Let $V$ be a finite-dimensional real vector space of dimension $n$. A bilinear form $B: V \times V \rightarrow \mathbb{R}$ is nondegenerate if for all $\mathbf{v} \neq 0$ in $V$, there is some $\mathbf{w} \in V$ with $B(\mathbf{v}, \mathbf{w}) \neq 0$. For $\mathbf{v} \in V$, define $\langle\mathbf{v}\rangle^{\perp}=\{\mathbf{w} \in V \mid B(\mathbf{v}, \mathbf{w})=0\}$. Assuming $B$ is nondegenerate, show that $V=\langle\mathbf{v}\rangle \oplus\langle\mathbf{v}\rangle^{\perp}$ whenever $B(\mathbf{v}, \mathbf{v}) \neq 0$.

Suppose that $B$ is a nondegenerate, symmetric bilinear form on $V$. Prove that there is a basis $\left\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\right\}$ of $V$ with $B\left(\mathbf{v}_{i}, \mathbf{v}_{j}\right)=0$ for $i \neq j$. [If you use the fact that symmetric matrices are diagonalizable, you must prove it.]

Define the signature of a quadratic form. Explain how to determine the signature of the quadratic form associated to $B$ from the basis you constructed above.

A linear subspace $V^{\prime} \subset V$ is said to be isotropic if $B(\mathbf{v}, \mathbf{w})=0$ for all $\mathbf{v}, \mathbf{w} \in V^{\prime}$. Show that if $B$ is nondegenerate, the maximal dimension of an isotropic subspace of $V$ is $(n-|\sigma|) / 2$, where $\sigma$ is the signature of the quadratic form associated to $B$.