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Given an arbitrary $(x_1, \dots, x_n) \in [0, 1]^n$, is there any name/known results for the following $n \times n$ matrix (which is constructed by iterating $(x_1 \to \dots \to x_n \to x_1 \to \dots)$ in circle and taking products along the way) $$M = \begin{bmatrix} 1 & x_1 & (x_1 x_2) & (x_1 x_2 x_3) & \dots & (x_1 \dots x_{n - 1}) \\ (x_2 \dots x_n) & 1 & x_2 & (x_2 x_3) & \dots & (x_2 \dots x_{n - 1}) \\ (x_3 \dots x_n) & (x_3 \dots x_n x_1) & 1 & x_3 & \dots & (x_3 \dots x_{n - 1}) \\ \dots & \dots & \dots & \dots & \dots & \dots & \\ x_n & (x_n x_1) & (x_n x_1 x_2) & (x_n x_1 x_2 x_3) & \dots & 1 \\ \end{bmatrix}? $$ Thanks!

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    $\begingroup$ A quick Mathematica calculation (using Mat[n_]:=Table[Product[Subscript[x,Mod[k,n,1]], {k,i,If[j<i,j+n,j] - 1}], {i,n}, {j,n}]) reveals that $$ \det(M_n) = (1-x_1 x_2 \ldots x_n)^{n-1}\,. $$ $\endgroup$
    – Fred Hucht
    Commented Mar 13 at 21:37
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    $\begingroup$ Moreover, the answer obtained by @FredHucht admits an immediate proof. Subtract from the first line $x_1$ times the second, develop along the first row, and do the same with the remaining matrix. Every time you earn $1-x_1x_2\cdots x_n$ as a factor. $\endgroup$ Commented Mar 14 at 7:17

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