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Deflation.

The partial generalized Schur form can be obtained in a number of successive steps. Suppose that we have the partial generalized Schur form $AQ_{k-1}={Z}_{k-1}R^A_{k-1}$ and $BQ_{k-1}={Z}_{k-1}R^B_{k-1}$. We want to expand this partial generalized Schur form with the new right Schur vector ${q}$ and the left Schur vector ${z}$ to

\begin{displaymath}
A\onebytwo{Q_{k-1}}{{q}}=\onebytwo{{Z}_{k-1}}{{z}}
\left[\b...
...{@{\,}cc@{\,}}
R^A_{k-1}& {a}\\
0& \alpha\end{array}\right]
\end{displaymath}

and

\begin{displaymath}
B\onebytwo{Q_{k-1}}{q}=\onebytwo{{Z}_{k-1}}{{z}}
\left[\be...
...@{\,}cc@{\,}}
R^B_{k-1}& {b}\\
0& \beta \end{array}\right].
\end{displaymath}

The new generalized Schur pair $((\alpha,\beta), {q})$ satisfies

\begin{displaymath}
Q^\ast_{k-1}{q}=0
\quad \mbox{and} \quad
(\beta A-\alpha B){q}-{Z}_{k-1}(\beta {a} - \alpha {b})=0,
\end{displaymath}

or, since $\beta {a} - \alpha {b}={Z}^\ast_{k-1}(\beta A-\alpha B)q$,

\begin{displaymath}
Q^\ast_{k-1}q=0\quad\mbox{and}\quad
\left(I-{Z}_{k-1}{Z}_{k-...
...ight)(\beta A-\alpha B)
\left(I-Q_{k-1}Q_{k-1}^\ast\right)q=0.
\end{displaymath}

The vectors ${a}$ and ${b}$ can be computed from

\begin{displaymath}
{a}={Z}_{k-1}^\ast Aq\quad\mbox{and}\quad {b}={Z}_{k-1}^\ast Bq.
\end{displaymath}

Hence, the generalized Schur pair $((\alpha,\beta),q)$ is an eigenpair of the deflated matrix pair
\begin{displaymath}
\begin{array}{l}
\left(\left(I-{Z}_{k-1}{Z}_{k-1}^\ast\rig...
...ight)B
\left(I-Q_{k-1}Q_{k-1}^\ast\right)\right).
\end{array}\end{displaymath} (228)

This eigenproblem can be solved again with the Jacobi-Davidson process that we have outlined in §8.4.1. In that process we construct vectors $v_i$ that are orthogonal to $Q_{k-1}$ and vectors $w_i$ that are orthogonal to ${Z}_{k-1}$. This simplifies the computation of the interaction matrices $M^A$ and $M^B$, associated with the deflated operators:
\begin{displaymath}
\left\{\begin{array}{l}
M^A\equiv W^\ast\left(I-{Z}_{k-1}{...
...\ast\right) V=W^\ast B V,\rule{0pt}{3.5ex}
\end{array}\right.
\end{displaymath} (229)

and $M^A$ and $M^B$ can be simply computed as $W^\ast A V$ and $W^\ast B V$, respectively.


next up previous contents index
Next: Restart. Up: Deflation and Restart Previous: Deflation and Restart   Contents   Index
Susan Blackford 2000-11-20