The Arnoldi and Lanczos procedures are memory-intensive for large k because all previous basis vectors must be stored (in Arnoldi) or because re-orthogonalisation requires them (in Lanczos). One practical remedy is:

Question

BlackTom AI · Accepted Answer

Restarted Arnoldi / Lanczos: run k steps, extract the best Ritz vector as a new starting vector, discard the rest, and restart — keeping storage bounded at O(kn) for a fixed restart length k.

1) Relevant knowledge points and concepts
- Arnoldi builds an orthonormal basis {v1,...,vm} of a Krylov subspace; Lanczos is the symmetric analogue. Both accumulate basis vectors, so storage scales with the number of steps m times the vector length n.
- Ritz vectors/values are approximate eigenvectors/values extracted from the small Hessenberg (Arnoldi) or tridiagonal (Lanczos) matrix.
- A “restart” means limiting m to a fixed k, using the best Ritz vector(s) to form a new starting vector, and repeating the process so storage and work per cycle remain bounded.

2) Step-by-step reasoning
- If you let the method run to m iterations, Arnoldi stores m basis vectors → storage O(mn). For very large m this is prohibitive.
- Restarted methods cap m at k: run k iterations, compute Ritz pairs, keep one (or a small set) of Ritz vectors that best approximate desired eigenvectors, and reinitialize the process with them.
- By discarding the other k−1 basis vectors you free memory and start a new cycle; each cycle only needs to store k vectors.

3) Calculations / intermediate steps
- One basis vector has n entries → memory cost ~ n per vector.
- Storing k basis vectors costs O(k n).
- Without restart, storing m vectors costs O(m n). With restart and fixed k, memory is bounded by O(k n) regardless of how many cycles are run.

4) Why the correct answer is correct
- Restarting directly addresses the memory growth: it truncates the space, retains useful spectral information via Ritz vectors, and reuses it. Thus the method keeps storage bounded and still converges over repeated cycles toward desired eigenpairs.

5) Why other options are incorrect
- a. Conjugate Gradient (CG) is for SPD linear systems (not general eigenproblems) and while it has short recurrences, it does not generally replace Arnoldi/Lanczos for eigenvalue computations; the option’s claim “same cost but no memory issue” is misleading.
- c. Increasing tolerance may reduce iterations but sacrifices accuracy and does not fundamentally solve the memory-growth mechanism of storing many basis vectors.
- d. Switching to LU decomposition is impractical for large sparse problems: LU causes fill-in and typically requires far more memory and work than restarted Krylov methods; it’s not a remedy for memory in iterative eigenvalue solvers.

6) Conclusion
The restarted Arnoldi/Lanczos strategy (option b) is the practical remedy that bounds storage to O(k n); this matches the provided correct answer.

The Arnoldi and Lanczos procedures are memory-intensive for large k because all previous basis vectors must be stored (in Arnoldi) or because re-orthogonalisation requires them (in Lanczos). One practical remedy is:单项选择题

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