How statement and proof provenance work
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A shift near the wanted eigenvalue suppresses its component under repeated implicit restarts
Example
Let , start from , and seek the largest eigenvalue . Use an implicit restart from dimension to with the fixed shift rule on each cycle, residual tolerance , and restart budget at least when considering cycles. This shift rule is allowed by the procedure; it is not claimed to be a good selection rule.
Supply the following finite-return projected solver for this example. For a Hermitian matrix , if , return success with the ordered pairs and . If , set
and return success with the ordered pairs and . Return unavailable on other shapes or non-Hermitian inputs. Each displayed vector is nonzero, direct multiplication verifies its eigenpair equation, and the two factors reproduce the quadratic characteristic polynomial. This supplied finite-dimensional example provider is not a general eigensolver or a new restart method.
Verification
Given: The exact diagonal matrix, starting vector, target, and shift above.
[F1] Exact implicit Arnoldi restart applies its shift polynomial and preserves an Arnoldi residual identity: for , the exact implicit restart retains the one-dimensional filtered Krylov space generated by .
[F2] Eigenvalues, eigenvectors, eigenspaces , and the spectrum of an endomorphism: an eigenvector is a nonzero vector satisfying .
Every projected in this example is Hermitian because , so the supplied two-dimensional solver returns valid complete projected data in finite exact work. By [F1], . The squared component weights are proportional to , so the Rayleigh quotient of the restarted vector is , whereas the initial quotient is . The wanted component-to- component ratio falls from 1 to 1/19.
Repeating the same exact restart multiplies the component by and the component by at every cycle. Hence after cycles their magnitude ratio is , which tends to zero. This shift choice suppresses the wanted eigendirection and gives no general convergence claim.
For each finite , the restart vector is proportional to , with all coordinates nonzero because none of vanishes. If a Ritz residual were zero, its lifted vector would satisfy ; the diagonal equations and distinct force to be proportional to some . No lies in : representing one would require for the other two distinct eigenvalues, hence . Also are independent because their coordinate determinant is the product of the three nonzero coordinates and the Vandermonde determinant for . Thus no pair is accepted at tolerance zero and the space is not invariant before each scheduled restart.
In a separate boundary run with shift , the filter has and , so the wanted component is removed exactly in one filtering step. The filtered vector is still nonzero, so this is not filtered breakdown; it is a shift-selection failure.
Depends on
- Finite-budget thick and implicitly restarted Arnoldi with residual acceptance and locking
- Exact implicit Arnoldi restart applies its shift polynomial and preserves an Arnoldi residual identity
- Eigenvalues, eigenvectors, eigenspaces $E_\lambda(T)=\ker(T-\lambda I)$, and the spectrum $\sigma_F(T)$ of an endomorphism
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
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Sources
- Yousef Saad, Numerical Methods for Large Eigenvalue Problems, second edition, §7.3.1 (standard reference, not scraped)
- Peter Arbenz, Lecture Notes on Solving Large Scale Eigenvalue Problems, §11.2 (standard reference, not scraped)