PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31

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For Hermitian matrices, the Rayleigh quotient and residual converge with the expected rates along power iteration

Statement

Let $A$ be Hermitian with simple dominant eigenvalue $λ_{1}$ , let $q_{1}$ be a unit eigenvector, and let $(x_{k})$ be the power iteration from a start vector with nonzero $q_{1}$ -component. Put

$μ_{k} : = ⟨ A x_{k}, x_{k} ⟩, r_{k} : = A x_{k} - μ_{k} x_{k} .$

Then

$∣ μ_{k} - λ_{1} ∣ = O ⁣ ({∣ \frac{λ_{2}}{λ_{1}} ∣}^{2 k}), ∥ r_{k} ∥_{2} = O ⁣ ({∣ \frac{λ_{2}}{λ_{1}} ∣}^{k}) .$

In particular, $μ_{k} \to λ_{1}$ and $r_{k} \to 0$ .

Facts & Assumptions

Given: A Hermitian matrix $A$ with simple dominant eigenpair $(λ_{1}, q_{1})$ and a valid power iteration $(x_{k})$ .

[L1]

Power iteration converges projectively to the dominant eigendirection at rate $O (∣ λ_{2} / λ_{1} ∣^{k})$ (If a diagonalisable matrix has a simple eigenvalue of strictly largest modulus and the start vector has a nonzero component in that eigendirection, power iteration converges projectively at the eigenvalue-ratio rate).

[L2]

Hermitian means $A^{*} = A$ , so if $z ⊥ q_{1}$ then $⟨ A z, q_{1} ⟩ = ⟨ z, A q_{1} ⟩ = λ_{1} ⟨ z, q_{1} ⟩ = 0$ (Self-adjoint and normal endomorphisms of a finite-dimensional real or complex inner product space).

[L3]

The power iteration is the normalised recurrence $x_{k + 1} = A x_{k} / ∥ A x_{k} ∥_{2}$ , with Rayleigh estimates from the same iterates (Power iteration with normalisation and Rayleigh-quotient eigenvalue estimates).

[L4]

A real self-adjoint or complex Hermitian matrix has an orthonormal eigenbasis and is therefore diagonalisable (Real spectral theorem: a self-adjoint endomorphism of a finite-dimensional real 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

Let $ρ : = ∣ λ_{2} / λ_{1} ∣ < 1$ . By [L4], the Hermitian matrix is diagonalisable, so [L1] applies. After choosing phases $α_{k}$ of modulus one, one has $∥ α_{k} x_{k} - q_{1} ∥_{2} = O (ρ^{k}) .$ Write $α_{k} x_{k} = c_{k} q_{1} + z_{k}$ with $z_{k} ⊥ q_{1}$ . Then $∣ c_{k} - 1 ∣ + ∥ z_{k} ∥_{2} = O (ρ^{k})$ .

L1L4algebra

2.1

By [L2], the orthogonal complement $q_{1}^{⊥}$ is $A$ -invariant. Since $x_{k}$ is unit and $μ_{k} = ⟨ A (α_{k} x_{k}), α_{k} x_{k} ⟩$ , the cross terms vanish: $μ_{k} = ∣ c_{k} ∣^{2} λ_{1} + ⟨ A z_{k}, z_{k} ⟩ .$ Using $∣ c_{k} ∣^{2} + ∥ z_{k} ∥_{2}^{2} = 1$ , this becomes $μ_{k} - λ_{1} = ⟨ A z_{k}, z_{k} ⟩ - λ_{1} ∥ z_{k} ∥_{2}^{2} .$ Therefore $∣ μ_{k} - λ_{1} ∣ \leq (∥ A ∥_{2} + ∣ λ_{1} ∣) ∥ z_{k} ∥_{2}^{2} = O (ρ^{2 k}) .$

L2step 1.1algebra

3.1

Using $A q_{1} = λ_{1} q_{1}$ , one has $α_{k} r_{k} = A (α_{k} x_{k}) - μ_{k} (α_{k} x_{k}) = c_{k} (λ_{1} - μ_{k}) q_{1} + (A - μ_{k} I) z_{k} .$ Since $∣ α_{k} ∣ = 1$ , $∥ r_{k} ∥_{2} \leq ∣ c_{k} ∣ ∣ λ_{1} - μ_{k} ∣ + (∥ A ∥_{2} + ∣ μ_{k} ∣) ∥ z_{k} ∥_{2} .$ Step 2.1 and step 1.1 give $∥ r_{k} ∥_{2} = O (ρ^{k}) .$

step 1.1step 2.1algebra

4.1

The displayed bounds force $μ_{k} \to λ_{1}$ and $r_{k} \to 0$ .

step 2.1step 3.1L3∎

Depends on

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Sources

Andrew Stuart and Jochen Voss, Matrix Analysis and Algorithms (standard reference, not scraped)
Per-Olof Persson, The QR Algorithm I (standard reference, not scraped)