[proofplan] We proceed by induction on $n = \dim V$. If $\phi = 0$, any basis gives the zero diagonal matrix. Otherwise, we find a vector $v$ with $\phi(v, v) \neq 0$ (using $\mathrm{Char}\,\mathbb{F} \neq 2$ to handle the case where all diagonal entries of $\phi$ on basis vectors vanish but $\phi$ is nonzero), decompose $V = \langle v \rangle \oplus v^\perp$, and apply the inductive hypothesis to $v^\perp$. [/proofplan] [step:Handle the base case and the zero form] If $n = 0$ or $\phi = 0$, any basis gives the zero diagonal matrix. [/step] [step:Find a non-isotropic vector $v$ with $\phi(v, v) \neq 0$] Assume $\phi \neq 0$. Choose any basis $(e_1, \dots, e_n)$ for $V$. If $\phi(e_i, e_i) \neq 0$ for some $i$, take $v = e_i$. If $\phi(e_i, e_i) = 0$ for all $i$ but $\phi(e_i, e_j) \neq 0$ for some $i \neq j$, set $v = e_i + e_j$. Then: \begin{align*} \phi(v, v) = \phi(e_i, e_i) + 2\phi(e_i, e_j) + \phi(e_j, e_j) = 2\phi(e_i, e_j) \neq 0. \end{align*} The factor $2\phi(e_i, e_j) \neq 0$ uses $\mathrm{Char}\,\mathbb{F} \neq 2$. [guided] Why must there exist some basis pair $(i, j)$ with $\phi(e_i, e_j) \neq 0$? Because $\phi \neq 0$ means $\phi(x, y) \neq 0$ for some $x, y \in V$. Expanding in any basis, $\phi(x, y) = \sum_{i,j} x_i y_j \phi(e_i, e_j)$, so at least one $\phi(e_i, e_j) \neq 0$. If all diagonal entries $\phi(e_i, e_i)$ vanish, some off-diagonal entry $\phi(e_i, e_j) \neq 0$ with $i \neq j$. We exploit this by computing: \begin{align*} \phi(e_i + e_j, e_i + e_j) = \phi(e_i, e_i) + \phi(e_i, e_j) + \phi(e_j, e_i) + \phi(e_j, e_j) = 0 + 2\phi(e_i, e_j) + 0. \end{align*} The factor of $2$ appears because $\phi$ is symmetric: $\phi(e_j, e_i) = \phi(e_i, e_j)$. Since $\mathrm{Char}\,\mathbb{F} \neq 2$ and $\phi(e_i, e_j) \neq 0$, the product $2\phi(e_i, e_j) \neq 0$. This is the only place in the proof where $\mathrm{Char}\,\mathbb{F} \neq 2$ is used. Over a field of characteristic $2$, a symmetric bilinear form need not be diagonalisable (since the off-diagonal cross-terms cannot be detected by evaluating $\phi$ on a single vector). [/guided] [/step] [step:Decompose $V = \langle v \rangle \oplus v^\perp$ using $\phi(v, v) \neq 0$] Set $e_1 = v$ and define $W = v^\perp = \{w \in V : \phi(v, w) = 0\}$. The map $w \mapsto \phi(v, w)$ is a nonzero linear functional on $V$ (nonzero because $\phi(v, v) \neq 0$). By [Rank-Nullity](/theorems/384), $\dim W = n - 1$. [claim:Direct Sum Decomposition] $V = \langle v \rangle \oplus W$. [/claim] [proof] If $u \in \langle v \rangle \cap W$, then $u = cv$ for some $c \in \mathbb{F}$, and $\phi(v, cv) = c\,\phi(v, v) = 0$. Since $\phi(v, v) \neq 0$, we have $c = 0$, so $u = \mathbf{0}$. Therefore $\langle v \rangle \cap W = \{\mathbf{0}\}$. Since $\dim\langle v \rangle + \dim W = 1 + (n - 1) = n = \dim V$, the sum $\langle v \rangle + W$ has dimension $n$, giving $V = \langle v \rangle \oplus W$. [/proof] [/step] [step:Apply the inductive hypothesis to $\phi|_{W \times W}$ and assemble the diagonal basis] The restriction $\phi|_{W \times W}$ is a symmetric bilinear form on $W$ with $\dim W = n - 1$. By the inductive hypothesis, there exists a basis $(e_2, \dots, e_n)$ for $W$ in which $\phi|_W$ is diagonal. Then $(e_1, e_2, \dots, e_n)$ is a basis for $V$. For $j \geq 2$: $\phi(e_1, e_j) = 0$ since $e_j \in W = v^\perp$. For $2 \leq i \neq j \leq n$: $\phi(e_i, e_j) = 0$ by the inductive hypothesis. So the matrix of $\phi$ with respect to $(e_1, \dots, e_n)$ is $\mathrm{diag}(\phi(e_1, e_1), \dots, \phi(e_n, e_n))$. [/step]