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For Hermitian positive-definite $A$ , the exact solution of $A x = b$ is the unique minimizer of the quadratic energy

Statement

Let $A \in M_{n} (C)$ be Hermitian positive definite, let $b \in C^{n}$ , and let $ϕ$ be the quadratic energy from The quadratic energy $ϕ (x) = \frac{1}{2} ⟨ A x, x ⟩ - Re ⟨ b, x ⟩$ of a Hermitian positive-definite linear system. Then the system $A x = b$ has a unique solution $x_{*}$ , and for every $x \in C^{n}$ one has

$ϕ (x) - ϕ (x_{*}) = \frac{1}{2} ∥ x - x_{*} ∥_{A}^{2} .$

In particular, $x_{*}$ is the unique minimizer of $ϕ$ .

Facts & Assumptions

Given: A Hermitian positive-definite matrix $A \in M_{n} (C)$ , a vector $b \in C^{n}$ , and the quadratic energy $ϕ (x)$ .

[F1]

The quadratic energy is $ϕ (x) = \frac{1}{2} ⟨ A x, x ⟩ - Re ⟨ b, x ⟩$ (The quadratic energy $ϕ (x) = \frac{1}{2} ⟨ A x, x ⟩ - Re ⟨ b, x ⟩$ of a Hermitian positive-definite linear system).

[L1]

The energy norm is defined by $∥ u ∥_{A}^{2} = ⟨ A u, u ⟩$ for Hermitian positive-definite $A$ (The energy inner product $⟨ u, v ⟩_{A} = ⟨ A u, v ⟩$ and energy norm $∥ u ∥_{A}$ for a Hermitian positive-definite matrix).

[L2]

A matrix is Hermitian positive definite if and only if it has a Cholesky factorization $A = L L^{*}$ with positive diagonal, and that factor is unique (A matrix admits a Cholesky factorisation with positive diagonal exactly when it is Hermitian positive definite, and that factor is unique).

Proof

technique · direct

1.1

By [L2], there is a lower-triangular matrix $L$ with positive diagonal such that $A = L L^{*}$ . Hence $L$ and $L^{*}$ are invertible, so $A$ is invertible and the system $A x = b$ has the unique solution $x_{*} : = A^{- 1} b$ .

L2algebra

2.1

Since $A x_{*} = b$ , [F1] gives $ϕ (x) - ϕ (x_{*}) = \frac{1}{2} ⟨ A (x - x_{*}), x - x_{*} ⟩ + Re ⁣ (\frac{1}{2} ⟨ A x_{*}, x - x_{*} ⟩ - \frac{1}{2} ⟨ A (x - x_{*}), x_{*} ⟩) .$ Because $A$ is Hermitian, $⟨ A x_{*}, x - x_{*} ⟩ = ⟨ x_{*}, A (x - x_{*}) ⟩$ is the complex conjugate of $⟨ A (x - x_{*}), x_{*} ⟩$ , so the real part in parentheses vanishes. Therefore $ϕ (x) - ϕ (x_{*}) = \frac{1}{2} ⟨ A (x - x_{*}), x - x_{*} ⟩ = \frac{1}{2} ∥ x - x_{*} ∥_{A}^{2} .$

F1L1step 1.1algebra

3.1

By [L1], the quantity $∥ x - x_{*} ∥_{A}^{2}$ is nonnegative and is zero only when $x = x_{*}$ . Step 2.1 therefore shows $ϕ (x) \geq ϕ (x_{*})$ for every $x$ , with equality only at $x_{*}$ . Hence $x_{*}$ is the unique minimizer of $ϕ$ .

L1step 2.1∎

Depends on

Used by

CG is the Krylov Galerkin method, and the mth iterate uniquely minimizes the A-norm of the error over x₀+Kₘ(A,r₀) Theorem

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Sources

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

For Hermitian positive-definite AA, the exact solution of Ax=bAx=b is the unique minimizer of the quadratic energy