ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

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Projection onto a finite-dimensional subspace by a Gram matrix

Example

Assume the Axiom of Countable Choice. Let $H$ be a real or complex Hilbert space, let $v_{1}, \dots, v_{n}$ be a linearly independent finite list in $H$ , put $M = span {v_{1}, \dots, v_{n}}$ , and for $x \in H$ set

$G_{i j} = ⟨ v_{i}, v_{j} ⟩, b_{i} = ⟨ x, v_{i} ⟩ (1 \leq i, j \leq n) .$

Then $M$ is closed, and the Hilbert projection of $x$ onto $M$ is

$P_{M} x = \sum_{j = 1}^{n} c_{j} v_{j}, where c \in K^{n} is the unique solution of G^{T} c = b .$

The result does not depend on the chosen independent spanning list: any other such list produces the same vector $P_{M} x$ and its own unique coefficient vector solving the corresponding system.

Facts & Assumptions

[A1]

The Gram matrix of an independent list satisfies $\det G > 0$ , and a square matrix is invertible exactly when its determinant is nonzero (A Gram determinant is nonnegative and is positive exactly when the vector list is linearly independent, The Gram matrix $G (v_{0}, \dots, v_{r - 1}) = (⟨ v_{i}, v_{j} ⟩)_{i, j < r}$ and Gram determinant, with empty value $1$ , A finite-dimensional linear operator over a field is invertible if and only if its determinant is nonzero).

[A2]

A finite-dimensional subspace of a normed space is closed, and the Hilbert projection $P_{M}$ is characterised by $P_{M} x \in M$ and $x - P_{M} x \in M^{⊥}$ (A finite-dimensional normed subspace is closed, The Hilbert orthogonal projection onto a closed subspace).

[A3]

$S^{⊥} = {v : ⟨ v, s ⟩ = 0 for all s \in S}$ and the pairing is linear in the first argument and conjugate-linear in the second (Orthogonality and the orthogonal complement, The Hilbert orthogonal projection onto a closed subspace).

[A4]

Countable Choice is the hypothesis under which the Hilbert projection is defined (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Verification

technique · direct

Given: Countable Choice, a Hilbert space $H$ , an independent list $v_{1}, \dots, v_{n} \in H$ , its span $M$ and a vector $x \in H$ .

1.1

The Gram matrix $G$ is invertible by [A1], so $G^{T}$ is invertible and $c = (G^{T})^{- 1} b$ is the unique solution of $G^{T} c = b$ ; and $M$ is closed by [A2].

A1A2A4

2.1

With $m = \sum_{j} c_{j} v_{j} \in M$ one has $⟨ x - m, v_{i} ⟩ = ⟨ x, v_{i} ⟩ - \sum_{j} c_{j} ⟨ v_{j}, v_{i} ⟩ = b_{i} - (G^{T} c)_{i} = 0$ for every $i$ , and hence $⟨ x - m, w ⟩ = 0$ for every $w = \sum_{i} a_{i} v_{i} \in M$ by conjugate-linearity in the second argument.

A3step 1.1algebra

3.1

Therefore $m \in M$ and $x - m \in M^{⊥}$ , so $m$ satisfies the two defining properties of the Hilbert projection and $P_{M} x = m = \sum_{j} c_{j} v_{j}$ .

step 1.1step 2.1A2

4.1

Basis independence and uniqueness: if $w_{1}, \dots, w_{n}$ is another independent list with the same span $M$ , its Gram matrix again has nonzero determinant and the same argument gives $P_{M} x$ as a linear combination of the $w_{i}$ with the unique coefficient vector solving the corresponding system; since $P_{M} x \in M$ is the same vector, the two displayed formulas agree, and the coefficient vector is unique because $G^{T}$ is invertible.

step 3.1A1A2∎

Depends on

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Sources

Theo Bühler and Dietmar Salamon, Functional Analysis, §1.3.3, pp.38–41 (standard reference, not scraped)
Andrew Lin and Casey Rodriguez, MIT 18.102 Introduction to Functional Analysis, Definition 182 and Proposition 183 (standard reference, not scraped)