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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-08-29
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Residual bounds on forward error for a nonsingular linear system

Statement

Let n1, let pQ with p1, let AGLn(R), let bRn with b0, let x=A1b, and let x^Rn be an approximate solution with residual r:=bAx^ and error e:=xx^. Then r=Ae and e=A1r, and the following bounds hold:

  1. Absolute bounds. ep    A1prp,rp    Apep.
  2. Relative bounds. 1κp(A)rpbp    epxp    κp(A)rpbp, where κp(A) is the condition number of The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system.

In words: a small residual bounds the forward error below by rp/Ap, but bounds it above only up to the condition number κp(A).

Facts & Assumptions

Given: An invertible matrix AGLn(R) with n1, b0, x=A1b, x^Rn, r=bAx^ and e=xx^.

[L1]

A1A=In (Invertible matrices and the general linear group GLn(F)), and matrix multiplication distributes over addition, so A(xx^)=AxAx^ (Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication).

[L2]

Compatibility: MypMpyp (Induced matrix norms are compatible with matrix-vector multiplication, submultiplicative, and satisfy ||I|| = 1).

[L3]

κp(A)=ApA1p (The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system).

Proof

technique · direct
1.1

Since Ax=b, the residual is r=bAx^=AxAx^=A(xx^)=Ae, using distributivity of [L1].

givenL1algebra
2.1

Multiplying r=Ae on the left by A1 and using [L1] gives e=A1r.

step 1.1L1algebra
2.2

The lower absolute bound. Applying [L2] to A at the vector e in step 1.1 gives rp=AepApep.

step 1.1L2algebra
3.1

The upper absolute bound. Applying [L2] to A1 at the vector r in step 2.1 gives ep=A1rpA1prp.

step 2.1L2algebra
3.2

The lower relative bound. From x=A1b and [L2], xpA1pbp, so 1/bpA1p/xp; combining with step 2.2 gives rp/bpApA1pep/xp=κp(A)ep/xp by [L3], which rearranges to the lower bound on ep/xp.

step 2.2L2L3algebra
4.1

The upper relative bound. From b=Ax and [L2], bpApxp, and b0 forces xp>0, so 1/xpAp/bp; multiplying by step 3.1 gives ep/xpA1pAprp/bp=κp(A)rp/bp by [L3].

step 3.1L2L3algebra
5.1

Claim 1 is steps 3.1 and 2.2, and claim 2 is steps 4.1 and 3.2.

step 3.1step 2.2step 4.1step 3.2

Depends on

Used by

Dependency tree · two levels

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Sources