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A successful LOBPCG Rayleigh-Ritz update cannot increase the selected Ritz values

Statement

Let $A = A^{*}$ and $B = B^{*} > 0$ . Suppose one successful update of Finite-budget block LOBPCG for a Hermitian-definite pencil with explicit rank-loss status starts from a $q$ -dimensional $B$ -orthonormal block $X_{k}$ with selected Ritz values $θ_{1} (X_{k}) \leq \dots \leq θ_{q} (X_{k})$ . The update forms a trial space $S_{k} = span (X_{k}, W_{k}, D_{k})$ , retains the old block, and selects the $q$ smallest generalized Ritz values on $S_{k}$ . Then $θ_{j} (X_{k + 1}) \leq θ_{j} (X_{k}) (1 \leq j \leq q) .$ This assertion applies only to an update that reaches the Rayleigh-Ritz step; the algorithm's rank-loss and budget-exhausted exits make no update claim.

Facts & Assumptions

Given: One successful LOBPCG update for a Hermitian-definite pair, with the old block retained in its trial space.

[F1]

Finite-budget block LOBPCG for a Hermitian-definite pencil with explicit rank-loss status: the trial block is $Y_{k} = [X_{k}, W_{k}, D_{k}]$ , so it retains $range (X_{k})$ .

[F2]

Finite-budget block LOBPCG for a Hermitian-definite pencil with explicit rank-loss status: a successful projected solve selects the $q$ smallest Ritz values on the full trial span and uses their Ritz vectors as $X_{k + 1}$ .

[L1]

Generalized Ritz values obey min-max bounds and are monotone under trial-space inclusion: if $S \subseteq T$ , then $θ_{j} (T) \leq θ_{j} (S)$ for each index available in $S$ .

Proof

technique · direct

1.1givenF1F2algebra

On the successful branch, [F1] constructs $S_{k}$ with $range (X_{k}) \subseteq S_{k}$ , and [F2] sets the new block to the q smallest Ritz vectors on $S_{k}$ . Hence $θ_{j} (X_{k + 1}) = θ_{j} (S_{k})$ and $θ_{j} (X_{k}) = θ_{j} (range (X_{k}))$ .

2.1L1step 1.1algebra∎

Apply [L1] to $range (X_{k}) \subseteq S_{k}$ . It gives $θ_{j} (S_{k}) \leq θ_{j} (range (X_{k}))$ for every $j \leq q$ ; substituting the equalities from step 1.1 proves the claim.

Depends on

Used by

A two-vector LOBPCG block resolves a clustered pair in one finite Rayleigh-Ritz update Example

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Peter Arbenz, Lecture Notes on Solving Large Scale Eigenvalue Problems, §13.6 (standard reference, not scraped)
Andrew Knyazev, Recent implementations, applications, and extensions of LOBPCG, §1 (standard reference, not scraped)