---
tag: EEFAGXWAHA
kind: proof
name: Proof of the classification of complex quadratic forms by rescaling a diagonal form
msc: ["11E04", 15A63]
proves: 0CEX9BRXRZ
uses: [0Z3FEWZ8HA, 0HZNE5QZDJ, 0TX2YH2FQS, 0P24YW565F, 0SPYAGE7T0, 0KWXYZPJKN, 0EWQEYRVA8, 02FDWY55AS, 09GK5PS824, 0JP0W07Y9A, 0R8X9XA83A, 0VTQTPQGAV, 0EVHJFTG9Z, 096V7H13H9]
sources:
  - url: https://kconrad.math.uconn.edu/blurbs/linmultialg/bilinearform.pdf
    locator: Theorem 6.19
---

## Proof

The 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 $Q(2v) = 4Q(v)$.

1. 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 $\mathcal{B}'$.
2. 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 $P P' = I$.
3. 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 $M = P'^T M' P'$.
4. 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]] $r$.
5. 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 $Q(\sum_i y_i f_i) = \sum_i a_i y_i^2$.
6. Renumber the basis so that $a_i \ne 0$ for $i \le s$ and $a_i = 0$ for $i > s$, for some $s$ with $0 \le s \le n$.
7. 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 $c_i \ne 0$.
8. Let $e_i = f_i / c_i$ for $i \le s$ and $e_i = f_i$ for $i > s$. Rescaling basis vectors by nonzero numbers gives again a [[basis|0VTQTPQGAV]].
9. By step 5, $Q(\sum_i x_i e_i) = \sum_{i \le s} a_i (x_i / c_i)^2 = x_1^2 + \cdots + x_s^2$.
10. 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 $i \ne j$.
11. Also $B(e_i, e_i) = Q(e_i)$, which is $1$ for $i \le s$ and $0$ for $i > s$. So the matrix of $B$ in the basis $(e_i)$ is diagonal with $s$ ones and $n - s$ zeros.
12. 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.
13. 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.
14. 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 $T \colon V \to V$.
15. Suppose $Q$ and $Q'$ both have rank $r$. By step 13, there are bases $(e_i)$ and $(e'_i)$ with $Q(\sum_i x_i e_i) = \sum_{i \le r} x_i^2 = Q'(\sum_i x_i e'_i)$.
16. 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.
17. 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.
18. Conversely, let $Q' = Q \circ T$ with $T$ invertible and linear. Then the form associated with $Q'$ is $B'(u, v) = B(Tu, Tv)$.
19. 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]].
20. 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 $(T b_i)$.
21. By step 4, these two matrices have the same rank. So equivalent forms have the same rank, and the two conditions are equivalent.

## Idea

Diagonalise $Q$ 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.