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

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The Krylov subspace consists exactly of the vectors $p (A) b$ for zero polynomials or polynomials of degree less than $m$

Statement

Let $A$ be a square matrix over $R$ or $C$ , let $b$ be a vector of matching size, and let $m \geq 1$ . Then

$K_{m} (A, b) = {p (A) b : p = 0 or \deg p < m} .$

Facts & Assumptions

Given: A square matrix $A$ over $R$ or $C$ , a vector $b$ of matching size, and an integer $m \geq 1$ .

[F1]

For $m \geq 1$ , the Krylov subspace is $K_{m} (A, b) = span {b, A b, \dots, A^{m - 1} b}$ (The Krylov subspace $K_{m} (A, b) = span {b, A b, \dots, A^{m - 1} b}$ ).

Proof

technique · direct

1.1

If $v \in K_{m} (A, b)$ , then [F1] gives scalars $c_{0}, \dots, c_{m - 1}$ with $v = \sum_{j = 0}^{m - 1} c_{j} A^{j} b .$ With $p (z) : = \sum_{j = 0}^{m - 1} c_{j} z^{j}$ , either $p = 0$ or $\deg p < m$ , and $v = p (A) b$ .

F1constructalgebra

2.1

Conversely, $p = 0$ gives $p (A) b = 0 \in K_{m} (A, b)$ . If instead $\deg p < m$ and $p (z) = \sum_{j = 0}^{m - 1} c_{j} z^{j}$ , then $p (A) b = \sum_{j = 0}^{m - 1} c_{j} A^{j} b,$ which lies in the span from [F1]. Hence every displayed $p (A) b$ lies in $K_{m} (A, b)$ .

F1algebra∎

Depends on

The Krylov subspace $K_m(A,b)=\operatorname{span}\{b,Ab,\dots,A^{m-1}b\}$

Used by

Dependency tree · two levels

2 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

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)

The Krylov subspace consists exactly of the vectors p(A)bp(A)b for zero polynomials or polynomials of degree less than mm