Diagonalisation of a quadratic form

Statement

Let $F$ be a field of characteristic not $2$, and let $V$ be a vector space over $F$ of finite dimension $n$. Let $Q$ be a quadratic form on $V$. Then $V$ has a basis $(e_1, \dots, e_n)$ and there are $a_1, \dots, a_n \in F$ such that, for all $x_1, \dots, x_n \in F$, $$Q(x_1 e_1 + \cdots + x_n e_n) = a_1 x_1^2 + a_2 x_2^2 + \cdots + a_n x_n^2.$$

Remarks

The basis is orthogonal for the associated symmetric bilinear form, and $a_i = Q(e_i)$. In matrix terms, each symmetric matrix $A$ over $F$ has an invertible $P$ such that $P^T A P$ is diagonal. The numbers $a_i$ are not unique; for example, each $a_i$ can be multiplied by a nonzero square. Over $\R$, the orthogonal diagonalisation of the principal axis theorem is a special case.

Explanations

L1 human

Every quadratic form becomes a sum of squares with coefficients after a good change of variables. For example, $x^2 + 2xy = (x + y)^2 - y^2$. With $u = x + y$ and $v = y$, the form becomes $u^2 - v^2$.