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The Lanczos recursion is constructed to make the basis $V$ orthogonal, but this is true only for infinite precision computation. In the algorithm we only make sure that the new vector $v_{j+1}$ is orthogonal to working precision to the two latest vectors $v_{j-1}$ and $v_j$, and orthogonality to the earlier vectors follows from the symmetry of $A$ and the recursion (4.10). As soon as one eigenvalue converges, i.e., the Ritz pair has a small residual ([*]), all the basis vectors $v_j$ get perturbations in the direction of the eigenspace of the converged eigenvalue. As a result of this, a duplicate copy of that eigenvalue will soon show up in the tridiagonal matrix $T$. Paige [347] was the first to discover this; the reader is referred to the monograph [353] for a detailed discussion.

Let us consider three different strategies to handle this.


Susan Blackford 2000-11-20