# Something about the Roothaan equation

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?

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.
• The biggest problem is, that $\mathbf{F} \mathbf{C} = \mathbf{S} \mathbf{C} \pmb{\varepsilon}$ is not an eigenvalue problem. – Martin - マーチン Jul 5 '18 at 8:39