---
tag: 04GYPXDXNH
kind: statement
role: theorem
name: Riesz representation theorem in finite dimensions
msc: [46C05, 15A04]
uses: [01EK923SKP, 0RT96J4D3H]
sources:
  - url: https://en.wikipedia.org/wiki/Riesz_representation_theorem
    locator: Example in finite dimensions using matrix transformations
---

## Statement

Let $V$ be a finite-dimensional inner product space over $\R$ or $\C$, with the inner product linear in the first argument.
Then, for each [[linear form|07YQ1C78G0]] $f \colon V \to \mathbb{K}$, there is exactly one vector $y \in V$ with
$$f(x) = \langle x, y \rangle \qquad \text{for all } x \in V.$$

## Remarks

For an orthonormal basis $e_1, \dots, e_n$, the vector is $y = \sum_i \overline{f(e_i)}\, e_i$.
The map $y \mapsto \langle \cdot, y \rangle$ from $V$ to its dual is bijective; it is linear over $\R$ and conjugate linear over $\C$.
The statement holds for continuous linear forms on every Hilbert space.

## Explanations

### L1 human

In an inner product space, every linear form is an inner product with a fixed vector.
On $\R^3$, the form $x \mapsto 2x_1 - x_2 + 5x_3$ is the dot product with $(2, -1, 5)$.
So in finite dimensions, vectors and linear forms correspond one to one.