TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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Finite-dimensional Riesz representation: every functional is uniquely $v \mapsto ⟨ v, w ⟩$

Statement

Let $V$ be a finite-dimensional real or complex inner product space, with the inner product linear in its first argument. For every linear functional $f : V \to F$ , there is a unique $w \in V$ such that

$f (v) = ⟨ v, w ⟩ for every v \in V .$

The map $w \mapsto (v \mapsto ⟨ v, w ⟩)$ is a conjugate-linear bijection from $V$ to its algebraic dual. This includes $V = 0$ .

Facts & Assumptions

Given: A finite-dimensional inner product space $V$ and a linear functional $f$ .

[L1]

The space $V$ has a finite orthonormal basis (Every finite-dimensional real or complex inner product space has an orthonormal basis).

[L2]

An orthonormal basis $(e_{i})$ gives $v = \sum_{i} ⟨ v, e_{i} ⟩ e_{i}$ (Bessel's inequality for a finite orthonormal list and Parseval's identity for an orthonormal basis).

[L3]

If $⟨ v, z ⟩ = 0$ for every $v$ , then $z = 0$ (Inner products separate vectors, and the induced norm is homogeneous: $∥ λ v ∥ = ∣ λ ∣ ∥ v ∥$ ).

[L4]

The algebraic dual consists of all linear functionals from $V$ to its scalar field (Linear functionals and the algebraic dual $V^{*} = L (V, F)$ ).

Proof

technique · direct

1.1L1choose

Choose an orthonormal basis $(e_{0}, \dots, e_{n - 1})$ by [L1] and define $w = \sum_{i < n} \overline{f (e_{i})} e_{i}$ .

2.1step 1.1L2L4algebra

For $v \in V$ , [L2] and linearity of $f$ give $f (v) = \sum_{i} ⟨ v, e_{i} ⟩ f (e_{i})$ . Conjugate-linearity in the second argument makes the right side equal to $⟨ v, w ⟩$ .

3.1step 2.1L3

If $w^{'}$ is another representative, then $⟨ v, w - w^{'} ⟩ = 0$ for every $v$ , so [L3] gives $w = w^{'}$ .

4.1step 1.1step 2.1step 3.1L4∎

The assignment $w \mapsto ⟨ \cdot, w ⟩$ is conjugate-linear because the inner product is conjugate-linear in its second argument. Existence makes it surjective and uniqueness makes it injective. When $V = 0$ , the chosen basis and both sums are empty and the same argument applies.

Depends on

Used by

Dependency tree · two levels

16 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

Sheldon Axler, Linear Algebra Done Right, 4th ed., result 6.42 (standard reference, not scraped)

Finite-dimensional Riesz representation: every functional is uniquely v↦⟨v,w⟩v↦⟨v,w⟩