Determinant of a certain Toeplitz matrix

Question

Compute the following determinant $$\begin{vmatrix} x & 1 & 2 & 3 & \cdots & n-1 & n\\ 1 & x & 1 & 2 & \cdots & n-2 & n-1\\ 2 & 1 & x & 1 & \cdots & n-3 & n-2\\ 3 & 2 & 1 & x & \cdots & n-4 & n-3\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ n-1 & n-2 & n-3 & n-4 & \cdots & x & 1\\ n & n-1 & n-2 & n-3 & \cdots & 1 &x \end{vmatrix}$$

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I tried the following. I subtracted the second row from the first, the third from the second, the fourth from the third, and so on. I got:

\begin{vmatrix} x-1 & 1-x & 1 & 1 & \cdots & 1 & 1\\ -1 & x-1 & 1-x & 1 & \cdots & 1 & 1\\ -1 & -1 & x-1 & 1-x & \cdots & 1 & 1\\ 3 & 2 & 1 & x & \cdots & n-4 & n-3\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ -1 & -1 & -1 & -1 & \cdots & x-1 & 1-x\\ n & n-1 & n-2 & n-3 & \cdots & 1 &x \end{vmatrix}

I did the same thing with the columns. I subtracted the second row from the first, the third from the second, the fourth from the third, and so on. And I got:

\begin{vmatrix} 2x-2 & -x & 0 & 1 & \cdots & 0 & 1\\ -x & 2x-2 & -x & 1 & \cdots & 0 & 1\\ -2 & -x & 2x-2 & 1-x & \cdots & 0 & 1\\ 1 & 1 & 1-x & x & \cdots & -1 & n-3\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ -2 & -2 & -2 & -1 & \cdots & 2x-2 & 1-x\\ 1 & 1 & 1 & n-3 & \cdots & 1-x &x \end{vmatrix}

I hope I didn’t make a mistake somewhere. With this part I don't know what to do next. I don't know if I'm doing it right. Thank you in advance !

Answer 1

Here are the first few determinants with the help of WA: $$ \begin{array}{rl} n & \text{determinant} \\ 1 & x^2 - 1 \\ 2 & x^3 - 6 x + 4 \\ 3 & x^4 - 20 x^2 + 32 x - 12 \\ 4 & x^5 - 50 x^3 + 140 x^2 - 120 x + 32 \\ 5 & x^6 - 105 x^4 + 448 x^3 - 648 x^2 + 384 x - 80 \\ 6 & x^7 - 196 x^5 + 1176 x^4 - 2520 x^3 + 2464 x^2 - 1120 x + 192 \\ 7 & x^8 - 336 x^6 + 2688 x^5 - 7920 x^4 + 11264 x^3 - 8320 x^2 + 3072 x - 448 \end{array} $$ There are some patterns for the coefficients but I don't see a complete pattern:

• The polynomial has degree is $n+1$

• The coefficient of $x^{n+1}$ is $1$

• The coefficient of $x^{n}$ is $0$

• The coefficient of $x^{n-1}$ is $-$A002415$(n+1)$

• The independent term is $(-1)^n$A001787$(n)$

OEIS doesn't have the sequence of coefficients of $x^{n-2}$ nor of $x$.

I don't expect a nice closed form in monomial form. A recurrence is more probable.

Answer 2

Partial answer

Let's define the following $[0,n] \times [0,n]$ matrices (diagonal, subdiagonal, and all-ones Lower triangular) $$ \eqalign{ & {\bf I}_{\,n} = \left( {\matrix{ 1 & {} & {} & {} \cr {} & 1 & {} & {} \cr {} & {} & \ddots & {} \cr {} & {} & {} & 1 \cr } } \right)\quad {\bf E}_{\,n} = \left( {\matrix{ {} & {} & {} & {} \cr 1 & {} & {} & {} \cr {} & \ddots & {} & {} \cr {} & {} & 1 & {} \cr } } \right) \cr & {\bf S}_{\,n} = \left( {{\bf I}_{\,n} - {\bf E}_{\,n} } \right)^{\, - \,{\bf 1}} = \left( {\matrix{ 1 & {} & {} & {} \cr 1 & 1 & {} & {} \cr \vdots & \vdots & \ddots & {} \cr 1 & 1 & 1 & 1 \cr } } \right) \cr} $$ where the zeros have been omitted to make more clear the ones-pattern, and let's denote with an over-bar the transpose.

It is then easy to see that our matrix reads as $$ {\bf M}_{\,n} (x) = x\,{\bf I}_{\,n} + {\bf S}_{\,n} ^{\,{\bf 2}} - {\bf S}_{\,n} + \overline {{\bf S}_{\,n} } ^{\,{\bf 2}} - \overline {{\bf S}_{\,n} } $$

Since the determinant of ${\bf S}_{\,n}$ is unitary, we can multiply by its inverse to get a simpler matrix $$ \eqalign{ & {\bf N}_{\,n} (x) = {\bf S}_{\,n} ^{\, - {\bf 1}} \,{\bf M}_{\,n} (x)\;\overline {{\bf S}_{\,n} } ^{\, - {\bf 1}} = \cr & = x\,{\bf S}_{\,n} ^{\, - {\bf 1}} \overline {{\bf S}_{\,n} } ^{\, - {\bf 1}} + {\bf S}_{\,n} \overline {{\bf S}_{\,n} } ^{\, - {\bf 1}} - \overline {{\bf S}_{\,n} } ^{\, - {\bf 1}} + {\bf S}_{\,n} ^{\, - {\bf 1}} \overline {{\bf S}_{\,n} } - {\bf S}_{\,n} ^{\, - {\bf 1}} = \cr & = x\,{\bf S}_{\,n} ^{\, - {\bf 1}} \overline {{\bf S}_{\,n} } ^{\, - {\bf 1}} + \left( {{\bf S}_{\,n} \overline {{\bf S}_{\,n} } ^{\, - {\bf 1}} + {\bf S}_{\,n} ^{\, - {\bf 1}} \overline {{\bf S}_{\,n} } } \right) - \left( {\overline {{\bf S}_{\,n} } ^{\, - {\bf 1}} + {\bf S}_{\,n} ^{\, - {\bf 1}} } \right) \cr} $$

Now ${\bf N}_{\,n} (x) $ has the following structure $$ {\bf N}_{\,n} (x) = \left( {\matrix{ x & {1 - x} & 1 & 1 & \cdots & 1 \cr {1 - x} & {2\left( {x - 1} \right)} & { - x} & 0 & \cdots & 0 \cr 1 & { - x} & {2\left( {x - 1} \right)} & { - x} & \ddots & 0 \cr 1 & 0 & { - x} & {2\left( {x - 1} \right)} & \ddots & 0 \cr \vdots & \vdots & \ddots & \ddots & \ddots & { - x} \cr 1 & 0 & 0 & \cdots & { - x} & {2\left( {x - 1} \right)} \cr } } \right) $$ and developing the determinat along e.g. the last column it seems that it might be possible to develop a recursive relation for it.