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For $n\in\Bbb{N}$, define three matrices $A_n(x,y), B_n$ and $M_n$ as follows:

(a) the $n\times n$ tridiagonal matrix $A_n(x,y)$ with main diagonal all $y$'s, superdiagonal all $x$'s and subdiagonal all $-x$'s. For example, $$ A_4(x,y)=\begin{pmatrix} y&x&0&0\\-x&y&x&0\\0&-x&y&x \\0&0&-x&y\end{pmatrix}. $$

(b) the $n\times n$ antidigonal matrix $B_n$ consisting of all $1$'s. For example, $$B_4=\begin{pmatrix} 0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0\end{pmatrix}.$$

(c) the $n^2\times n^2$ block-matrix $M_n=A_n(B_n,A_n(1,1))$ or using the Kronecker product $M_n=A_n(1,0)\otimes B_n+I_n\otimes A_n(1,1)$.

Question. What is the determinant of $M_n$?

UPDATE. For even indices, I conjecture that

$$\det(M_{2n})=\prod_{j,k=1}^n\left[1+4\cos^2\left(\frac{j\pi}{2n+1}\right)+4\cos^2\left(\frac{k\pi}{2n+1}\right)\right]^2.$$

This would confirm what Philipp Lampe's "perfect square" claim.

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    $\begingroup$ I’m guessing you have an amazing conjectural answer? $\endgroup$ Sep 15, 2018 at 4:44
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    $\begingroup$ I conjecture that the determinant of $M_n$ is a perfect square. I conjecture that the zeros of the characteristic polynomial of $M_n$ are complex numbers with real part $1$. Here is some numerical experimentation supporting the statements. $\endgroup$ Sep 17, 2018 at 9:50
  • $\begingroup$ @AnthonyQuas: Working on it. $\endgroup$ Sep 17, 2018 at 13:32
  • $\begingroup$ @PhilippLampe: That is cool, thanks Perhaps proving your conjecture would also be interesting. $\endgroup$ Sep 17, 2018 at 13:33
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    $\begingroup$ Let $A=A_n(1,0)$, which has eigenvalues $2i\cos(j\pi/(n+1))$. Flip the order of the Kronecker products to get a block matrix where all blocks are polynomial in $A$ and therefore commute. The determinant is $\det(f(A))$ where $f$ is the characteristic polynomial of $-A_n(1,1)B_n$, which appears to be given by OEIS A152063. The determinant would then be the resultant of $f$ and the characteristic polynomial of $A$. $\endgroup$
    – MTyson
    Sep 18, 2018 at 15:12

1 Answer 1

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Flip the order of the Kronecker products to get $M'=A_n(I_n,I_n)+B_n\otimes T_n$, where $T_n=A_n(1,0)$. Note that $\det M=\det M'$. Since all blocks are polynomial in $A$, they commute, and therefore the determinant of $M'$ is $\det(f(T_n))$, where $f(x)=\det(A_n(1,1)+xB_n)$. That is, $f(x)=\det(B_n)\det(x+A_n(1,1)B_n)$ is $(-1)^{n\choose 2}$ times the characteristic polynomial of $H=-A_n(1,1)B_n$.

Let $t_n(x)$ be the characteristic polynomial of $T_n$. By repeated cofactor expansion on the first row, $t_n(x)=xt_{n-1}(x)+t_{n-2}(x)$. The initial conditions then imply that the $t_n$ is the $(n+1)^{th}$ Fibonacci polynomial. The roots of $t_n$ are $2i\cos(k\pi/(n+1))$ for $k=1,\dots,n$.

The eigenvalues of $H$ are worked out in "The eigenvalues of some anti-tridiagonal Hankel matrices". When $n$ is odd they are $1$ and $\pm\sqrt{3+2\cos(\frac{2k\pi}{n+1})}$ for $k=1,\dots,\frac{n-1}{2}$. When $n$ is even they are $\pm\sqrt{1+4\cos^2(\frac{(2k+1)\pi}{n+1})}$ for $k=0,\dots,\frac{n}{2}-1$.

By a quick diagonalization argument $\det(f(T_n))$ is the resultant of $f$ and $t_n$. This plus some trig gives $$ \det(M_{2n})=\prod_{j,k=1}^n\left[1+4\cos^2\left(\frac{j\pi}{2n+1}\right)+4\cos^2\left(\frac{k\pi}{2n+1}\right)\right]^2 $$ and $$ \det(M_{2n-1})=\prod_{j=1}^{n-1}\left[1+4\cos^2\left(\frac{j\pi}{2n}\right)\right]^2\prod_{k=1}^{n-1}\left[1+4\cos^2\left(\frac{j\pi}{2n}\right)+4\cos^2\left(\frac{k\pi}{2n}\right)\right]^2. $$

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