 Solution

First work over a field of characteristic different from two. If a nonzero symmetric bilinear form had $\varphi (v,v)=0$ for every vector, polarization would givea contradiction. Hence choose $e$ with $\varphi (e,e)\ne 0$. Every $v$ decomposes uniquely aswith the second summand orthogonal to $e$. Induct on the orthogonal complement to obtain a diagonal basis. If the restricted form becomes zero, any basis of that remaining space completes the diagonalization. In this basis the rank of a quadratic form is exactly the number of nonzero diagonal coefficients; the remaining basis vectors span its radical of a bilinear form.

Over the reals, rescale the nonzero basis vectors so the diagonal entries are $p$ copies of $+1$, $q$ copies of $-1$ and $z=n-p-q$ zeros. Define the signature of a quadratic form by $\sigma =p-q$, so $r=p+q$. To prove it is basis independent, note that $p$ is the largest dimension of a linear subspace on which the form is positive-definite. The positive coordinate linear subspace attains dimension $p$; on any other positive linear subspace, projection to those positive coordinates is injective, since its kernel has nonpositive quadratic value. Thus its dimension is at most $p$. The same argument for the negative form characterizes $q$. This proves Sylvester's law of inertia and that the signature is well-defined.

Let $e_i$ and $f_i$ be positive and negative normalized basis vectors, and let $Z$ be the radical of a bilinear form. The span of $Z$ and $e_i+f_i$ for $1\le i\le \min (p,q)$ is a totally isotropic subspace: all mutual bilinear pairings vanish. Its dimension is $z+\min (p,q)$. Conversely, project any totally isotropic subspace $U$ to the nondegenerate positive-plus-negative quotient. The projected linear subspace is still null, and its projections to both the positive and negative coordinate spaces are injective: a vector with one component zero would have strictly signed quadratic value unless it vanished. Its dimension is therefore at most $\min (p,q)$, while the kernel of the quotient map on $U$ has dimension at most $z$. HenceThis proves both construction and sharpness of the maximum dimension of a totally isotropic subspace.

For the five-variable example, the symmetric coefficient matrix isIts cyclic structure makes the vectors $(1,{\zeta }^k,{\zeta }^{2k},{\zeta }^{3k},{\zeta }^{4k})$, $\zeta =e^{2\pi i/5}$, eigenvectors with eigenvalues ${\zeta }^k+{\zeta }^{-k}=2\cos (2\pi k/5)$. Thus the eigenvalues areTheir product is $2$, and three are positive and two negative. ThereforeThe real and imaginary parts of each conjugate pair of eigenvectors give the corresponding real two-dimensional eigenspaces, consistent with the real inertia calculation.