Classification of complex quadratic forms

Statement

Let $V$ be a complex vector space of finite dimension $n$, and let $Q$ be a quadratic form on $V$. Let $r$ be the rank of the symmetric matrix of $Q$ in a basis. Then $V$ has a basis $(e_1, \dots, e_n)$ such that, for all $x_1, \dots, x_n \in \C$, $$Q(x_1 e_1 + \cdots + x_n e_n) = x_1^2 + x_2^2 + \cdots + x_r^2.$$ So two quadratic forms on $V$ are equivalent if and only if their matrices have the same rank.

Remarks

The proof starts from a diagonal form $a_1 x_1^2 + \cdots + a_n x_n^2$ and replaces $e_i$ by $e_i / \sqrt{a_i}$ for each $a_i \ne 0$. The same result holds over every algebraically closed field of characteristic not $2$. Over $\R$, this fails: $x^2$ and $-x^2$ have the same rank but are not equivalent.

Explanations

L1 human

Over the complex numbers, every nonzero number is a square. So all coefficients of a diagonal form can be made equal to $1$. Only the number of squares remains, and it is the rank.