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Why we cannot solve Roothaan equation $\mathbf{F} \mathbf{C} = \mathbf{S} \mathbf{C} \mathbf{\epsilon}$ by just move S matrix to the left, as $\mathbf{S}^{-1} \mathbf{F} \mathbf{C} = \mathbf{C} \mathbf{\epsilon}$ as a new matrix $\mathbf{F}'$, and then solve the eigenvalue problem $\mathbf{F}' \mathbf{C} = \mathbf{C} \mathbf{\epsilon}$ why we cannot do that?

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We could. Mathematically there is nothing wrong with that - the overlap matrix is positive definite and therefore must have an inverse, and so your manipulation is perfectly correct mathematically. The problem is that $\mathbf{S}^{-1} \mathbf{F}$ is not a symmetric (or in the more general case Hermitian) matrix. Computationally it is MUCH more difficult to find accurate eigenvalues and eigenvectors of a general non-symmetric matrix than a symmetric one. Thus, when solving the problem on a computer, if one can cast one's algorithm in terms of diagonalisation of a symmetric rather than a non-symmetric matrix you are likely to get more correct answers, quicker. And this is what the standard manipulations achieve.

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    $\begingroup$ The biggest problem is, that $\mathbf{F} \mathbf{C} = \mathbf{S} \mathbf{C} \pmb{\varepsilon}$ is not an eigenvalue problem. $\endgroup$ – Martin - マーチン Jul 5 '18 at 8:39
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    $\begingroup$ It is a generalized eigenvalue problem, see e.g. en.wikipedia.org/wiki/… , fourier.eng.hmc.edu/e161/lectures/algebra/node7.html , epubs.siam.org/doi/abs/10.1137/1.9780898717808.ch6 . Algorithms do exist to solve it directly, and may be more efficient than reducing it to the standard eigenvalue problem, for instance the QZ algorithm, but in Qchem reduction to the standard form via some factorisation of S is the most common $\endgroup$ – Ian Bush Jul 5 '18 at 8:53
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    $\begingroup$ Sorry, but I don't understand where you are going with this. I know that it looks like an eigenvalue problem, but since the Fock operator, and therefore the Fock matrix, depend on their own solutions, it is not. You will in any case have to solve this equation iterative, a transformation of the matrices involved won't change that. $\endgroup$ – Martin - マーチン Jul 5 '18 at 9:17
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    $\begingroup$ I'm assuming that what is being asked here is about each step of the SCF procedure, i.e. for a given F how do we generate our next set of eigenvectors, not for the procedure as a whole - which as you say is a non-linear eigenvalue problem and requires an iterative solution $\endgroup$ – Ian Bush Jul 5 '18 at 9:28
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    $\begingroup$ Ah, that clears up quite a bit, and I agree. When I read solving the Roothaan equation, however, I immediately think about finding the self-consistent version of the equation. The question might be a bit unclear in that regard. $\endgroup$ – Martin - マーチン Jul 5 '18 at 9:38

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