lanczos

name	lanczos
origin	workshop
description	Compute the dominant k singular values of a LARGE SPARSE matrix or a MATRIX-FREE linear operator (available only through matvec) using Lanczos (scipy.sparse.linalg.svds). Use when the object is sparse or is only available as an apply()/matvec — never densify it. For a small dense matrix, use the qr-iteration skill instead.

Lanczos skill

Use this skill when:

• The user has a large sparse matrix and wants its dominant singular values.
• The user has a matrix-free operator (only apply(v) / matvec available), e.g. an implicit PDE solve.

Do not densify the object. For a small dense matrix, use the qr-iteration skill.

How to invoke

# sparse matrix
python .claude/skills/lanczos/lanczos_svd.py --problem problem.py --source sparse --k 6

# matrix-free operator
python .claude/skills/lanczos/lanczos_svd.py --problem problem.py --source operator --k 6

The --problem module must expose get_sparse() (a scipy sparse matrix) and/or get_operator() (an object with .apply and .shape). --source selects which; --k sets how many dominant singular values.

Output looks like:

matrix-free operator (1600, 1600): dominant 6 singular values via Lanczos (svds)
  [0.8352 0.6699 0.6699 0.5593 0.5043 0.5043]

Interpreting the output

• svds returns the dominant k singular values via implicitly restarted Lanczos; each iteration costs one matvec (for the operator, one apply = one sparse solve).
• If the helper aborts with "this object is dense," route it to qr-iteration instead.

Extending

Add a --which SM option for the smallest singular values (useful for conditioning), or accept rmatvec separately so nonsymmetric operators are handled correctly (this skill assumes a symmetric operator, rmatvec = apply).