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  "body": "\n## Proof\n\nThe symmetric matrix of $Q$ in a basis is the [[matrix|0P24YW565F]] of the [[associated symmetric bilinear form|0TX2YH2FQS]] $B(u, v) = \\frac12 \\bigl( Q(u+v) - Q(u) - Q(v) \\bigr)$. Recall that $B(v, v) = Q(v)$, since $Q(2v) = 4Q(v)$.\n\n1. First, the rank does not depend on the basis. Let $\\mathcal{B}$ and $\\mathcal{B}'$ be bases, with matrices $M$ and $M'$ of $B$. Let $P$ be the [[change of basis matrix|0KWXYZPJKN]] from $\\mathcal{B}$ to $\\mathcal{B}'$.\n2. Let $P'$ be the change of basis matrix from $\\mathcal{B}'$ to $\\mathcal{B}$. Substituting one expansion into the other and comparing coordinates in $\\mathcal{B}$ gives $P P' = I$.\n3. By the [[change of basis rule|0SPYAGE7T0]], $M' = P^T M P$. By the [[transpose of a product|09GK5PS824]], $P'^T P^T = (P P')^T = I$, so $M = P'^T M' P'$.\n4. By the [[rank of a product|02FDWY55AS]], $\\rank M' \\le \\rank M$ and $\\rank M \\le \\rank M'$. So all matrices of $Q$ have the same [[rank|0EWQEYRVA8]] $r$.\n5. By the [[diagonalisation of a quadratic form|0Z3FEWZ8HA]], $V$ has a basis $f_1, \\dots, f_n$ and there are $a_i \\in \\C$ with $Q(\\sum_i y_i f_i) = \\sum_i a_i y_i^2$.\n6. Renumber the basis so that $a_i \\ne 0$ for $i \\le s$ and $a_i = 0$ for $i \u003e s$, for some $s$ with $0 \\le s \\le n$.\n7. For $i \\le s$, the polynomial $t^2 - a_i$ has a root $c_i \\in \\C$ by the [[fundamental theorem of algebra|0JP0W07Y9A]]. Then $c_i^2 = a_i$, so $c_i \\ne 0$.\n8. Let $e_i = f_i / c_i$ for $i \\le s$ and $e_i = f_i$ for $i \u003e s$. Rescaling basis vectors by nonzero numbers gives again a [[basis|0VTQTPQGAV]].\n9. By step 5, $Q(\\sum_i x_i e_i) = \\sum_{i \\le s} a_i (x_i / c_i)^2 = x_1^2 + \\cdots + x_s^2$.\n10. By step 9, $Q(e_i + e_j) = Q(e_i) + Q(e_j)$ for $i \\ne j$. So $B(e_i, e_j) = 0$ for $i \\ne j$.\n11. Also $B(e_i, e_i) = Q(e_i)$, which is $1$ for $i \\le s$ and $0$ for $i \u003e s$. So the matrix of $B$ in the basis $(e_i)$ is diagonal with $s$ ones and $n - s$ zeros.\n12. The column space of this matrix is spanned by the first $s$ standard basis vectors, which are independent. So its rank is $s$, and $s = r$ by step 4.\n13. By steps 9 and 12, $Q(x_1 e_1 + \\cdots + x_n e_n) = x_1^2 + \\cdots + x_r^2$. This proves the first claim.\n14. Two quadratic forms $Q$ and $Q'$ on $V$ are equivalent when $Q'(v) = Q(Tv)$ for all $v \\in V$, for some invertible linear map $T \\colon V \\to V$.\n15. Suppose $Q$ and $Q'$ both have rank $r$. By step 13, there are bases $(e_i)$ and $(e'_i)$ with $Q(\\sum_i x_i e_i) = \\sum_{i \\le r} x_i^2 = Q'(\\sum_i x_i e'_i)$.\n16. By [[the extension from a basis|0R8X9XA83A]], there are linear maps $T$ and $S$ with $T e'_i = e_i$ and $S e_i = e'_i$. Then $ST$ and $TS$ fix the bases, so they are the identity, and $T$ is invertible.\n17. By step 15, $Q(T \\sum_i x_i e'_i) = Q(\\sum_i x_i e_i) = Q'(\\sum_i x_i e'_i)$. So $Q' = Q \\circ T$, and $Q$ and $Q'$ are equivalent.\n18. Conversely, let $Q' = Q \\circ T$ with $T$ invertible and linear. Then the form associated with $Q'$ is $B'(u, v) = B(Tu, Tv)$.\n19. Let $(b_i)$ be a basis of $V$. The vectors $T b_i$ are [[linearly independent|0EVHJFTG9Z]], since $T$ is injective. There are $n$ of them, so they form a [[basis|096V7H13H9]].\n20. By step 18, the matrix of $Q'$ in $(b_i)$ has entries $B(T b_i, T b_j)$. It is the matrix of $Q$ in the basis $(T b_i)$.\n21. By step 4, these two matrices have the same rank. So equivalent forms have the same rank, and the two conditions are equivalent.\n\n## Idea\n\nDiagonalise $Q$ first. Over $\\C$ every nonzero coefficient $a_i$ is a square $c_i^2$, so dividing $f_i$ by $c_i$ turns it into $1$. The number of ones that remain is the rank of the matrix, which does not change under a change of basis.\n",
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