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CG obeys the Chebyshev $A$ -norm bound in terms of the spectral condition number $κ_{2} (A)$

Statement

Let $A \in M_{n} (C)$ be Hermitian positive definite, let $x_{*}$ solve $A x = b$ , let $e_{m} : = x_{*} - x_{m}$ be the error in the $m$ th exact-arithmetic CG iterate, let $0 < λ_{\min} \leq λ_{\max}$ be the smallest and largest eigenvalues of $A$ , and let

$κ : = \frac{λ_{\max}}{λ_{\min}} .$

Then for every $m \geq 0$ ,

$∥ e_{m} ∥_{A} \leq 2 {(\frac{\sqrt{κ} - 1}{\sqrt{κ} + 1})}^{m} ∥ e_{0} ∥_{A} .$

Facts & Assumptions

Given: A Hermitian positive-definite matrix $A$ , the exact solution $x_{*}$ , the CG iterates $x_{m}$ , the errors $e_{m} = x_{*} - x_{m}$ , and the extremal eigenvalues $0 < λ_{\min} \leq λ_{\max}$ of $A$ .

[L1]

The $m$ th CG iterate uniquely minimizes the $A$ -norm of the error over $x_{0} + K_{m} (A, r_{0})$ (CG is the Krylov Galerkin method, and the $m$ th iterate uniquely minimizes the $A$ -norm of the error over $x_{0} + K_{m} (A, r_{0})$ ).

[L2]

In an orthonormal basis, a Hermitian matrix is normal (In an orthonormal basis, self-adjoint means conjugate-transpose symmetry and normal means commuting with the conjugate transpose).

[L3]

A normal endomorphism of a finite-dimensional complex inner-product space has an orthonormal eigenbasis (Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely).

Proof

technique · direct

1.1

Every vector in $x_{0} + K_{m} (A, r_{0})$ has the form $x_{0} + A z (A) e_{0}$ for some polynomial $z$ of degree at most $m - 1$ , because $r_{0} = A e_{0}$ . Hence every attainable error has the form $e_{0} - A z (A) e_{0} = p (A) e_{0},$ where $p (0) = 1$ and $\deg p \leq m$ . By [L1], the actual CG error therefore satisfies $∥ e_{m} ∥_{A} \leq \min_{\begin{array}{c} p (0) = 1 \\ \deg p \leq m \end{array}} ∥ p (A) e_{0} ∥_{A} .$

L1algebra

2.1

Since $A$ is Hermitian, [L2] and [L3] give an orthonormal basis of eigenvectors, so $A = U Λ U^{*}$ with $U$ unitary and $Λ = diag (λ_{1}, \dots, λ_{n})$ , where each $λ_{j} > 0$ . Writing $U^{*} e_{0} = c$ , one gets $∥ p (A) e_{0} ∥_{A}^{2} = ⟨ A p (A) e_{0}, p (A) e_{0} ⟩ = \sum_{j = 1}^{n} λ_{j} ∣ p (λ_{j}) ∣^{2} ∣ c_{j} ∣^{2} \leq \max_{λ \in [λ_{\min}, λ_{\max}]} ∣ p (λ) ∣^{2} ∥ e_{0} ∥_{A}^{2} .$ Thus $∥ e_{m} ∥_{A} \leq \min_{\begin{array}{c} p (0) = 1 \\ \deg p \leq m \end{array}} \max_{λ \in [λ_{\min}, λ_{\max}]} ∣ p (λ) ∣ ∥ e_{0} ∥_{A} .$

L2L3step 1.1algebra

3.1

Let $t (λ) : = \frac{λ_{\max} + λ_{\min} - 2 λ}{λ_{\max} - λ_{\min}}, p_{m} (λ) : = \frac{T_{m} (t (λ))}{T_{m} ⁣ (\frac{λ_{\max} + λ_{\min}}{λ_{\max} - λ_{\min}})},$ where $T_{m}$ is the $m$ th Chebyshev polynomial. Then $p_{m} (0) = 1$ , $\deg p_{m} = m$ , and $∣ T_{m} (t) ∣ \leq 1$ on $[- 1, 1]$ , so $\max_{λ \in [λ_{\min}, λ_{\max}]} ∣ p_{m} (λ) ∣ \leq \frac{1}{∣ T_{m} ⁣ (\frac{λ_{\max} + λ_{\min}}{λ_{\max} - λ_{\min}}) ∣} .$ With $κ = λ_{\max} / λ_{\min}$ , the denominator equals $T_{m} ((κ + 1) / (κ - 1))$ . For $s > 1$ , $T_{m} (s) = \frac{(s + \sqrt{s^{2} - 1})^{m} + (s - \sqrt{s^{2} - 1})^{m}}{2} \geq \frac{(s + \sqrt{s^{2} - 1})^{m}}{2},$ and at $s = (κ + 1) / (κ - 1)$ one has $\frac{1}{s + \sqrt{s^{2} - 1}} = \frac{\sqrt{κ} - 1}{\sqrt{κ} + 1} .$ Therefore $\max_{λ \in [λ_{\min}, λ_{\max}]} ∣ p_{m} (λ) ∣ \leq 2 {(\frac{\sqrt{κ} - 1}{\sqrt{κ} + 1})}^{m} .$ Combining this polynomial with step 2.1 gives the stated bound.

step 2.1algebra∎

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Sources

Jonathan Richard Shewchuk, An Introduction to the Conjugate Gradient Method Without the Agonizing Pain (standard reference, not scraped)

CG obeys the Chebyshev AA-norm bound in terms of the spectral condition number κ2(A)κ2​(A)