Parseval's identity

Statement

Let $V$ be a finite-dimensional inner product space over $\R$ or $\C$, and let $(e_1, \dots, e_n)$ be an orthonormal basis of $V$. Then, for all $x, y \in V$, $$\langle x, y \rangle = \sum_{i=1}^{n} \langle x, e_i \rangle \overline{\langle y, e_i \rangle} \qquad \text{and} \qquad \norm{x}^2 = \sum_{i=1}^{n} \abs{\langle x, e_i \rangle}^2.$$

Remarks

It is the case of equality in Bessel's inequality. In coordinates of an orthonormal basis, the inner product becomes the standard inner product of $\mathbb{K}^n$. For Fourier series, the same identity holds with infinite sums, for square integrable functions.

Explanations

L1 human

In an orthonormal basis, the inner product of two vectors is the standard inner product of their coordinates. In particular, the squared length is the sum of the squared coordinates. This is the Pythagorean theorem in $n$ dimensions.