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Gaussian weights equal the total mass times a Jacobi eigenvector's first-coordinate square

Statement

Let $(a, b, m, 2 m - 1, μ)$ be moment data and let $Q_{m}$ have nodes $x_{i}$ and weights $w_{i}$ as in The m-node Gaussian quadrature rule for a positive measure. Let $J_{m}$ be the Jacobi matrix of The Jacobi matrix spectrum is the set of Gauss nodes, and let $v_{i}$ be any unit eigenvector of $J_{m}$ for eigenvalue $x_{i}$ . Then $w_{i} = μ ([a, b]) ∣ v_{i, 1} ∣^{2} .$

Facts & Assumptions

Given: Moment data, its Gaussian rule, its Jacobi matrix, and a unit eigenvector $v_{i}$ for node $x_{i}$ .

[F1]

The m-node Gaussian quadrature rule for a positive measure: $w_{i} = \int ℓ_{i} d μ$ and the nodes are the roots of $p_{m}$ .

[F2]

Gaussian quadrature is exact through degree 2m−1 and has positive weights: the rule is exact through degree $2 m - 1$ and $w_{i} > 0$ .

[F3]

Monic orthogonal polynomials exist uniquely through the quadrature order and Monic orthogonal polynomials satisfy a three-term recurrence: $p_{0} = 1$ , the $p_{j}$ are pairwise orthogonal, and their squared norms $h_{j}$ are positive.

[F4]

The Jacobi matrix spectrum is the set of Gauss nodes: the vectors $(p_{0} (x_{i}) / \sqrt{h_{0}}, \dots, p_{m - 1} (x_{i}) / \sqrt{h_{m - 1}})^{T}$ are nonzero eigenvectors of $J_{m}$ with eigenvalue $x_{i}$ , the nodes are distinct, and every eigenvalue is simple.

[F5]

Real spectral theorem: a self-adjoint endomorphism of a finite-dimensional real inner product space has an orthonormal eigenbasis: a real self-adjoint endomorphism has an orthonormal eigenbasis.

[F6]

Cardinal interpolation functions are polynomials of the stated degrees: $ℓ_{i}$ has degree at most $m - 1$ and $ℓ_{i} (x_{k}) = δ_{i k}$ .

[F7]

The polynomial moment pairing and Monic orthogonal polynomials satisfy a three-term recurrence define $h_{0} = \int p_{0}^{2} d μ$ . The integral of a nonnegative simple function and The nonnegative integral agrees with the simple integral on simple functions give $\int 1 d μ = μ ([a, b])$ by the one-cell simple function on the whole interval.

Proof

1.1F3F4construct

Put $q_{j} = p_{j} / \sqrt{h_{j}}$ for $j = 0, \dots, m - 1$ . By [F3], they are orthonormal. Their degrees are $0, \dots, m - 1$ with nonzero leading coefficients, so successive subtraction of top terms shows they span the degree-at-most- $m - 1$ polynomial space and are a basis. For a fixed node $x_{i}$ , set $z_{i} = (q_{0} (x_{i}), \dots, q_{m - 1} (x_{i}))^{T}$ . By [F4], $z_{i}$ is a nonzero eigenvector of $J_{m}$ for $x_{i}$ .

1.2F4F5algebraconstruct

For $i \neq k$ , self-adjointness gives $x_{i} ⟨ z_{i}, z_{k} ⟩ = ⟨ J_{m} z_{i}, z_{k} ⟩ = ⟨ z_{i}, J_{m} z_{k} ⟩ = x_{k} ⟨ z_{i}, z_{k} ⟩$ . Since the nodes are distinct by [F4], their difference is nonzero and the pairing is zero. Normalize these vectors to $u_{i} = z_{i} / ∥ z_{i} ∥_{2}$ ; their first coordinates satisfy $∣ u_{i, 1} ∣^{2} = \frac{q_{0} (x_{i})^{2}}{\sum_{j = 0}^{m - 1} q_{j} (x_{i})^{2}} .$ The simplicity assertion in [F4], together with the orthonormal eigenbasis [F5], makes each eigenspace one-dimensional. Therefore the arbitrary unit eigenvector $v_{i}$ in the Statement is $u_{i}$ or $- u_{i}$ , and its first-coordinate square is the same.

1.3F1F2F6algebra

Expand $ℓ_{i}$ in the orthonormal basis $q_{0}, \dots, q_{m - 1}$ . Its coefficient of $q_{j}$ is $\int ℓ_{i} q_{j} d μ = Q_{m} (ℓ_{i} q_{j}) = w_{i} q_{j} (x_{i}),$ because $\deg (ℓ_{i} q_{j}) \leq 2 m - 2$ and $ℓ_{i} (x_{k}) = δ_{i k}$ by [F6].

2.1F2F6step 1.3algebra

By [F6], $ℓ_{i}^{2}$ has degree at most $2 m - 2$ and $Q_{m} (ℓ_{i}^{2}) = w_{i}$ , so exactness [F2] gives $\int ℓ_{i}^{2} d μ = w_{i}$ . Taking squared norms of the expansion in step 1.3 gives $w_{i} = \int ℓ_{i}^{2} d μ = w_{i}^{2} \sum_{j = 0}^{m - 1} q_{j} (x_{i})^{2} .$ Since $w_{i} > 0$ by [F2], it follows that $w_{i} = 1 / \sum_{j} q_{j} (x_{i})^{2}$ .

3.1F1F3F7step 1.2step 2.1algebra∎

Since $p_{0} = 1$ by [F3], [F7] gives $h_{0} = \int p_{0}^{2} d μ = \int 1 d μ = μ ([a, b])$ ; thus one has $q_{0} (x_{i})^{2} = 1 / μ ([a, b])$ . Substituting this and step 2.1 into step 1.2 yields $∣ v_{i, 1} ∣^{2} = w_{i} / μ ([a, b])$ , which is the claimed formula.

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Sources

John A. Gubner, Gaussian Quadrature and the Eigenvalue Problem, §§2–5, 2020 (standard reference, not scraped)
MATH 373 Lecture Notes, Lecture 14: Gaussian Quadrature, §§10.2–10.5 (standard reference, not scraped)
Walter Gautschi, The Interplay Between Classical Analysis and (Numerical) Linear Algebra, §2.1 (standard reference, not scraped)