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Let $n=2m$ be an even number. Let us construct $n\times n$ symmetric matrices $S_n$ in the following way. The entries are indeterminates $X_1,\ldots,X_{n-1}$. We choose a $1$-factorization of the complete graph over $[\![1,n]\!]$ (which exists because $n$ is even). The $1$-factors (these are perfect matchings) are labelled $F_1,\ldots,F_{n-1}$. The indeterminate $X_r$ is attributed to the $(ij)$-entry whenever the edge $(ij)$ belongs to $F_r$. On the diagonal, we put zeros. In other words, every line and every column of $S$ contains a zero and all the indeterminates exactly once. You may call this a symmetric Latin square. For instance $$S_2=\begin{pmatrix} 0 & X \\ X & 0 \end{pmatrix},\qquad S_4=\begin{pmatrix} 0 & X & Y & Z \\ X & 0 & Z & Y \\ Y & Z & 0 & X \\ Z & Y & X & 0 \end{pmatrix}.$$ It seems that the determinant $D_n$ of $S_n$, a homogeneous polynomial in the indeterminates, is symmetric. One verifies easily that for $n=2$ or $4$, $D_n$ splits: $$D_2=-X^2,\qquad D_4=(X+Y+Z)(X-Y-Z)(Y-Z-X)(Z-X-Y).$$ Actually, the characteristic polynomials $P_n(T,X_1,\ldots,X_{n-1})$ split for $n=2$ or $4$ : $$P_2=(T-X)(T+X),\qquad P_4=(T-X-Y-Z)(T+X+Y-Z)(T+Y+Z-X)(T+Z+X-Y).$$

Is it true that $D_n$ or $P_n$ always split into linear factors ?

Notice that $P_n$ has always the factor $T-X_1-\cdots-X_{n-1}$.

For $n\ge6$, the $1$-factorization is not unique. There are already 6240 of them for $n=8$. Does the answer depend on the choice of the $1$-factorization ?

Edit. Patrik's negative answer raises a far-reaching question. The original question ressembles vaguely that of the splitting of the determinant associated with the Cayley graph of a finite group $G$. In the latter case, we know that the determinant splits into linear factors if and only if $G$ is abelian. For a general group, the nature of the splitting is given by Frobenius' Theorem about the irreducible characters of $G$. In the present situation, I suspect that another algebraic theory will tell us what are the degrees of the irreducible factors of $P_n$. When $n$ is small, these degrees are small, because the underlying algebra must be trivial ($n=2$ or $4$) or very simple ($n=8$).

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    $\begingroup$ Possibly the work of K W Johnson (and others) on Latin square determinants would be of interest, e.g., Latin square determinants, in Algebraic, extremal and metric combinatorics, 1986 (Montreal, PQ, 1986), 146–154, London Math. Soc. Lecture Note Ser., 131, Cambridge Univ. Press, Cambridge, 1988, MR1052664 (91h:05026). $\endgroup$ Sep 17, 2015 at 23:10

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No it's not true in general. I claim that the following matrix

\begin{bmatrix} t&x&y&z&w&u&v&r\\ x&t&z&y&r&w&u&v\\ y&z&t&x&v&r&w&u\\ z&y&x&t&u&v&r&w\\ w&r&v&u&t&x&y&z\\ u&w&r&v&x&t&z&y\\ v&u&w&r&y&z&t&x\\ r&v&u&w&z&y&x&t \end{bmatrix}

is of the desired form (when $t=0$) and has determinant

$[(r-u)^2-(t-y)^2+(v - w + x - z)^2]\cdot $
$[(r-u)^2-(t-y)^2+(v - w - x + z)^2]\cdot$ $(r + u + t + y + v + w + x + z) (r + u -t - y + v + w - x - z)$ $(r + u + t + y - v - w - x - z)(r + u -t - y - v - w + x + z)$.

Realize $K_8$ as two copies of $K_4$ and connect all vertices (one with vertex set $\{1,2,3,4\}$ and one with vertex set $\{1',2',3',4'\}$). The 1-factors $x,y,z$ come from uniting corresponding matchings of $K_4$ and $K_4'$, and the edges between $K_4$ and $K_4'$ are decomposed into the 1-factors $w,u,v,r$ given by $\{11',22',33',44'\},\{12',23',34',41'\},\{13',24',31',42'\},\{14',21',32',43'\}$ respectively, completing the 1-factorization. The determinant computation was done using a computer, so unfortunately I have no particular understanding of what is going on.

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    $\begingroup$ Nice ! This answer raises questions ... $\endgroup$ Sep 17, 2015 at 16:34
  • $\begingroup$ @DenisSerre Indeed. For example: What can the degrees of the irreducible factors tell us about the automorphism group of the 1-factorization? Supposedly the bigger the degrees, the smaller the automorphism group. $\endgroup$
    – Wolfgang
    Jan 26, 2017 at 20:38

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