TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

In exact arithmetic, CG terminates no later than the relative grade and hence in at most $n$ steps

Statement

Let $A \in M_{n} (C)$ be Hermitian positive definite, let $A x = b$ , let $x_{0}$ be an initial guess, let $r_{0} = b - A x_{0}$ , and let $ν = ν (A, r_{0})$ be the grade of $r_{0}$ . Then exact-arithmetic conjugate gradients reaches the exact solution no later than step $ν$ . In particular, $ν \leq n$ , so CG terminates in at most $n$ steps.

Facts & Assumptions

Given: A Hermitian positive-definite system $A x = b$ , an initial guess $x_{0}$ , the initial residual $r_{0} = b - A x_{0}$ , the grade $ν = ν (A, r_{0})$ , and the exact solution $x_{*}$ .

[F1]

The grade $ν (A, r_{0})$ is the degree of the monic polynomial of least degree that annihilates $r_{0}$ , and it is $0$ when $r_{0} = 0$ (The grade of a start vector and its relative minimal polynomial).

[L1]

If $A$ is invertible, then the exact solution of $A e = r_{0}$ lies in $K_{ν} (A, r_{0})$ (If $A$ is invertible, the exact solution of $A x = b$ lies in the Krylov space at the grade).

[L2]

The $m$ th CG iterate uniquely minimizes the $A$ -norm of the error over $x_{0} + K_{m} (A, r_{0})$ (CG is the Krylov Galerkin method, and the $m$ th iterate uniquely minimizes the $A$ -norm of the error over $x_{0} + K_{m} (A, r_{0})$ ).

Proof

technique · direct

1.1F1L1algebra

If $r_{0} = 0$ , then $A x_{0} = b$ , so CG has already reached the exact solution at step $0 = ν$ by [F1]. Assume now that $r_{0} \neq 0$ . The initial error $e_{0} : = x_{*} - x_{0}$ satisfies $A e_{0} = r_{0}$ , so [L1] gives $e_{0} \in K_{ν} (A, r_{0})$ . Therefore $x_{*} = x_{0} + e_{0} \in x_{0} + K_{ν} (A, r_{0}) .$

2.1L2step 1.1

If CG has already reached $x_{*}$ at some step $m < ν$ , then it has certainly terminated no later than step $ν$ . Otherwise the recurrence defines the $ν$ th iterate, so [L2] applies with $m = ν$ . Since $x_{*}$ itself belongs to the admissible affine space from step 1.1 and has zero error, the minimal $A$ -norm error over that space is $0$ . Hence the unique minimizer is $x_{ν} = x_{*}$ , so CG terminates no later than step $ν$ .

3.1F1step 2.1algebra∎

Because $A$ is $n \times n$ , the $n + 1$ vectors $r_{0}, A r_{0}, \dots, A^{n} r_{0}$ are linearly dependent. Thus some nonzero polynomial of degree at most $n$ annihilates $r_{0}$ , and the minimal possible degree in [F1] therefore satisfies $ν \leq n$ . Combining this with step 2.1 gives termination in at most $n$ steps.

Depends on

Used by

Dependency tree · two levels

10 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

Magnus R. Hestenes and Eduard Stiefel, Methods of Conjugate Gradients for Solving Linear Systems (standard reference, not scraped)

In exact arithmetic, CG terminates no later than the relative grade and hence in at most nn steps