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  "body": "\n## Statement\n\nLet $V$ be a complex vector space of finite dimension $n$, and let $Q$ be a quadratic form on $V$.\nLet $r$ be the [[rank|0ZC9YTXZRY]] of the symmetric matrix of $Q$ in a basis.\nThen $V$ has a basis $(e_1, \\dots, e_n)$ such that, for all $x_1, \\dots, x_n \\in \\C$,\n$$Q(x_1 e_1 + \\cdots + x_n e_n) = x_1^2 + x_2^2 + \\cdots + x_r^2.$$\nSo two quadratic forms on $V$ are equivalent if and only if their matrices have the same rank.\n\n## Remarks\n\nThe proof starts from a [[diagonal form|0Z3FEWZ8HA]] $a_1 x_1^2 + \\cdots + a_n x_n^2$ and replaces $e_i$ by $e_i / \\sqrt{a_i}$ for each $a_i \\ne 0$.\nThe same result holds over every algebraically closed field of characteristic not $2$.\nOver $\\R$, this fails: $x^2$ and $-x^2$ have the same rank but are not equivalent.\n\n## Explanations\n\n### L1 human\n\nOver the complex numbers, every nonzero number is a square.\nSo all coefficients of a diagonal form can be made equal to $1$.\nOnly the number of squares remains, and it is the rank.\n",
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