Riesz representation theorem in finite dimensions

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 $f \colon V \to \mathbb{K}$, 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.