This question is related to the matrices described in Deyi Chen's recent MO post (look at some examples there). The main difference: we are asking for a determinant evaluation instead of a permanent, plus we have added variables. Define the sequence $$f_n=\mathrm{det}\left[x_{i-j}\cdot\operatorname{sgn} \left(\tan\frac{(i+j)\pi}{2n+1} \right)\right]_{1\le i, \,j\le 2n},$$ where $n\geq1$ and $\operatorname{sgn}$ is the sign-function; i.e. $\operatorname{sgn}(y)=1$ if $y>0$; $\operatorname{sgn}(y)=-1$ if $y<0$; $\operatorname{sgn}(0)=0$.

If we set all $x_k=1$, user44191 offers an alternative description of the matrix: $n−1$ antidiagonals of $1$'s, $n$ antidiagonals of $-1$'s, an antidiagonal of $0$'s, $n$ antidiagonals of $1$'s, and $n-1$ antidiagonals of $-1$'s, in the given order.

QUESTION 1. Is this true? The determinant equals $f_n=(-1)^nP(\dots,x_{-2},x_{-1},x_0,x_1,x_2,\dots)^2$ for some squared polynomial $P$ of several variables (coefficients in $\mathbb{Z}$) so that each of its monomials is of the form $x_{i_1}x_{i_2}\cdots x_{i_m}$ with $i_1+i_2+\cdots+i_m=0$.

QUESTION 2 (specialization). Is this true? If we set $x_k=1$, for all $k$, then $f_n=(-1)^n$.

Example. For instance, $$f_2= \det\left[ \begin {array}{cccc} x_0&-x_{-1}&-x_{-2}&0\\ -x_1&-x_0&0&x_{-2} \\ -x_2&0&x_0&x_{-1}\\ 0&x_2&x_1&-x_0 \end {array} \right] =(x_{-2}x_2-x_{-1}x_1-x_0^2)^2.$$

Example. The matrix for $n=3$ and its determinant: $$ \left[ \begin {array}{cccccc} x_0&x_{-1}&-x_{-2}&-x_{-3}&-x_{-4}&0\\ x_1&-x_0&-x_{-1}&-x_{-2}&0&x_{-4}\\ -x_2&-x_1&-x_0&0&x_{-2}&x_{-3}\\ -x_3&-x_2&0&x_0&x_{-1}&x_{-2}\\ -x_4&0&x_2&x_1&x_0&-x_{-1}\\ 0&x_4&x_3&x_2&-x_1&-x_0 \end {array} \right], $$ \begin{align*} f_3&=-(x_{-4}x_0x_4 - x_{-4}x_{1}x_3 + x_{-4}x_2^2 - x_{-3}x_{-1}x_4 + x_{-3}x_0x_3 \\ & \qquad + x_{-3}x_1x_2 + x_{-2}^2x_4 + x_{-2}x_{-1}x_3 - 2x_{-2}x_0x_2 \\ & \qquad - x_{-2}x_1^2 - x_{-1}^2x_2 - x_0^3)^2. \end{align*}

  • $\begingroup$ In my preprint On some determinants involving the tangent function posted to arXiv two years ago (see arxiv.org/abs/1901.04837) , I proved that $$\det\left[\tan\pi\frac{j+k}{2n+1}\right]_{1\le j,k\le 2n}=(-1)^n(2n+1)^{2n-1}.$$ The method there might be helpful to your questions. $\endgroup$ Aug 28, 2021 at 0:32
  • $\begingroup$ Determinants are relatively easier to handle than permanents since there are many techniques for evaluating determinants. $\endgroup$ Aug 28, 2021 at 0:43
  • $\begingroup$ Note: You can replace the $m$ in Question 1 by an $n$. $\endgroup$ Aug 28, 2021 at 0:53
  • $\begingroup$ Question 2 seems easy: For $\sigma\in S_{2n}$ let $\bar{\sigma}(j)=2n+1-\sigma(j)$ for all $j=1,\ldots,2n$. Then $$\mathrm{sign}(\sigma)\times\mathrm{sign}(\bar{\sigma})=(-1)^{2n(2n+1)/2}=(-1)^n.$$ So it suffices to show that $$\det\left[\mathrm{sgn}\left(\tan\pi\frac{j-k}{2n+1}\right)\right]_{1\le j,k\le 2n}=1.$$ This should be easy as the determiant is close to a circular determinant. $\endgroup$ Aug 28, 2021 at 1:56

1 Answer 1


to Question 1: Yes.

To prove this, let me fix a positive integer $n$ and denote your matrix (whose determinant $f_{n}$ is) by $A$. The notation $\left[ k\right] $ shall be used for the set $\left\{ 1,2,\ldots,k\right\} $ whenever $k$ is an integer. The notation $M_{i,j}$ will be used for the $\left( i,j\right) $-th entry of any matrix $M$. Thus, \begin{equation} A_{i,j}=x_{i-j}\cdot\operatorname*{sgn}\left( \tan\dfrac{\left( i+j\right) \pi}{2n+1}\right) \label{eq.darij1.1} \tag{1} \end{equation} for any $i,j\in\left[ 2n\right] $.

Let $B$ be the $2n\times2n$-matrix obtained by "turning $A$ upside down", i.e., reversing the order of the rows of $A$. Explicitly, this means that \begin{equation} B_{i,j}=A_{2n+1-i,j}\qquad\text{for all }i,j\in\left[ 2n\right] . \label{eq.darij1.2} \tag{2} \end{equation} We note that $B$ can be obtained from $A$ by $n$ row-swaps (i.e., by $n$ steps, where each step swaps a pair of rows). Indeed, all we need to do is to swap the $1$-st and the last row, then to swap the $2$-nd and the $2$-nd-to-last row, etc., until we reach the middle of the matrix. Since each of these swaps multiplies the determinant by $-1$, this entails that \begin{equation} \det B=\left( -1\right) ^{n}\det A. \label{eq.darij1.3} \tag{3} \end{equation}

Now, I claim that the matrix $B$ is alternating -- i.e., that \begin{equation} B_{i,i}=0\qquad\text{for all }i\in\left[ 2n\right] \label{eq.darij1.4} \tag{4} \end{equation} and \begin{equation} B_{i,j}=-B_{j,i}\qquad\text{for all }i,j\in\left[ 2n\right] . \label{eq.darij1.5} \tag{5} \end{equation}

Indeed, in order to prove \eqref{eq.darij1.4}, it suffices to observe that \begin{align*} B_{i,i} & =A_{2n+1-i,i}=x_{\left( 2n+1-i\right) -i}\cdot\operatorname*{sgn} \left( \tan\dfrac{\left( \left( 2n+1-i\right) +i\right) \pi} {2n+1}\right) \\ & =x_{2n+1-2i}\cdot\underbrace{\operatorname*{sgn}\left( \tan\dfrac{\left( 2n+1\right) \pi}{2n+1}\right) }_{=\operatorname*{sgn}\left( \tan\pi\right) =\operatorname*{sgn}0=0}=0. \end{align*} The proof of \eqref{eq.darij1.5} is not much harder (using the fact that $\dfrac{\left( 2n+1-i+j\right) \pi}{2n+1}=\pi-\dfrac{\left( i-j\right) \pi}{2n+1}$ and therefore \begin{align} \tan\dfrac{\left( 2n+1-i+j\right) \pi}{2n+1}=\tan\left( \pi-\dfrac{\left( i-j\right) \pi}{2n+1}\right) =-\tan\dfrac{\left( i-j\right) \pi}{2n+1}, \end{align} and furthermore $\tan$ is an odd function).

Thus, we know that the matrix $B$ is alternating. Hence, as for any alternating $2n\times2n$-matrix, its determinant is the square of its Pfaffian. In other words, \begin{equation} \det B=\left( \operatorname*{Pf}B\right) ^{2}, \label{eq.darij1.6} \tag{6} \end{equation} where $\operatorname*{Pf}B$ denotes the Pfaffian of $B$. The latter Pfaffian is a polynomial in the entries of the matrix with coefficients in $\mathbb{Z} $. Since the entries of the matrix belong to $\mathbb{Z}\left[ \ldots ,x_{-2},x_{-1},x_{0},x_{1},x_{2},\ldots\right] $, we thus conclude that the Pfaffian belongs to $\mathbb{Z}\left[ \ldots,x_{-2},x_{-1},x_{0},x_{1} ,x_{2},\ldots\right] $ as well. In other words, \begin{equation} \operatorname*{Pf}B\in\mathbb{Z}\left[ \ldots,x_{-2},x_{-1},x_{0},x_{1} ,x_{2},\ldots\right] . \label{eq.darij1.7} \tag{7} \end{equation}

Now, \eqref{eq.darij1.3} yields \begin{align} \det A=\left( -1\right) ^{n}\det B=\left( -1\right) ^{n}\left( \operatorname*{Pf}B\right) ^{2} \end{align} (by \eqref{eq.darij1.6}). Because of \eqref{eq.darij1.7}, this shows that $\det A$ equals $\left( -1\right) ^{n}\cdot P^{2}$ for some polynomial $P\in\mathbb{Z}\left[ \ldots,x_{-2},x_{-1},x_{0},x_{1},x_{2},\ldots\right] $ (namely, for $P=\operatorname*{Pf}B$), exactly as claimed in Question 1.

In order to complete the answer to Question 1, we now need to show that each monomial in $P=\operatorname*{Pf}B$ is of the form $x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}$ with $i_{1}+i_{2}+\cdots+i_{n}=0$. This can be done in various ways, but the easiest is probably the following: Let us equip the polynomial ring $\mathbb{Z}\left[ \ldots,x_{-2},x_{-1},x_{0},x_{1},x_{2},\ldots\right] $ with a $\mathbb{Z}$-grading in which each indeterminate $x_{i}$ is homogeneous of degree $i$. Now, recall the explicit formula for the Pfaffian as a sum over all perfect matchings on the set $\left[ 2n\right] $ (see Definition 3 in Michel Goemans, 18.438 in Spring 2014, Lectures 4 and 6, or any good textbook on Pfaffians). If \begin{equation} M=\left\{ \left\{ a_{1},b_{1}\right\} ,\left\{ a_{2},b_{2}\right\} ,\ldots,\left\{ a_{n},b_{n}\right\} \right\} \label{eq.darij1.9o} \tag{9} \end{equation} is such a perfect matching, then the corresponding addend in $\operatorname*{Pf}B$ is \begin{equation} \pm B_{a_{1},b_{1}}B_{a_{2},b_{2}}\cdots B_{a_{n},b_{n}}. \label{eq.darij1.9} \tag{10} \end{equation} Each of the $n$ factors $B_{a_{i},b_{i}}$ in this product can be rewritten as \begin{align} B_{a_{i},b_{i}}=A_{2n+1-a_{i},b_{i}}=x_{\left( 2n+1-a_{i}\right) -b_{i} }\cdot\left( 1\text{ or }-1\text{ or }0\right) , \end{align} and thus (using our weird grading) is homogeneous of degree $\left( 2n+1-a_{i}\right) -b_{i}=2n+1-a_{i}-b_{i}$. Hence, the entire product \eqref{eq.darij1.9} is homogeneous of degree \begin{align*} \sum_{i=1}^{n}\left( 2n+1-a_{i}-b_{i}\right) & =n\left( 2n+1\right) -\underbrace{\sum_{i=1}^{n}\left( a_{i}+b_{i}\right) } _{\substack{=1+2+\cdots+2n\\\text{(since \eqref{eq.darij1.9o} is a}\\\text{perfect matching of }\left[ 2n\right] \text{)}}}\\ & =n\left( 2n+1\right) -\left( 1+2+\cdots+2n\right) =0. \end{align*} This means that this product is a $\mathbb{Z}$-linear combination of monomials of the form $x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}$ with $i_{1}+i_{2} +\cdots+i_{n}=0$. Clearly, the same must therefore holds for the polynomial $\operatorname*{Pf}B$ (since this polynomial is a sum of such products). This concludes the answer to Question 1.

Answering Question 2 requires proving that $\det B=1$ when all $x_{i}$ are set to $1$. This should be easy given that $\operatorname*{sgn}\left( \tan \dfrac{\left( i+j\right) \pi}{2n+1}\right) $ can be explicitly computed (and the matrix $B$ becomes a circulant when all $x_{i}$ are $1$); but it's late here and I have too many things on my list until the quarter begins. Sorry!


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