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Residual bounds on forward error for a nonsingular linear system

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 $x = A^{- 1} b$ , and let $\hat{x} \in R^{n}$ be an approximate solution with residual $r : = b - A \hat{x}$ and error $e : = x - \hat{x}$ . Then $r = A e$ and $e = A^{- 1} r$ , and the following bounds hold:

Absolute bounds. $∥ e ∥_{p} \leq ∥ A^{- 1} ∥_{p} ∥ r ∥_{p}, ∥ r ∥_{p} \leq ∥ A ∥_{p} ∥ e ∥_{p} .$
Relative bounds. $\frac{1}{κ_{p} (A)} \frac{∥ r ∥_{p}}{∥ b ∥_{p}} \leq \frac{∥ e ∥_{p}}{∥ x ∥_{p}} \leq κ_{p} (A) \frac{∥ r ∥_{p}}{∥ b ∥_{p}},$ where $κ_{p} (A)$ 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 $∥ r ∥_{p} / ∥ A ∥_{p}$ , but bounds it above only up to the condition number $κ_{p} (A)$ .

Facts & Assumptions

Given: An invertible matrix $A \in {GL}_{n} (R)$ with $n \geq 1$ , $b \neq 0$ , $x = A^{- 1} b$ , $\hat{x} \in R^{n}$ , $r = b - A \hat{x}$ and $e = x - \hat{x}$ .

[L1]

$A^{- 1} A = I_{n}$ (Invertible matrices and the general linear group ${GL}_{n} (F)$ ), and matrix multiplication distributes over addition, so $A (x - \hat{x}) = A x - A \hat{x}$ (Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication).

[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

1.1givenL1algebra

Since $A x = b$ , the residual is $r = b - A \hat{x} = A x - A \hat{x} = A (x - \hat{x}) = A e$ , using distributivity of [L1].

2.1step 1.1L1algebra

Multiplying $r = A e$ on the left by $A^{- 1}$ and using [L1] gives $e = A^{- 1} r$ .

2.2step 1.1L2algebra

The lower absolute bound. Applying [L2] to $A$ at the vector $e$ in step 1.1 gives $∥ r ∥_{p} = ∥ A e ∥_{p} \leq ∥ A ∥_{p} ∥ e ∥_{p}$ .

3.1step 2.1L2algebra

The upper absolute bound. Applying [L2] to $A^{- 1}$ at the vector $r$ in step 2.1 gives $∥ e ∥_{p} = ∥ A^{- 1} r ∥_{p} \leq ∥ A^{- 1} ∥_{p} ∥ r ∥_{p}$ .

3.2step 2.2L2L3algebra

The lower relative bound. From $x = A^{- 1} b$ and [L2], $∥ x ∥_{p} \leq ∥ A^{- 1} ∥_{p} ∥ b ∥_{p}$ , so $1 / ∥ b ∥_{p} \leq ∥ A^{- 1} ∥_{p} / ∥ x ∥_{p}$ ; combining with step 2.2 gives $∥ r ∥_{p} / ∥ b ∥_{p} \leq ∥ A ∥_{p} ∥ A^{- 1} ∥_{p} ∥ e ∥_{p} / ∥ x ∥_{p} = κ_{p} (A) ∥ e ∥_{p} / ∥ x ∥_{p}$ by [L3], which rearranges to the lower bound on $∥ e ∥_{p} / ∥ x ∥_{p}$ .

4.1step 3.1L2L3algebra

The upper relative bound. From $b = A x$ and [L2], $∥ b ∥_{p} \leq ∥ A ∥_{p} ∥ x ∥_{p}$ , and $b \neq 0$ forces $∥ x ∥_{p} > 0$ , so $1 / ∥ x ∥_{p} \leq ∥ A ∥_{p} / ∥ b ∥_{p}$ ; multiplying by step 3.1 gives $∥ e ∥_{p} / ∥ x ∥_{p} \leq ∥ A^{- 1} ∥_{p} ∥ A ∥_{p} ∥ r ∥_{p} / ∥ b ∥_{p} = κ_{p} (A) ∥ r ∥_{p} / ∥ b ∥_{p}$ by [L3].

5.1step 3.1step 2.2step 4.1step 3.2∎

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

Depends on

Used by

A tiny residual does not force a small forward error Counterexample
Residual formulas for normwise and componentwise backward error Proposition

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

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)