Rank $r$ solutions to the matrix equation $XAX^{T}=C, A$ alternate, over $\textrm {GF}(2^{y})$

Abstract:

Let ${\text {GF}}(q)$ denote a finite field of characteristic two. Let ${V_n}$ denote an n-dimensional vector space over ${\text {GF}}(q)$. An $n \times n$ symmetric matrix A over ${\text {GF}}(q)$ is said to be an alternate matrix if A has zero diagonal. Let A be an $n \times n$ alternate matrix over ${\text {GF}}(q)$ and let C be an $s \times s$ symmetric matrix over ${\text {GF}}(q)$. By using Albert’s canonical forms for symmetric matrices over fields of characteristic two, the number $N(A,C,n,s,r)$ of $s \times n$ matrices X of rank r over ${\text {GF}}(q)$ such that $XA{X^T} = C$ is determined. A symmetric bilinear form on ${V_n} \times {V_n}$ is said to be alternating if $f(x,x) = 0$, for each x in ${V_n}$. Let f be such a bilinear form. A basis $({x_1}, \ldots ,{x_\rho },{y_1}, \ldots ,{y_\rho }),n = 2\rho$, for ${V_n}$ is said to be a symplectic basis for ${V_n}$ if $f({x_i},{x_j}) = f({y_i},{y_j}) = 0$ and $f({x_i},{y_j}) = {\delta _{ij}}$, for each i, $j = 1,2, \ldots ,\rho$. In determining the number $N(A,C,n,s,r)$, it is shown that a symplectic basis for any subspace of ${V_n}$, can be extended to a symplectic basis for ${V_n}$. Furthermore, the number of ways to make such an extension is determined.