A relation between the determinant and the permanent on singular matrices
Let \(\mathbb{F}\) be a field with at least \(n\geq 3\) elements and characteristic \(\mathrm{char}\,\mathbb{F}\neq 2\), let \(M_n(\mathbb{F})\) be the vector space of \(n\times n\) matrices over~\(\mathbb{F}\), and let \(V\) be its subspace. The determinant is convertible into the permanent on \(V\) if there is a linear transformation \(T:V\to V\) such that \(\mathrm{per}\,T(X)=\det{X}\) for all \(X\in V\). \textit{M. Marcus} and \textit{H. Minc} [Ill. J. Math.~5, 376--381 (1961; Zbl 0104.00904)] proved non-convertibility when \(V=M_n(\mathbb{F})\) and \(\mathrm{char}\,\mathbb{F}=0\). \textit{M.~Purificação Coelho} and the first author of the present paper [Linear Multilinear Algebra 51, No. 2, 127--136 (2002; Zbl 1042.15007)] proved non-convertibility when \(V\) is the subspace of symmetric matrices. \textit{G. Dolinar} et al. [Eur. J. Comb. 32, No. 1, 116--132 (2011; Zbl 1209.15011)] proved that if \(V=M_n(\mathbb{F})\) and \(\mathbb{F}\) is finite with sufficiently large~\(n\), then \(T\) cannot be bijective. \smallskip In case of \(V=M_n(\mathbb{F})\), the present authors prove that \(T\) cannot satisfy \(\det{X}=0\Leftrightarrow\mathrm{per}\,T(X)=0\) for all \(X\in V\).
scientific article; zbMATH DE number 6153438