Jacobi-Davidson Method with Cayley Transform

The Jacobi-Davidson method is discussed in §7.12. The Jacobi-Davidson method differs from the Davidson method in that the linear system to be solved is projected onto the space orthogonal to the current Ritz vector. This leads to the solution of the correction equation

$$\left( I - \frac{B x_j x_j^{\ast}}{x_j^{\ast} B x_j} \right)... ..._j^{\ast}}{x_j^{\ast} x_j} \right) w_j = (A - \theta_j B) x_j,$$ (276)

where $(\theta_j,x_j)$ is a Ritz pair on iteration $j$. (Note that usually, a minus sign is put in front of the residual in the right-hand side.) Assume that $w_j$ is orthogonal to $x_j$, i.e., $( I - x_j x_j^{\ast} / x_j^{\ast} x_j) w_j = w_j$. When an inexact solver is used, we have a residual $s_j$ that satisfies

$$s_j = (A - \theta_j B) x_j - \left( I - \frac{B x_j x_j^{\ast} }{x_j^{\ast} B x_j}\right) (A - \theta_j B) w_j \ .$$

Note that the projection in front of $w_j$ is dropped since $w_j$ is assumed orthogonal to $x_j$. We can rewrite this into

$$\begin{eqnarray*} (A- \theta_j B) w_j & = & -\epsilon_j B x_j + (A - \theta_j B) x_j - s_j \\ & = & ( A - (\theta_j + \epsilon_j) B ) x_j - s_j, \end{eqnarray*}$$

where $\epsilon_j = -(x_j^{\ast} (A -\theta_j B) w_j) / ( x_j^{\ast} B x_j)$.

In words, the solution $w_j$ of the correction equation is obtained by the action of the Cayley transform $T_{\mathrm{C}} = (A-\theta_j B)^{-1} (A-(\theta_j+\epsilon_j)B)$ to the most recent Ritz vector. Example 11.2.1 in [411] shows that $\epsilon_j$ tends to zero on convergence. Both pole and zero of the Cayley transform lie close to the desired eigenvalue. This meets the conditions for good matching between eigenvectors of $Ax = \lambda Bx$ and $T_{\mathrm{IC}} u = \eta u$, motivated at the end of §11.2.2.

The following observation is a bit funny. Since $w_j = -\epsilon_j (A-\theta_j B)^{-1} B x_j + x_j$, when $s_j=0$, we have from $x_j^{\ast} w_j=0$ that

$$\theta_j + \epsilon_j = \theta_j + \frac{x_j^{\ast} x_j}{x_j^{\ast} (A-\theta_j B)^{-1} B x_j} \ .$$

So, the pole of the Cayley transform is the Rayleigh-Ritz quotient of $x_j$ and the zero is the harmonic Rayleigh-Ritz quotient with target $\theta_j$.