TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31

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The dimensions of the Krylov spaces grow by one until the grade and then stabilize

Statement

Let $ν = ν (A, b)$ be the grade of $b$ relative to $A$ . Then:

for every integer $m$ with $0 \leq m < ν$ , one has $\dim K_{m} (A, b) = m$ ;
for every integer $m \geq ν$ , one has $K_{m} (A, b) = K_{ν} (A, b)$ .

Equivalently, the dimensions increase by one at each step until the grade and then stop changing.

Facts & Assumptions

Given: A square matrix $A$ , a vector $b$ , and its grade $ν = ν (A, b)$ .

[F1]

The relative minimal polynomial $q_{A, b}$ is monic of degree $ν$ and satisfies $q_{A, b} (A) b = 0$ (The grade of a start vector and its relative minimal polynomial).

[L1]

For every $m \geq 1$ , $K_{m} (A, b) = {p (A) b : \deg p < m}$ (The Krylov subspace consists exactly of the vectors $p (A) b$ for zero polynomials or polynomials of degree less than $m$ ).

Proof

technique · direct

1.1

If $0 \leq m < ν$ and $\sum_{j = 0}^{m - 1} c_{j} A^{j} b = 0$ , then the polynomial $p (z) = \sum_{j = 0}^{m - 1} c_{j} z^{j}$ satisfies $\deg p < m < ν$ and $p (A) b = 0$ . By minimality in [F1], this forces $p = 0$ , so $b, A b, \dots, A^{m - 1} b$ are linearly independent. Therefore $\dim K_{m} (A, b) = m$ .

F1L1algebra

2.1

Writing $q_{A, b} (z) = a_{0} + a_{1} z + \dots + a_{ν - 1} z^{ν - 1} + z^{ν}$ , the relation $q_{A, b} (A) b = 0$ from [F1] shows that $A^{ν} b \in K_{ν} (A, b)$ . Multiplying the same relation by $A^{k}$ for each $k \geq 0$ gives $A^{ν + k} b \in K_{ν} (A, b)$ as well. Hence every generator of $K_{m} (A, b)$ for $m \geq ν$ already lies in $K_{ν} (A, b)$ , so $K_{m} (A, b) = K_{ν} (A, b)$ .

F1algebra∎

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Sources

Yousef Saad, CSCI 8314 Set 13: Krylov subspace methods; Introduction; Krylov subspaces; Gram-Schmidt process (review); The Arnoldi process; FOM and GMRES (standard reference, not scraped)