PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

For a Hermitian matrix, the eigenvectors are the stationary points of the Rayleigh quotient and twice the residual is its constrained gradient

Statement

Let $A$ be Hermitian and let $x$ be a unit vector. Write

$ρ (x) : = \frac{⟨ A x, x ⟩}{⟨ x, x ⟩}, r (x) : = A x - ρ (x) x .$

Then:

For every tangent vector $h$ to the unit sphere at $x$ , $D ρ (x) [h] = 2 Re ⟨ r (x), h ⟩ .$
The stationary points of $ρ_{A}$ on the unit sphere are exactly the unit eigenvectors of $A$ .

Thus $2 r (x)$ is the constrained gradient of the Rayleigh quotient for the standard real Riemannian metric on the unit sphere.

Facts & Assumptions

Given: A Hermitian matrix $A$ , a unit vector $x$ , and a tangent vector $h$ with $Re ⟨ x, h ⟩ = 0$ .

[L1]

Hermitian means $A^{*} = A$ (Self-adjoint and normal endomorphisms of a finite-dimensional real or complex inner product space).

Proof

technique · direct

1.1

Because $x$ is unit and $h$ is tangent to the unit sphere at $x$ , one has $⟨ x, x ⟩ = 1, Re ⟨ x, h ⟩ = 0.$ Differentiating $ρ (x + t h) = \frac{⟨ A (x + t h), x + t h ⟩}{⟨ x + t h, x + t h ⟩}$ at $t = 0$ therefore gives $D ρ (x) [h] = ⟨ A h, x ⟩ + ⟨ A x, h ⟩ .$ By [L1], $⟨ A h, x ⟩ = ⟨ h, A x ⟩$ , so $D ρ (x) [h] = 2 Re ⟨ A x, h ⟩ = 2 Re ⟨ A x - ρ (x) x, h ⟩,$ because $Re ⟨ x, h ⟩ = 0$ .

L1algebra

2.1

Since $⟨ x, x ⟩ = 1$ , one has $⟨ r (x), x ⟩ = ⟨ A x, x ⟩ - ρ (x) ⟨ x, x ⟩ = 0.$ So $r (x)$ itself lies in the tangent space at $x$ .

step 1.1algebra

2.2

Conversely, if $A x = λ x$ , then $ρ (x) = λ$ and $r (x) = 0$ , so step 1.1 gives $D ρ (x) [h] = 0$ for every tangent vector $h$ . Thus $x$ is a stationary point.

step 1.1algebra

3.1

If $x$ is a stationary point, then step 1.1 gives $Re ⟨ r (x), h ⟩ = 0$ for every tangent vector $h$ . Since step 2.1 places $r (x)$ in that tangent space, choosing $h = r (x)$ yields $∥ r (x) ∥_{2}^{2} = 0.$ Hence $r (x) = 0$ and $A x = ρ (x) x$ .

step 1.1step 2.1algebra

4.1

Steps 1.1, 3.1, and 2.2 prove the gradient and stationary-point claims.

step 1.1step 3.1step 2.2∎

Depends on

Self-adjoint and normal endomorphisms of a finite-dimensional real or complex inner product space

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

2 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

Per-Olof Persson, The QR Algorithm I (standard reference, not scraped)