Gram-map image, full-rank fibers, and rank-stratum dimensions

A symmetric bilinear form on a finite-dimensional $k$-space has a basis with diagonal matrix: if the form is nonzero, polarization in characteristic different from two gives a vector $v$ with $B(v,v)\mathrm{\ne }0$; scale $v$ to norm one using a square root in $k$, split off $kv$ and induct on its orthogonal complement. For an alternating form, choose nonzero $v$, choose $w$ with $B(v,w)=1$, split off the nondegenerate plane they span, and induct on its orthogonal complement. Applying the same argument after quotienting by the radical shows that any rank-$r$ symmetric matrix is congruent to $\mathrm{diag}(I_r,0)$ and any rank-$r$ alternating matrix is congruent to $J_r\oplus 0$, with $r$ even in the latter case.

Conversely, every Gram matrix has rank at most $\dim V$, since it factors through the linear map $k^t\to V$ whose columns are the chosen vectors. For every canonical rank-$r$ matrix in step 1.1, choose $r$ vectors in $V$ giving that standard form and set the remaining vectors to zero. If $S$ is an invertible $t\times t$ matrix satisfying $S^{\mathsf{T}}GS=C$ for an arbitrary rank-$r$ matrix $G$ and its canonical matrix $C$, multiplying the chosen tuple by $S^{-1}$ gives Gram matrix $G$. This realizes every matrix in the rank-bounded image.

Let $f_1,f_2:k^t\to V$ be the surjective maps defined by two tuples with the same rank-$m$ Gram matrix $G$. For $c\in k^t$, $f_i(c)=0$ implies $Gc=0$; conversely $Gc=0$ implies $B(f_i(c),f_i(d))=0$ for every $d$, and surjectivity plus nondegeneracy gives $f_i(c)=0$. Thus $\ker f_1=\ker f_2$, so $T(f_1(c))=f_2(c)$ is a well-defined linear isomorphism. The equality of Gram matrices makes $T$ preserve the form. It is the unique isometry carrying the first tuple to the second, and a form-preserving map fixing a spanning tuple is the identity.

A rank-$r$ symmetric matrix is equivalent to its kernel, a $(t-r)$-plane, and a nondegenerate symmetric form on the $r$-dimensional quotient. The finitely many standard graph charts cover the Grassmannian of $(t-r)$-planes, and each chart has $r(t-r)$ coordinates. Over such a chart, symmetric forms on the quotient have $r(r+1)\mathrm{/}2$ free coordinates, and nondegeneracy is the nonvanishing of the determinant, an open condition. Thus this stratum has dimension $r(t-r)+r(r+1)\mathrm{/}2=rt-r(r-1)\mathrm{/}2$. In the alternating case the form space has dimension $r(r-1)\mathrm{/}2$ and nondegeneracy is again open, so the stratum dimension is $r(t-r)+r(r-1)\mathrm{/}2=rt-r(r+1)\mathrm{/}2$.