Abstract:
We give some necessary conditions for the maximality of $(0, 1)$-determinant. Let $\mathbf{M}$ be a nondegenerate $(0,1)$-matrix of order $n$. Denote by $\mathbf{A}$ the matrix of order $n+1$ which is obtained from $\mathbf{M}$ by adding the $(n+1)$th row $(0,0,\dots,0,1)$ and the $(n+1)$th column consisting of 1's. We prove that if $\mathbf{A}^{-1}=(l_{i,j})$ then for all $i=1,\dots,n$ we have $\sum\limits^{n+1}_{j=1}|l_{I,j}|\ge2$. Moreover, if $|\det(\mathbf{M})|$ is equal to the maximal value of a $(0,1)$-determinant of order $n$, then $\sum\limits^{n+1}_{j=1}|l_{I,j}|=2$ for all $i=1,\dots,n$.
Keywords:$(0,1)$-matrix with the maximal determinant, simplex, cube, axial diameter.
Fulltext:
PDF file (331 kB)