Generalized Eigenvalue Problem

The generalized eigenvalue problem is a mathematical problem that arises in many areas of science and engineering. It is a generalization of the standard eigenvalue problem, where we seek the eigenvalues and eigenvectors of a square matrix. In the generalized eigenvalue problem, we consider two square matrices $\mathbf{A}$ and $\mathbf{B}$, and we seek the eigenvalues $\lambda$ and eigenvectors $\mathbf{x}$ that satisfy the equation

\begin{equation}\label{eq:gen_eig} \mathbf{A}\mathbf{C}=\mathbf{B}\mathbf{C}\mathbf{\Lambda}, \end{equation}

where $\mathbf{C}$ is a matrix of eigenvectors and $\mathbf{\Lambda}$ is a diagonal matrix of eigenvalues. The quick way to solve the generalized eigenvalue problem is to transform it into a standard eigenvalue problem by multiplying both sides of the equation by the inverse of $\mathbf{B}$ as

\begin{equation} \mathbf{B}^{-1}\mathbf{A}\mathbf{C}=\mathbf{C}\mathbf{\Lambda}. \end{equation}

This method is not always numerically stable, especially when the matrices $\mathbf{A}$ and $\mathbf{B}$ are ill-conditioned. Should you try to use this method for the Roothaan equations in the Hartree–Fock method, you would find that the solution is is not correct. A more stable approach is to modify the equation \eqref{eq:gen_eig} as

\begin{align} \mathbf{A}\mathbf{C}&=\mathbf{B}\mathbf{C}\mathbf{\Lambda}\nonumber \\
\mathbf{B}^{-\frac{1}{2}}\mathbf{A}\mathbf{C}&=\mathbf{B}^{\frac{1}{2}}\mathbf{C}\mathbf{\Lambda}\nonumber \\
\mathbf{B}^{-\frac{1}{2}}\mathbf{A}\mathbf{B}^{-\frac{1}{2}}\mathbf{B}^{\frac{1}{2}}\mathbf{C}&=\mathbf{B}^{\frac{1}{2}}\mathbf{C}\mathbf{\Lambda}, \end{align}

where we solve the standard eigenvalue problem for the matrix $\mathbf{B}^{-\frac{1}{2}}\mathbf{A}\mathbf{B}^{-\frac{1}{2}}$ and obtain the eigenvectors $\mathbf{B}^{\frac{1}{2}}\mathbf{C}$ and eigenvalues $\mathbf{\Lambda}$. The eigenvectors of the original problem are then given by $\mathbf{C}=\mathbf{B}^{-\frac{1}{2}}\mathbf{B}^{\frac{1}{2}}\mathbf{C}$ and the eigenvalues are the same.