LemmaStatement: AI-generatedProof: AI-generated

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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.
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Distinct complex exponentials have full interval Gram rank

Statement

Let $t > 0$ , let $z_{1}, \dots, z_{n}$ be pairwise distinct complex numbers, and let $a_{1}, \dots, a_{n}$ be nonzero complex numbers. The functions $a_{j} e^{z_{j} s}$ on $0 < s < t$ are linearly independent. Consequently the vectors $(a_{j} e^{z_{j} s})_{j = 1}^{n}$ , as $s$ varies in this interval, span $C^{n}$ . Repeated identical exponents contribute only one direction per distinct exponent after combining the corresponding nonzero coordinate vector.

Statement references

The exponential series of a bounded operator

Finite operator and ket conventions

The vector space F^X of all functions X → F with pointwise operations, and Fⁿ as the case X = n = {0, 1, …, n-1}

Linear independence: a finite list v : n → V is independent when ∑_i<n λᵢ vᵢ = 0_V forces every λᵢ = 0_F, and a subset S ⊆ V is independent when every injective finite list into S is independent

Linear combination of a finite list, and the span span(S) as the smallest linear subspace containing S

span(S) is exactly the set of linear combinations of finite lists of elements of S, and span(∅) = {0_V}

If dim_F V = n and U is a linear subspace of V, then U is finite-dimensional, dim_F U ≤ n, and dim_F U = n if and only if U = V

The induced length is a norm

Every finite-dimensional normed space is Banach

A linear map from a finite-dimensional normed space is bounded

Facts & Assumptions

[F1]

Give $C$ its standard coordinate inner product and ordered basis $(1)$ . Its induced length $∥ w ∥ = \sqrt{w \overline{w}} = ∣ w ∣$ is a complex norm by the inner-product norm corollary. With this one-vector ordered basis, the finite-dimensional completeness corollary makes it a complex Banach space. For fixed $z \in C$ , scalar multiplication $M_{z} (w) = z w$ is complex-linear, since $M_{z} (a u + b v) = a M_{z} (u) + b M_{z} (v)$ ; the finite-domain boundedness corollary therefore makes $M_{z} : C \to C$ bounded. Apply the bounded-operator exponential theorem with $X = C$ and $A = M_{z}$ . Its partial sums are multiplication by $\sum_{r = 0}^{N} (z s)^{r} / r!$ , and evaluating the operator-norm limit at $1$ identifies $E_{z} (s) 1 = e^{z s}$ . Evaluation is continuous because $∥ B 1 ∥ \leq ∥ B ∥ ∥ 1 ∥ = ∥ B ∥$ for a bounded operator $B$ . The derivative and group conclusions give $\frac{d^{k}}{d s^{k}} e^{z s} = z^{k} e^{z s}$ for every nonnegative integer $k$ and $e^{z s} e^{- z s} = 1$ , so $e^{z s} \neq 0$ . Finite operator and ket conventions The induced length is a norm Every finite-dimensional normed space is Banach A linear map from a finite-dimensional normed space is bounded The exponential series of a bounded operator

[F2]

Finite coordinate vectors, their specified bases, and linear pairings use the operator conventions. Finite operator and ket conventions

[F3]

The functions in the Statement belong to the complex vector space $C^{(0, t)}$ with pointwise operations and pointwise equality. Linear independence means that a vanishing finite linear combination forces every scalar coefficient to vanish. Span is the smallest containing linear subspace, equivalently the set of finite linear combinations. The vector space F^X of all functions X → F with pointwise operations, and Fⁿ as the case X = n = {0, 1, …, n-1} Linear independence: a finite list v : n → V is independent when ∑_i<n λᵢ vᵢ = 0_V forces every λᵢ = 0_F, and a subset S ⊆ V is independent when every injective finite list into S is independent Linear combination of a finite list, and the span span(S) as the smallest linear subspace containing S span(S) is exactly the set of linear combinations of finite lists of elements of S, and span(∅) = {0_V}

[F4]

A subspace of a finite-dimensional vector space has a finite basis, and any independent subset extends to a basis of the ambient finite-dimensional space, without a choice principle. If dim_F V = n and U is a linear subspace of V, then U is finite-dimensional, dim_F U ≤ n, and dim_F U = n if and only if U = V, Statement claims 1 and 3.

Proof

Given: The finite data and hypotheses in the Statement, with the function-space and coordinate conventions of [F2] and [F3].

Goal: Prove independence of the functions and the stated spans, including the repeated-exponent case.

1.1F1F3algebra

The scalar hypotheses and specialization established in [F1] apply to each $z_{j}$ . If $\sum_{j} c_{j} a_{j} e^{z_{j} s} = 0$ throughout the interval, differentiate $k = 0, \dots, n - 1$ times at an interior point $s_{0}$ . This yields $\sum_{j} d_{j} z_{j}^{k} = 0$ , where $d_{j} = c_{j} a_{j} e^{z_{j} s_{0}}$ . For each $l$ , the polynomial $p_{l} (x) = \prod_{j \neq l} (x - z_{j}) / (z_{l} - z_{j})$ has degree at most $n - 1$ and values $p_{l} (z_{j}) = δ_{l j}$ . Taking the corresponding linear combination of these equations gives $d_{l} = 0$ . Since $a_{l} e^{z_{l} s_{0}} \neq 0$ , every $c_{l} = 0$ , which is independence in $C^{(0, t)}$ .

2.1F2F3F4step 1.1algebra

Put $v (s) = (a_{j} e^{z_{j} s})_{j = 1}^{n}$ and $W = span {v (s) : 0 < s < t}$ . If $W$ were proper, [F4] would give a basis $b_{1}, \dots, b_{r}$ of $W$ extending to a basis $b_{1}, \dots, b_{n}$ of $C^{n}$ , with $r < n$ . For $x = \sum_{i = 1}^{n} λ_{i} b_{i}$ define $ℓ (x) = λ_{r + 1}$ . Independence of the basis makes these coordinates unique; comparison of coordinates under addition and scalar multiplication makes $ℓ$ linear. It vanishes on $W$ and satisfies $ℓ (b_{r + 1}) = 1$ . In the standard coordinate basis $(e_{j})$ , set $c_{j} = ℓ (e_{j})$ ; then $ℓ (x) = \sum_{j} c_{j} x_{j}$ and some $c_{j}$ is nonzero. But $ℓ (v (s)) = \sum_{j} c_{j} a_{j} e^{z_{j} s} = 0$ for all $s$ , contradicting step 1.1. Thus $W = C^{n}$ .

3.1F2F3step 2.1algebra∎

For a finite list of exponents allowing repetitions, let $w_{1}, \dots, w_{m}$ be its distinct values, and let $u_{l}$ have coordinate $a_{j}$ when $z_{j} = w_{l}$ and zero otherwise. Each $u_{l}$ is nonzero, and their disjoint coordinate supports make them independent: in $\sum_{l} α_{l} u_{l} = 0$ , a nonzero coordinate in the $l$ th group forces $α_{l} = 0$ . The profile vector is $\sum_{l} e^{w_{l} s} u_{l}$ . By step 2.1 with all amplitudes equal to one, the vectors $(e^{w_{l} s})_{l = 1}^{m}$ span $C^{m}$ . The linear map $(x_{l})_{l} \mapsto \sum_{l} x_{l} u_{l}$ therefore carries their span onto $span {u_{1}, \dots, u_{m}}$ , and is injective by independence. Hence the profile span has exactly one direction per distinct exponent, as asserted.

Depends on

Used by · 1

One-jump span in the nine-dimensional lowering model

Dependency tree · 2 levels

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