Sep 10, 2010 at 5:02 PM, Charles R Harris <[email protected] > wrote: > > > On Fri, Sep 10, 2010 at 2:39 PM, Jose Borreguero <[email protected]>wrote: > >> Dear Numpy users, >> >> I have to solve for Z in the following equation Z^(-1) = A^(-1) - B^(-1), >> where A and B are covariance matrices with zero determinant. >> >> I have never used pseudoinverse matrixes, could anybody please point to me >> any cautions I have to take when solving this equation for Z? The brute >> force approach linalg.pinv( linalg.pinv(A) - lingal.pinv(B) ) gives me a >> matrix with all entries equal to 'infinity'. >> >> > Similar sorts of equations turn up in Kalman filters. You can also try > tricks like Z = B * (B - A)^-1 * A . Where does this problem come from? > There might be a better formulation. > > Chuck > > > _______________________________________________ > NumPy-Discussion mailing list > [email protected] > http://mail.scipy.org/mailman/listinfo/numpy-discussion > >
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