[proofplan] By the Diagonalisation of Symmetric Bilinear Forms theorem, we may choose a basis in which $\phi$ has a diagonal matrix. Over an algebraically closed field, every nonzero diagonal entry has a square root, so we rescale each basis vector to normalise the nonzero entries to $1$. The rank is invariant under congruence, so the canonical form is determined by the rank alone. [/proofplan] [step:Diagonalise and reorder by rank] By the [Diagonalisation of Symmetric Bilinear Forms](/theorems/427), there exists a basis $(e_1, \dots, e_n)$ in which the matrix of $\phi$ is $\mathrm{diag}(d_1, \dots, d_n)$. After reordering, assume $d_1, \dots, d_r \neq 0$ and $d_{r+1} = \cdots = d_n = 0$, where $r = \mathrm{rank}(\phi)$. [/step] [step:Rescale to normalise nonzero diagonal entries to $1$ using square roots] Since $\mathbb{F}$ is algebraically closed, every nonzero element of $\mathbb{F}$ has a square root: for each $i \leq r$, choose $c_i \in \mathbb{F}^\times$ with $c_i^2 = d_i$. Define $f_i = c_i^{-1}e_i$ for $i = 1, \dots, r$ and $f_i = e_i$ for $i > r$. Then: \begin{align*} \phi(f_i, f_i) = c_i^{-2}\,\phi(e_i, e_i) = c_i^{-2} \cdot c_i^2 = 1 \quad \text{for } i \leq r, \end{align*} and $\phi(f_i, f_j) = 0$ for $i \neq j$ since the $e_i$ were already orthogonal. The matrix of $\phi$ with respect to $(f_1, \dots, f_n)$ is $\mathrm{diag}(\underbrace{1, \dots, 1}_{r}, \underbrace{0, \dots, 0}_{n-r})$. [guided] The algebraic closure hypothesis is consumed here. In a general field, a nonzero element need not have a square root (for instance, $2$ has no square root in $\mathbb{Q}$). Over an algebraically closed field $\mathbb{F}$, the polynomial $x^2 - d_i$ splits for every $d_i \in \mathbb{F}^\times$, so we can choose $c_i$ with $c_i^2 = d_i$. The rescaling $f_i = c_i^{-1}e_i$ is designed so that \begin{align*} \phi(f_i, f_i) = \phi(c_i^{-1}e_i, c_i^{-1}e_i) = c_i^{-2} \phi(e_i, e_i) = c_i^{-2} \cdot d_i = c_i^{-2} \cdot c_i^2 = 1. \end{align*} The off-diagonal entries remain zero because the original basis was already $\phi$-orthogonal, and rescaling individual basis vectors does not affect orthogonality: $\phi(f_i, f_j) = c_i^{-1} c_j^{-1} \phi(e_i, e_j) = 0$ for $i \neq j$. [/guided] [/step] [step:Conclude that the rank is the only congruence invariant] The rank of a bilinear form is invariant under change of basis: if $P$ is an invertible matrix, $\mathrm{rank}(P^\top A P) = \mathrm{rank}(A)$. Since any symmetric bilinear form of rank $r$ can be brought to $\mathrm{diag}(I_r, 0)$, two symmetric bilinear forms over an algebraically closed field are congruent if and only if they have the same rank. [/step]