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Exercise 2172Linear algebra › Inner product spacesL1Added 3 Nov 2026

Riesz representation theorem (finite-dimensional)

#Riesz representation#linear forms#inner product

Statement

Let EE be a finite-dimensional inner product space, and φ:E→R\varphi:E\to\mathbb{R} a linear form. Show that there exists a unique u∈Eu\in E such that
φ(v)=⟨v,u⟩\varphi(v)=\langle v,u\rangle
for all v∈Ev\in E.

Hint

Choose an orthonormal basis (ei)(e_i) and set u=∑iφ(ei)eiu=\sum_i\varphi(e_i)e_i.

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Solution

Existence. Let (e1,…,en)(e_1,\ldots,e_n) be an orthonormal basis of EE. Define u=∑iφ(ei)eiu=\sum_i\varphi(e_i)e_i. For any v=∑jvjejv=\sum_j v_j e_j:
⟨v,u⟩=∑jvj⟨ej,u⟩=∑jvjφ(ej)=φ(∑jvjej)=φ(v).\langle v,u\rangle=\sum_j v_j\langle e_j,u\rangle=\sum_j v_j\varphi(e_j)=\varphi\left(\sum_j v_j e_j\right)=\varphi(v).

Uniqueness. If ⟨v,u⟩=⟨v,u′⟩\langle v,u\rangle=\langle v,u'\rangle for all vv, then ⟨v,u−u′⟩=0\langle v,u-u'\rangle=0 for all vv. Taking v=u−u′v=u-u': ∥u−u′∥2=0\|u-u'\|^2=0, so u=u′u=u'.

Conclusion. Every linear form on a finite-dimensional inner product space is represented by a unique vector.

Matrix form. In matrix terms, a linear form φ(v)=aTv\varphi(v)=a^Tv is represented by the vector u=au=a (in an orthonormal basis).