PropositionStatement: Literature-sourcedProof: AI-adapted

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.

The standard formulas $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ and $\sum_{k < n} x_{k} \overline{y_{k}}$ on $C^{n}$ are inner products

Statement

For $x, y \in R^{n}$ and $z, w \in C^{n}$ , the formulas

$⟨ x, y ⟩_{R^{n}} = \sum_{k < n} x_{k} y_{k}, ⟨ z, w ⟩_{C^{n}} = \sum_{k < n} z_{k} \overline{w_{k}}$

define inner products, linear in the first argument. At $n = 0$ , the unique pairing on the zero space is an inner product.

Facts & Assumptions

Given: A natural number $n$ and the two displayed coordinate pairings.

[L1]

Finite products in a commutative monoid have an empty value and may be read additively as finite sums (The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity).

[L2]

Complex conjugation preserves sums and products, and $z \overline{z} = ∣ z ∣^{2} \geq 0$ , with equality exactly when $z = 0$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[L3]

The standard unit vectors form an ordered basis of $F^{n}$ , including $n = 0$ (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[L4]

An inner product is linear in the first argument, conjugate symmetric, positive on the diagonal, and definite (Real and complex inner product spaces, with the inner product linear in the first argument).

[L5]

The reals form a totally ordered field, so addition preserves order; the square of a nonzero real number is positive (The reals form a totally ordered field, Squares of nonzero elements are positive).

Proof

technique · direct

1.1L1L2algebra

Distributivity of the finite sums in [L1] gives linearity in the first variable. In the complex case [L2] gives conjugate-linearity in the second and conjugate symmetry; in the real case conjugation is the identity.

1.2L1L2L3L5inductionalgebra

For a finite list of nonnegative real numbers $a_{0}, \dots, a_{n - 1}$ , put $S_{0} = 0$ and $S_{j + 1} = S_{j} + a_{j}$ as in [L1]. By [L5], if $S_{j} \geq 0$ and $a_{j} \geq 0$ , then $S_{j + 1} \geq S_{j}$ and $S_{j + 1} \geq a_{j}$ . Hence $S_{j + 1} = 0$ holds exactly when both $S_{j} = 0$ and $a_{j} = 0$ . Starting with the empty sum, induction proves $S_{n} \geq 0$ and $S_{n} = 0$ exactly when every $a_{k} = 0$ . On the diagonal, take $a_{k} = x_{k}^{2}$ in the real formula and $a_{k} = ∣ z_{k} ∣^{2}$ in the complex formula. By [L5], $x_{k}^{2}$ is nonnegative and equals zero exactly when $x_{k} = 0$ ; [L2] gives the corresponding assertion for $∣ z_{k} ∣^{2}$ . Thus both diagonal pairings are real and nonnegative, and each vanishes exactly when all coordinates vanish, which by [L3] means the vector is zero.

2.1step 1.1step 1.2L1L3L4∎

Hence all the axioms in [L4] hold. When $n = 0$ , the sum is empty and equals $0$ , while the zero vector is the only vector, so definiteness is valid.

Depends on

Dependency tree · 2 levels

Sources

Sheldon Axler, Linear Algebra Done Right, 4th ed., §6A, Example 6.3

The standard formulas ⟨x,y⟩=∑k<nxkyk⟨x,y⟩=∑k<n​xk​yk​ on RnRn and ∑k<nxkyk‾∑k<n​xk​yk​​ on CnCn are inner products