TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-08-29

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Matrix perturbations in Ax = b obey a first-order condition-number bound

Statement

Let $n \geq 1$ , let $p \in Q$ with $p \geq 1$ , let $A \in {GL}_{n} (R)$ , let $b \in R^{n}$ with $b \neq 0$ , let $δ A \in M_{n} (R)$ , let $x = A^{- 1} b$ , and suppose $δ x \in R^{n}$ satisfies $(A + δ A) (x + δ x) = b$ . Put

$c : = ∥ A^{- 1} ∥_{p} ∥ δ A ∥_{p} = κ_{p} (A) \frac{∥ δ A ∥_{p}}{∥ A ∥_{p}} .$

If $c < 1$ then

$\frac{∥ δ x ∥_{p}}{∥ x ∥_{p}} \leq \frac{c}{1 - c} .$

To first order in the matrix perturbation,

$\frac{∥ δ x ∥_{p}}{∥ x ∥_{p}} ≲ κ_{p} (A) \frac{∥ δ A ∥_{p}}{∥ A ∥_{p}},$

because $c / (1 - c) = c + c^{2} / (1 - c) = c + O (c^{2})$ as $c \to 0$ : the denominator $1 - c$ contributes only a second-order correction when the relative matrix perturbation is small.

Facts & Assumptions

Given: An invertible matrix $A \in {GL}_{n} (R)$ with $n \geq 1$ , vectors $b \neq 0$ , $x = A^{- 1} b$ , and $δ x$ with $(A + δ A) (x + δ x) = b$ for a matrix $δ A$ ; write $c = ∥ A^{- 1} ∥_{p} ∥ δ A ∥_{p}$ .

[L1]

$A^{- 1} A = I_{n}$ (Invertible matrices and the general linear group ${GL}_{n} (F)$ ).

[L2]

Compatibility: $∥ M y ∥_{p} \leq ∥ M ∥_{p} ∥ y ∥_{p}$ (Induced matrix norms are compatible with matrix-vector multiplication, submultiplicative, and satisfy ||I|| = 1).

[L3]

$κ_{p} (A) = ∥ A ∥_{p} ∥ A^{- 1} ∥_{p}$ (The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system).

Proof

technique · direct

1.1givenalgebra

Expanding the perturbed equation and subtracting $A x = b$ gives $A δ x + δ A x + δ A δ x = 0$ , so $A δ x = - δ A (x + δ x)$ .

2.1step 1.1L1algebra

Multiplying step 1.1 on the left by $A^{- 1}$ and using [L1] gives $δ x = - A^{- 1} δ A (x + δ x)$ .

3.1step 2.1L2algebra

Taking norms in step 2.1 and applying [L2] twice gives $∥ δ x ∥_{p} = ∥ A^{- 1} δ A (x + δ x) ∥_{p} \leq ∥ A^{- 1} ∥_{p} ∥ δ A ∥_{p} ∥ x + δ x ∥_{p} = c ∥ x + δ x ∥_{p}$ .

4.1step 3.1algebra

The triangle inequality of the vector $p$ -norm gives $∥ x + δ x ∥_{p} \leq ∥ x ∥_{p} + ∥ δ x ∥_{p}$ , so step 3.1 yields $∥ δ x ∥_{p} \leq c ∥ x ∥_{p} + c ∥ δ x ∥_{p}$ .

5.1step 4.1givenalgebra

Under the hypothesis $c < 1$ the factor $1 - c$ is positive, and rearranging step 4.1 gives $(1 - c) ∥ δ x ∥_{p} \leq c ∥ x ∥_{p}$ , hence $∥ δ x ∥_{p} / ∥ x ∥_{p} \leq c / (1 - c)$ , the claimed exact bound; $x \neq 0$ because $b \neq 0$ and $A$ is invertible.

6.1step 5.1L3algebra

The first-order form. Writing $c = κ_{p} (A) ∥ δ A ∥_{p} / ∥ A ∥_{p}$ by [L3], one has $c / (1 - c) = c + c^{2} / (1 - c)$ , so the exact bound of step 5.1 reads $∥ δ x ∥_{p} / ∥ x ∥_{p} \leq κ_{p} (A) ∥ δ A ∥_{p} / ∥ A ∥_{p} + O (c^{2})$ with the error term tending to zero quadratically in $c$ ; to first order the amplification is $κ_{p} (A)$ times the relative matrix perturbation.

7.1step 5.1step 6.1∎

Step 5.1 proves the exact bound and step 6.1 the first-order reading of it.

Depends on

Used by

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Sources

L. N. Trefethen and D. Bau III, Numerical Linear Algebra, Lecture 12 (standard reference, not scraped)
James Demmel, Math 221 Lecture 3: Vector Norms, Matrix Norms, and Condition Numbers (standard reference, not scraped)