Riesz Representation in Finite Dimensions

Derivation

Riesz Representation in Finite Dimensions

D-078 Home PU-104 Threads symmetry Depends on Cauchy–Schwarz and the Triangle Inequality, Invariance of Dimension of a Vector Space

Statement

Let V be a finite-dimensional inner-product space over F (either ℝ or ℂ), with inner product ⟨·,·⟩ linear in its first argument and conjugate-linear in its second. Then for every linear functional φ : V → F there exists one and only one vector u ∈ V such that φ(x) = ⟨x, u⟩ for all x ∈ V. The map φ ↦ u is a conjugate-linear bijection from the dual space V* onto V.

Why it matters

The theorem is the reason a finite-dimensional space and its dual are “the same” once an inner product is fixed: every measurement you can make with a covector is secretly a projection onto some concrete vector. This is what lets physicists raise and lower indices, identify bra ⟨ψ| with a ket |ψ⟩, and write forces as gradients of potentials rather than as abstract 1-forms.

It also isolates exactly what geometry adds. A bare vector space has a dual that is merely isomorphic to it after choosing a basis; the inner product upgrades that to a canonical (basis-independent up to conjugation) identification. Symmetry arguments — the thread this page belongs to — exploit precisely this: an invariant bilinear or Hermitian form turns group actions on V into matching actions on V*.

Assumptions

The space is finite-dimensional.Without it an orthonormal basis need not exist and the finite sum defining u becomes an infinite series that may not converge; the theorem then requires completeness (a Hilbert space) and continuity of φ.

⟨·,·⟩ is a genuine inner product — conjugate-symmetric and positive-definite.Drop positive-definiteness (a degenerate or indefinite form, as in Minkowski space) and the representer may fail to be unique, or the radical of the form may leave some functionals unrepresentable.

φ is linear.An affine, conjugate-linear, or nonlinear map has no single representing vector; conjugate-linear functionals are represented by ⟨u, ·⟩ instead, a mirror statement.

A fixed convention for which slot is linear.If instead the inner product is linear in the second slot (the physics bra–ket convention), the construction uses u = Σᵢ φ(eᵢ) eᵢ without the conjugate; get the convention wrong and every complex phase in u flips sign.

Derivation

∃ orthonormal basis {e₁, …, eₙ}, ⟨eᵢ, e⫺⟩ = δᵢ⫺

Finite dimension guarantees a finite basis (dimension invariance), and Gram–Schmidt orthonormalises it. B

x = Σᵢ xᵢ eᵢ, xᵢ = ⟨x, eᵢ⟩

Every vector expands in the basis; orthonormality makes the coordinate the projection onto eᵢ. A

φ(x) = Σᵢ xᵢ φ(eᵢ)

Linearity of φ distributes over the finite sum. A

φ(x) = Σᵢ ⟨x, eᵢ⟩ φ(eᵢ)

Substitute the coordinate from step 2. Symbols only — no numbers yet. A

⟨x, eᵢ⟩ φ(eᵢ) = ⟨x, φ(eᵢ)̅ eᵢ⟩

The inner product is conjugate-linear in its second slot, so pulling the scalar φ(eᵢ) in requires conjugating it. B

φ(x) = ⟨x, Σᵢ φ(eᵢ)̅ eᵢ⟩ ≡ ⟨x, u⟩, u := Σᵢ φ(eᵢ)̅ eᵢ

Additivity in the second slot collapses the sum into one argument; this defines a concrete u and proves existence. B

⟨x, u⟩ = ⟨x, u′⟩ ∀x ⇒ ⟨x, u − u′⟩ = 0; x = u − u′ ⇒ ‖u − u′‖² = 0

Positive-definiteness forces u − u′ = 0: the representer is unique. C

Result

φ(x) = ⟨x, u⟩ ∀x ∈ V, u = Σᵢ₁ⁿ φ(eᵢ)̅ eᵢ (unique)

Reading. Every “linear question” you can ask of vectors — every scalar-valued linear machine φ — is answered by taking the inner product against one fixed vector u. The dual space collapses onto the space itself: functionals are vectors, viewed through the geometry. In an orthonormal basis u’s components are the conjugated values φ(eᵢ)̅ that φ takes on the basis.

Units check. The identity is dimensionally consistent slot by slot: u is built from φ(eᵢ) times unit vectors eᵢ, so [u] = [φ(e)]·[e]. Then [⟨x, u⟩] = [x]·[u] = [x]·[φ(e)]·[e], and since [x] ∼ 1/[e] (a coordinate divided into unit vectors) this reduces to [φ(x)], matching the left side.

Limiting cases

Real field F = ℝ: the conjugation is trivial, u = Σᵢ φ(eᵢ) eᵢ, and φ ↦ u is an ordinary linear isomorphism V* ≅ V.
Zero functional φ = 0: the unique representer is u = 0, consistent with ⟨x, 0⟩ = 0.
One dimension: φ(x) = c x and u = c̅/⟨e,e⟩ · e; with a unit vector u = c̅ e.
Standard basis of ℝⁿ or ℂⁿ: φ(x) = a·x as a row vector times a column; u is the conjugate-transpose of that row.

Breaks when

Infinite dimensions without completeness. On the incomplete space of finitely-supported sequences with ⟨x,y⟩ = Σ xᵢ y̅ᵢ, the functional φ(x) = Σ xᵢ/i is well-defined and linear, but its candidate representer u = (1/1, 1/2, 1/3, …) is not in the space — no representing vector exists. Riesz then demands a complete (Hilbert) space and a bounded functional.
Degenerate or indefinite form. If ⟨·,·⟩ has a nonzero radical (some v ≠ 0 with ⟨v,·⟩ = 0), then representers are non-unique (add any radical vector) and functionals not vanishing on the radical are unrepresentable. In an indefinite metric like Minkowski space the theorem still gives existence but the “raising an index” map carries sign flips.
Non-linear or conjugate-linear φ. A conjugate-linear functional over ℂ cannot be written as ⟨x,u⟩ for any fixed u; it needs the mirror form ⟨u,x⟩.

Failure modes

Dropping the conjugate. Writing u = Σ φ(eᵢ) eᵢ in the complex case; this represents the wrong functional (it gives φ(x)̅’s partner) and only accidentally works when all φ(eᵢ) are real.
Using a non-orthonormal basis in the formula. The closed form u = Σ φ(eᵢ)̅ eᵢ holds only for an orthonormal basis; for a general basis one must solve the Gram-matrix system G ū = φ̄.
Conflating u with the coefficient row. Students identify u with the tuple (φ(eᵢ)) as a covector, forgetting it is a genuine vector whose expansion needs conjugation and the metric.
Assuming u is basis-independent as a formula. The vector u is canonical, but its component formula changes when the inner product changes; changing only the basis (not the metric) leaves u fixed as a geometric object.
Forgetting positive-definiteness in the uniqueness step. Concluding u = u′ from ⟨x, u−u′⟩ = 0 without noting that only a definite form makes x = u−u′ force the norm to vanish.

Discussion

The theorem is the finite-dimensional heart of a hierarchy. In a Hilbert space the same statement holds for every bounded functional, and the boundedness is exactly what finite dimension supplies for free (all linear functionals on a finite-dimensional normed space are continuous). Read this way, Riesz–Fréchet is not a new fact in infinite dimensions but the surviving fragment of a finite-dimensional triviality, rescued by completeness.

Physically, the map φ ↦ u is the abstract origin of the metric’s index-raising operator gᵢ⫺. A covector φ⫺ (a gradient, a momentum 1-form, a bra) is turned into a contravariant vector uᵢ = gᵢ⫺ φ⫺ precisely by the inner product. In Dirac notation the correspondence |ψ⟩ ↔ ⟨ψ| is the Riesz map, and its conjugate-linearity is why the bra carries the complex conjugate of the ket’s components.

At the level of naturality, the content is sharper than “V ≅ V*.” A bare finite-dimensional space is isomorphic to its dual but not naturally so: no isomorphism commutes with every linear map without extra data. The inner product is that data, and the Riesz map is natural with respect to the orthogonal (or unitary) group that preserves it — which is exactly why symmetry arguments transport freely between a representation and its dual once an invariant form is present. The double dual V** ≅ V, by contrast, is natural and needs no metric; Riesz is the one-step, geometry-dependent shadow of that two-step, geometry-free fact.

Common misconceptions. The theorem does not say the dual space equals the space — it gives a specific bijection that depends on the chosen inner product; a different metric gives a different u for the same φ. Nor does “finite-dimensional” merely simplify the proof: it is load-bearing, because it is what forces both the existence of an orthonormal basis and the automatic continuity of φ.

Worked examples

V = ℝ³ (standard inner product), φ(x) = 2x₁ − x₂ + 3x₃

Real space, so u = Σ φ(eᵢ) eᵢ with no conjugation. A

φ(e₁) = 2, φ(e₂) = −1, φ(e₃) = 3

Evaluate the functional on each basis vector. A

u = 2e₁ − e₂ + 3e₃ = (2, −1, 3)

Assemble the representer from the values. A

check: ⟨x, u⟩ = 2x₁ − x₂ + 3x₃ = φ(x) ✓

The dot product reproduces the functional. A

u = (2, −1, 3)

Reading. A real linear functional on ℝⁿ is just “dot with its coefficient vector”; the representer is the coefficient list itself.

Units check. If x carries units of metres and φ(x) of joules, the coefficients (hence u) carry J/m — a force-like covector, correctly.

V = ℂ² (standard Hermitian product, linear in 1st slot), φ(z) = z₁ + i z₂

Complex space: conjugation is now essential. A

φ(e₁) = 1, φ(e₂) = i

Evaluate on the orthonormal standard basis. A

u = φ(e₁)̅ e₁ + φ(e₂)̅ e₂ = 1·e₁ + (−i)·e₂ = (1, −i)

Conjugate each value before assembling u. B

check: ⟨z, u⟩ = z₁·1̅ + z₂·(−i)̅ = z₁ + i z₂ = φ(z) ✓

Conjugating u’s components in the second slot restores the original coefficients. B

u = (1, −i)

Reading. The representer’s components are the conjugates of the functional’s coefficients; the double conjugation inside the second slot of ⟨z, u⟩ cancels to return φ.

Units check. Dimensionless amplitudes in and out; the imaginary unit is a pure phase carrying no units, so u is dimensionless as required.

Problems

On ℝ² with the standard inner product, find the representer of φ(x) = 5x₁ − 2x₂.
Solution
Real case, orthonormal standard basis: u = (φ(e₁), φ(e₂)) = (5, −2). Check ⟨x,u⟩ = 5x₁ − 2x₂ = φ(x). So u = (5, −2).
On ℂ² (standard Hermitian product, linear in the first slot), find u for φ(z) = (2−i)z₁ + 3z₂.
Solution
Conjugate the coefficients: u = ((2−i)̅, 3̅) = (2+i, 3). Check ⟨z,u⟩ = z₁(2+i)̅ + z₂(3)̅ = (2−i)z₁ + 3z₂ = φ(z). So u = (2+i, 3).
On ℝ² equip the weighted inner product ⟨x,y⟩ = 2x₁y₁ + 3x₂y₂. Find the representer of φ(x) = x₁ + x₂. (Note the standard basis is orthogonal but not orthonormal here.)
Solution
Seek u = (u₁, u₂) with ⟨x,u⟩ = 2u₁x₁ + 3u₂x₂ = x₁ + x₂ for all x. Matching coefficients: 2u₁ = 1 and 3u₂ = 1, so u = (1/2, 1/3). This shows the representer depends on the metric: the same φ gives (1,1) under the standard product but (½, ⅓) here.
Let V = P₂, real polynomials of degree ≤ 2 on [0,1] with ⟨p,q⟩ = ∫₀¹ p(t)q(t) dt. Show the evaluation functional φ(p) = p(0) has a representer and describe how to compute it (you need not simplify the algebra fully).
Solution
V is finite-dimensional (dim 3) and φ is linear, so a unique u ∈ P₂ with p(0) = ∫₀¹ p(t)u(t) dt exists. Take the basis {1, t, t²} and Gram matrix Gᵢ⫺ = ∫₀¹ tⁱ⁺⫺ dt = 1/(i+j+1) (the 3×3 Hilbert matrix, rows/cols indexed 0,1,2). With u = a + bt + ct² the conditions φ(eᵢ) = ⟨eᵢ, u⟩ read G(a,b,c)ᵀ = (1,0,0)ᵀ since φ(1)=1, φ(t)=0, φ(t²)=0. Solving (a,b,c)ᵀ = G⁻¹(1,0,0)ᵀ gives the first column of the inverse Hilbert matrix: u(t) = 9 − 36t + 30t². (One verifies e.g. ∫₀¹ u\,dt = 9 − 18 + 10 = 1 = φ(1).)
Prove the Riesz map R : V* → V, R(φ) = u, is conjugate-linear and a bijection, and relate ‖φ‖ (operator norm) to ‖u‖.
Solution
Conjugate-linear: if φ ↦ u and ψ ↦ w, then for scalars α,β the functional αφ + βψ satisfies (αφ+βψ)(x) = α⟨x,u⟩ + β⟨x,w⟩ = ⟨x, α̅u + β̅w⟩, so R(αφ+βψ) = α̅u + β̅w — conjugate-linear. Injective: R(φ)=0 means ⟨x,0⟩ = 0 = φ(x) for all x, so φ = 0. Surjective: any u defines φ(x) = ⟨x,u⟩ ∈ V* with R(φ)=u. Hence a bijection. Norms: by Cauchy–Schwarz |φ(x)| = |⟨x,u⟩| ≤ ‖x‖‖u‖, giving ‖φ‖ ≤ ‖u‖; taking x = u gives |φ(u)| = ‖u‖², so ‖φ‖ ≥ ‖u‖. Therefore ‖φ‖ = ‖u‖ — the Riesz map is an isometry.