= Solution
The <polarization identity>[symmetric bilinear form associated with a quadratic form] is obtained by polarization:
$$
\boxed{\phi(u,v)=\frac12\bigl(q(u+v)-q(u)-q(v)\bigr)},
\qquad q(v)=\phi(v,v).
$$
To diagonalize it, argue by induction on $\dim V$. If $\phi(v,v)=0$ for every $v$, polarization gives $\phi=0$, so every basis works. Otherwise choose $e$ with $\phi(e,e)\ne0$. Then
$$
V=\operatorname{span}(e)\oplus e^\perp,
\qquad
v=\frac{\phi(v,e)}{\phi(e,e)}e+\left(v-\frac{\phi(v,e)}{\phi(e,e)}e\right),
$$
and diagonalize the restriction to $e^\perp$ inductively. Thus some basis gives a diagonal matrix with positive, negative, and zero entries. By <Sylvester's law of inertia>, their counts $n_+,n_-,n_0$ are basis-independent. In the convention relevant here, the <signature of a quadratic form> is
$$
\boxed{\operatorname{sig}(q)=n_+-n_-}.
$$
Suppose $R$ lies in the <radical of a bilinear form>, so $\phi(r,v)=0$ for all $r\in R,v\in V$. Then $q(r)=0$ and
$$
q(v+r)=q(v)+2\phi(v,r)+q(r)=q(v).
$$
Hence $q'(v+R)=q(v)$ is a well-defined quadratic form on the <quotient vector space> $V/R$. In an adapted diagonal basis, quotienting by $R$ merely removes zero diagonal directions, so $n_+$ and $n_-$, and therefore the signature, are unchanged.
Now let $W=\operatorname{span}(e,f)$. Its Gram matrix is
$$
\begin{pmatrix}0&1\\1&\phi(f,f)\end{pmatrix},
$$
which has determinant $-1$. Thus $W$ is nondegenerate and has one positive and one negative square: it is a <hyperbolic plane (quadratic form)> and has signature zero. Nondegeneracy gives the orthogonal direct sum
$$
\boxed{V=W\oplus U,\qquad U=W^\perp}.
$$
Signature is additive under orthogonal direct sums, so
$$
\boxed{\operatorname{sig}(q|_U)=\operatorname{sig}(q)-\operatorname{sig}(q|_W)=\operatorname{sig}(q).}
$$
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