An approach to the permanental-dominance conjecture (Q1330009)

Let \(A = (a_{ij})\) denote a complex \(n\)-square matrix. Then the Schur power matrix of \(A\) is the \(n!\)-square matrix \(\prod_A : = (\prod^n_{i = 1} a_{\sigma (i), \tau (i)})\), where \(\sigma, \tau \in S_n\) (the full symmetric group of order \(n)\). The maximum eigenvalue conjecture (MEC) states that, if \(A\) is positive semidefinite Hermitian, then \(\text{per} (A)\) is the largest eigenvalue of \(\prod_A\). It is noteworthy that the MEC is stronger than the well-known permanental dominance conjecture (PDC). This says that, if \(G\) is a subgroup of \(S_n\), and if \(\chi\) is any character of \(G\), then, for any positive semidefinite Hermitian \(n\)-square matrix \(A = (a_{ij})\), the generalized matrix function \[ d^G_\chi (A) = {1 \over \chi (id)} \sum_{\sigma \in G} \chi (\sigma) \prod^n_{i = 1} a_{i, \sigma (i)} \] is never larger than \(\text{per} (A)\). The PDC has attracted much attention in the recent years; a survey is given by \textit{R. Merris} [Current Trends in Matrix Theory, 213-223 (1987; Zbl 0671.15008)]. However, little is known about the MEC; a proof in the case \(n = 3\) was provided by \textit{R. Bapat} and \textit{V. Sunder} [Linear Algebra Appl. 76, 153-163 (1986; Zbl 0602.15008)]. In the present paper the following is proved: Should the MEC fail in the real case, then the smallest \(n\) for which it fails must be such that it fails at a singular matrix having certain properties (including zero row sums).