CHARACTERISTIC POLYNOMIALS OF THE MATRICES WITH $(\,j,k)$-ENTRY $q^{\,j\pm k}+t$

Abstract

We determine the characteristic polynomials of the matrices $[q^{\,j-k}+t]_{1\le \,j,k\le n}$ and $[q^{\,j+k}+t]_{1\le \,j,k\le n}$ for any complex number $q\not =0,1$. As an application, for complex numbers $a,b,c$ with $b\not =0$ and $a^2\not =4b$, and the sequence $(w_m)_{m\in \mathbb Z}$ with $w_{m+1}=aw_m-bw_{m-1}$ for all $m\in \mathbb Z$, we determine the exact value of $\det [w_{\,j-k}+c\delta _{jk}]_{1\le \,j,k\le n}$.

1. Introduction

For any integer $n\ge 3$ , we have the determinant identity

$$ \begin{align*} \det[\,j-k]_{1\le \,j,k\le n}=0 \end{align*} $$

since $(1-k)+(3-k)=2(2-k)$ for all $k=1,\ldots ,n$ . However, it is nontrivial to deter- mine the characteristic polynomial $\det [xI_n-(\,j-k)]_{1\le \,j,k\le n}$ of the matrix $[\,j-k]_{1\le \,j,k\le n}$ , where $I_n$ is the identity matrix of order n.

For $j,k\in \mathbb N=\{0,1,2,\ldots \}$ , the Kronecker symbol $\delta _{jk}$ takes the value $1$ or $0$ according to whether $j=k$ or not. In 2003, Cloitre [1] generated the sequence $\det [\,j-k+\delta _{jk}]_{1\le \,j,k\le n}$ $(n=1,2,3,\ldots )$ with the initial fifteen terms:

$$ \begin{align*} 1,\, 2,\, 7,\, 21,\, 51,\, 106,\, 197,\, 337,\, 541,\, 826,\, 1211,\, 1717,\, 2367,\, 3186,\, 4201. \end{align*} $$

In 2013, C. Baker added a comment to [1] in which he claimed that

(1.1) $$ \begin{align} \det[\,j-k+\delta_{jk}]_{1\le \,j,k\le n}=1+\frac{n^2(n^2-1)}{12} \end{align} $$

without any proof. It seems that Baker found the recurrence of the sequence using the Maple package gfun.

Recall that the q-analogue of an integer m is given by

$$ \begin{align*}[m]_q=\frac{q^m-1}{q-1}.\end{align*} $$

Note that $\lim _{q\to 1}[m]_q=m$ .

In our first theorem, we determine the characteristic polynomial of the matrix $[q^{\,j-k}+t]_{1\le \,j,k\le n}$ for any complex number $q\not =0,1$ .

Theorem 1.1. Let $n\ge 2$ be an integer and let $q\not =0,1$ be a complex number. Then the characteristic polynomial of the matrix $P=[q^{\,j-k}+t]_{1\le \,j,k\le n}$ is

(1.2) $$ \begin{align} \det(xI_n-P)=x^{n-2}(x^2-n(t+1)x+t(n^2-q^{1-n}[n]_q^2)). \end{align} $$

Putting $t=-1$ and replacing x by $(q-1)x$ in Theorem 1.1, we immediately obtain the following corollary.

Corollary 1.2. Let $n\ge 2$ be an integer and let $q\not =0,1$ be a complex number. For the matrix $P_q=[[\,j-k]_q]_{1\le \,j,k\le n}$ ,

$$ \begin{align*}\det(xI_n-P_q)=x^{n}+\frac{q^{1-n}[n]_q^2-n^2}{(q-1)^2}x^{n-2}. \end{align*} $$

Remark 1.3. Fix an integer $n\ge 2$ . Observe that

$$ \begin{align*}\lim_{q\to1}\frac{q^{1-n}[n]_q^2-n^2}{(q-1)^2} &=\lim_{t\to0}\frac{(t+1)^{1-n}(((t+1)^n-1)/t)^2-n^2}{t^2} \\&=\lim_{t\to0}\frac{(t+1)^{1-n}((\sum_{k=1}^n\binom nkt^{k-1})^2-n^2)+((t+1)^{1-n}-1)n^2}{t^2} \\&=\lim_{t\to0}\bigg(\frac{(n+\binom n2t+\binom n3t^2+\cdots)^2-n^2}{(t+1)^{n-1}t^2}+n^2\frac{1-(t+1)^{n-1}}{(t+1)^{n-1}t^2}\bigg) \\&=\binom n2^2+2n\binom n3+\lim_{t\to0}\bigg(2n\binom n2\frac{t^{-1}}{(t+1)^{n-1}}-n^2\frac{\sum_{k=1}^n\binom{n-1}kt^{k-2}}{(t+1)^{n-1}}\bigg) \\&=\binom n2^2+2n\binom n3-n^2\binom{n-1}2 =\frac{n^2(n^2-1)}{12}. \end{align*} $$

So, by Corollary 1.2,

(1.3) $$ \begin{align} \det[x\delta_{jk}-(\,j-k)]_{1\le \,j,k\le n}=x^n+\frac{n^2(n^2-1)}{12}x^{n-2}, \end{align} $$

which indicates that when $n>2$ , the n eigenvalues of $A_n=[\,j-k]_{1\le \,j,k\le n}$ are

$$ \begin{align*}\lambda_1=\frac{n\sqrt{n^2-1}}{2\sqrt3}\,i,\quad \lambda_2=-\frac{n\sqrt{n^2-1}}{2\sqrt3}\,i,\quad \lambda_3=\cdots=\lambda_n=0.\end{align*} $$

Note that (1.1) follows from (1.3) with $x=-1$ . Concerning the permanent of $A_n$ , motivated by [3, Conjecture 11.23], we conjecture that

$$ \begin{align*}\mathrm{per}(A_{p-1})\equiv3\pmod p\quad \text{and}\quad \mathrm{per}(A_p)\equiv 1+4p\pmod {p^2}\end{align*} $$

for any odd prime p. Inspired by (1.1), Sun [4] conjectured that for any positive integers m and n,

$$ \begin{align*}\det[(\,j-k)^m+\delta_{jk}]_{1\le \,j,k\le n}=1 + n^2(n^2-1)f(n)\end{align*} $$

for a certain polynomial $f(x)\in \mathbb Q[x]$ with $\deg f=(m+1)^2-4$ .

Applying Corollary 1.2 with $q=-1$ , we find that

$$ \begin{align*}\det(xI_n-P_{-1})=x^n+\frac{(-1)^{n-1}[n]_{-1}^2-n^2}4x^{n-2}\end{align*} $$

for any integer $n\ge 2$ . In particular,

$$ \begin{align*}\det\bigg[\frac{1-(-1)^{\,j-k}}2+\delta_{\,j,k}\bigg]_{1\le \,j,k\le n}=\frac{9-(-1)^n-2n^2}8.\end{align*} $$

Applying Theorem 1.1 with $(t,x)=(-1,-2)$ and $(1,-1)$ , we obtain the following result.

Corollary 1.4. For any positive integer n,

$$ \begin{align*}\det[2^{\,j-k}-1+2\delta_{jk}]_{1\le \,j,k\le n}=\frac{4^n-2^{n-1}n^2+1}2 \end{align*} $$

and

$$ \begin{align*}\det[2^{\,j-k}+1+\delta_{\,j,k}]_{1\le \,j,k\le n}=(n+1)^2-2^{1-n}(2^n-1)^2. \end{align*} $$

In contrast to Theorem 1.1, we also establish the following result.

Theorem 1.5. Let $n\ge 2$ be an integer and let $q\not =0,1$ be a complex number. Then the characteristic polynomial of the matrix $Q=[q^{\,j+k}+t]_{0\le \,j,k\le n-1}$ is

(1.4) $$ \begin{align}\det(xI_n-Q)=x^n-(nt+[n]_{q^2})x^{n-1}+(n[n]_{q^2}-[n]_q^2)tx^{n-2}. \end{align} $$

The identity (1.4) with $q=2$ and $x=t=-1$ yields the following corollary.

Corollary 1.6. For any positive integer n,

$$ \begin{align*}\det[2^{\,j+k}-1+\delta_{jk}]_{0\le \,j,k\le n-1}=(2^n-1)^2-(n-1)\frac{4^n+2}3. \end{align*} $$

For complex numbers a and $b\not =0$ , the Lucas sequence $u_m=u_m(a,b)$ $(m\in \mathbb Z)$ and its companion sequence $v_m=v_m(a,b)$ $(m\in \mathbb Z)$ are defined as follows:

$$ \begin{gather*}u_0=0,\ u_1=1\quad \text{and} \quad u_{k+1}=au_k-bu_{k-1}\quad \text{for all } k\in\mathbb Z; \\v_0=2,\ v_1=a\quad \text{and} \quad v_{k+1}=av_k-bv_{k-1}\quad \text{for all } k\in\mathbb Z. \end{gather*} $$

By the Binet formula,

$$ \begin{align*}(\alpha-\beta)u_m=\alpha^m-\beta^m \quad \text{and}\quad v_m=\alpha^m+\beta^m\quad \text{for all}\ m\in\mathbb Z,\end{align*} $$

where

(1.5) $$ \begin{align}\alpha=\frac{a+\sqrt{a^2-4b}}2\quad \text{and}\quad \beta=\frac{a-\sqrt{a^2-4b}}2\end{align} $$

are the two roots of the quadratic equation $x^2-ax+b=0$ . Clearly, $b^nu_{-n}=-u_n$ and $b^nv_{-n}=v_n$ for all $n\in \mathbb N$ . For any positive integer n, it is known that

$$ \begin{align*}u_n=\sum_{k=0}^{\lfloor(n-1)/2\rfloor}\binom{n-1-k}ka^{n-1-2k}(-b)^k \quad \text{and}\quad v_n=\sum_{k=0}^{\lfloor n/2\rfloor}\frac n{n-k}\binom{n-k}ka^{n-2k}(-b)^k\end{align*} $$

(see [5, page 10]), which can be easily proved by induction. Note also that $u_m(2,1)=m$ for all $m\in \mathbb Z$ .

For $P(z)=\sum _{k=0}^{n-1}a_kz^k\in \mathbb C[z]$ , it is known (see [2, Lemma 9]) that

$$ \begin{align*}\det[P(x_{\,j}+y_k)]_{1\le \,j,k\le n}=a_{n-1}^n\prod_{r=0}^{n-1}\binom{n-1}r\times\prod_{1\le \,j<k\le n}(x_j-x_k)(y_k-y_j).\end{align*} $$

Thus, for any integer $n\ge 3$ and complex numbers a and $b\not =0$ ,

$$ \begin{align*} (\alpha-\beta)^n\det[u_{\,j-k}(a,b)]_{1\le \,j,k\le n}=\det[v_{\,j-k}(a,b)]_{1\le \,j,k\le n}=0 \end{align*} $$

(where $\alpha $ and $\beta $ are given by (1.5)), since

$$ \begin{align*}\det[\alpha^{\,j-k}\pm\beta^{\,j-k}]_{1\le \,j,k\le n} =\prod_{k=1}^n\alpha^{-k}\times\prod_{\,j=1}^n\beta^{\,j}\times\det\bigg[\bigg(\frac{\alpha}{\beta}\bigg)^{\,j} \pm\bigg(\frac{\alpha}{\beta}\bigg)^k\bigg]_{1\le \,j,k\le n}=0. \end{align*} $$

As an application of Theorem 1.1, we obtain the following new result.

Theorem 1.7. Let a and $b\not =0$ be complex numbers with $a^2\not =4b$ . Let $(w_m)_{m\in \mathbb Z}$ be a sequence of complex numbers with $w_{k+1}=aw_k-bw_{k-1}$ for all $k\in \mathbb Z$ . For any complex number c and integer $n\ge 2$ ,

(1.6) $$ \begin{align}\det[w_{\,j-k}+c\delta_{jk}]_{1\le \,j,k\le n} =c^n+c^{n-1}nw_0+c^{n-2}(w_1^2-aw_0w_1+bw_0^2)\frac{b^{1-n}u_n(a,b)^2-n^2}{a^2-4b}. \end{align} $$

Remark 1.8. It would be hard to guess the exact formula for $\det [w_{\,j-k}+c\delta _{jk}]_{1\le \,j,k\le n}$ in Theorem 1.7 by looking at various numerical examples.

Corollary 1.9. Let $a,b,c$ be complex numbers with $b\not =0$ and $a^2\not =4b$ . For any integer $n\ge 2$ ,

$$ \begin{align*} \det[u_{\,j-k}(a,b)+c\delta_{jk}]_{1\le \,j,k\le n}=c^n+c^{n-2}\frac{b^{1-n}u_n(a,b)^2-n^2}{a^2-4b} \end{align*} $$

and

$$ \begin{align*} \det[v_{\,j-k}(a,b)+c\delta_{jk}]_{1\le \,j,k\le n}=c^{n-2}((n+c)^2-b^{1-n}u_n(a,b)^2). \end{align*} $$

For any $m\in \mathbb Z$ , $u_m(-1,1)$ coincides with the Legendre symbol $(\frac {m}{3})$ , and $v_m(1,-1)=\omega ^m+\bar \omega ^m$ , where $\omega $ denotes the cube root $(-1+\sqrt {-3})/2$ of unity. Applying Corollary 1.9 with $a=-1$ and $b=1$ , we get the following result.

Corollary 1.10. For any integer $n\ge 2$ and complex number c,

$$ \begin{align*} \det\bigg[\bigg(\frac{\,j-k}3\bigg)+c\delta_{\,j,k}\bigg]_{1\le \,j,k\le n}=c^n+c^{n-2}\bigg\lfloor\frac{n^2}3\bigg\rfloor. \end{align*} $$

Recall that $F_m=u_m(1,-1)$ ( $m\in \mathbb Z$ ) are the well-known Fibonacci numbers and $L_m=v_m(1,-1)$ $(m\in \mathbb Z$ ) are the Lucas numbers. Corollary 1.9 with $a=1$ and $b=-1$ yields the following result.

Corollary 1.11. For any integer $n\ge 2$ and complex number c,

$$ \begin{align*} \det[F_{\,j-k}+c\delta_{jk}]_{1\le \,j,k\le n}=c^n+\frac{c^{n-2}}5((-1)^{n-1}F_n^2-n^2) \end{align*} $$

and

$$ \begin{align*} \det[L_{\,j-k}+c\delta_{jk}]_{1\le \,j,k\le n}=c^{n-2}((n+c)^2+(-1)^nF_n^2). \end{align*} $$

Although we have Theorem 1.5 which is similar to Theorem 1.1, it seems impossible to use Theorem 1.5 to deduce a result similar to Theorem 1.7.

2. Proof of Theorem 1.1

Lemma 2.1. Let n be a positive integer, and let $q\neq 0$ and t be complex numbers with $n-[n]_q+t(q^{1-n}[n]_q-n)\not =0$ . Suppose that

(2.1) $$ \begin{align} \gamma=\frac{n(t+1)\pm\sqrt{n^2(t-1)^2+4tq^{1-n}[n]_q^2}}2 \quad \text{and}\quad y=\frac{\gamma-[n]_q-nt}{n-[n]_q+(q^{1-n}[n]_q-n)t}. \end{align} $$

Then, for any positive integer j,

(2.2) $$ \begin{align} \sum_{k=1}^n(q^{\,j-k}+t)(1+y(q^{k-n}-1))=\gamma(1+y(q^{\,j-n}-1)). \end{align} $$

Proof. As $\gamma ^2-n(t+1)\gamma +(n^2-q^{1-n}[n]_q^2)t=0,$

$$ \begin{align*}[n]_q(n-[n]_q+(q^{1-n}[n]_q-n)t)=(\gamma-[n]_q-nt)(\gamma-n+[n]_q)\end{align*} $$

and hence

(2.3) $$ \begin{align} (\gamma-n+[n]_q)y=[n]_q. \end{align} $$

For $j\in \{1,2,3,\ldots \}$ , set

$$ \begin{align*}\Delta_{\,j}=\sum_{k=1}^n(q^{\,j-k}+t)(1+y(q^{k-n}-1))-\gamma(1+y(q^{\,j-n}-1)).\end{align*} $$

Then, by (2.3),

$$ \begin{align*} \Delta_{\,j}-t(1+y(q^{k-n}-1))+\gamma(1-y) & = q^{\,j-n}\bigg(\sum_{k=1}^nq^{n-k}(1+y(q^{k-n}-1))-\gamma y\bigg) \\ & =q^{\,j-n}([n]_q(1-y)+ny-\gamma y)=0. \end{align*} $$

So $\Delta _1=\Delta _2=\cdots $ .

Next we show that $\Delta _n=0$ . Observe that

$$ \begin{align*} \sum_{k=1}^n(q^{n-k}+t)(1+(q^{k-n}-1)y) & =\sum_{k=1}^n(q^{n-k}(1-y)+t(1-y)+y+q^{k-n}ty) \\ & = [n]_q(1-y)+nt(1-y)+ny+q^{1-n}[n]_qty \\ & = [n]_q+nt+y(n-[n]_q+(q^{1-n}[n]_q-n)t) \\ & = \gamma=\gamma(1+y(q^{n-n}-1)) \end{align*} $$

by the definition of y. So $\Delta _n=0$ .

In view of the above, $\Delta _{j}=0$ for all $j=1,2,3,\ldots $ . This concludes the proof.

Proof of Theorem 1.1.

It is easy to verify the desired result for $n=2$ . Below we assume that $n\ge 3$ .

If $n-[n]_q$ and $q^{1-n}[n]_q-n$ are both zero, then $q^{n-1}=1$ and $n=[n]_q=1$ . As $n\ge 3$ , there are infinitely many $t\in \mathbb C$ such that

$$ \begin{align*}n-[n]_q+t(q^{1-n}[n]_q-n)\not=0 \quad\mbox{and}\quad n^2(t-1)^2+4tq^{1-n}[n]_q^2\not=0.\end{align*} $$

Take such a number t, and choose $\gamma $ and y as in (2.1). Then $\gamma $ given in (2.1) is an eigenvalue of the matrix $P=[q^{\,j-k}+t]_{1\le \,j,k\le n}$ , and the column vector $v=(v_1,\ldots ,v_n)^T$ with $v_k=1+y(q^{k-n}-1)$ is an eigenvector of P associated with the eigenvalue $\gamma $ . Note that $\gamma $ given by (2.1) has two different choices since $n^2(t-1)^2+4tq^{1-n}[n]_q^2\not =0$ .

Let $s\in \{3,\ldots ,n\}$ . For $1\le k\le n$ , let us define

$$ \begin{align*}v^{(s)}_k= \begin{cases} q^{2-s}[s-2]_q&\text{if}\ k=1, \\ -q^{2-s}[s-1]_q&\text{if}\ k=2, \\ \delta_{sk}&\text{if}\ 3\le k\le n. \end{cases}\end{align*} $$

It is easy to verify that

$$ \begin{align*}\sum_{k=1}^nv^{(s)}_k=0=\sum_{k=1}^nq^{\,j-k}v^{(s)}_k\quad \text{for all}\ j=1,\ldots,n.\end{align*} $$

Thus, $0$ is an eigenvalue of the matrix $P=[q^{\,j-k}+t]_{1\le \,j,k\le n}$ , and the column vector $v^{(s)}=(v^{(s)}_1,\ldots ,v^{(s)}_n)^T$ is an eigenvector of P associated with the eigenvalue $0$ .

If $\sum _{s=3}^n c_sv^{(s)}$ is the zero column vector for some $c_3,\ldots ,c_n\in \mathbb C$ , then for each $k=3,\ldots ,n$ ,

$$ \begin{align*}c_k=\sum_{s=3}^n c_s\delta_{sk}=\sum_{s=3}^n c_sv^{(s)}_k=0.\end{align*} $$

Thus, the $n-2$ column vectors $v^{(3)},\ldots ,v^{(n)}$ are linearly independent over $\mathbb C$ .

By the above, the n eigenvalues of the matrix $P=[q^{\,j-k}+t]_{1\le \,j,k\le n}$ are the two values of $\gamma $ given by (2.2) and $\lambda _3=\cdots =\lambda _n=0$ . Thus, the characteristic polynomial of P is

$$ \begin{align*}\det(xI_n-P)&=\bigg(x-\frac{n(t+1)}2-\frac{\sqrt{n^2(t-1)^2+4tq^{1-n}[n]_q^2}}2\bigg) \\&\quad\times\bigg(x-\frac{n(t+1)}2+\frac{\sqrt{n^2(t-1)^2+4tq^{1-n}[n]_q^2}}2\bigg)\prod_{s=3}^n(x-\lambda_s) \\&=x^{n-2}\bigg(\bigg(x-\frac{n(t+1)}2\bigg)^2-\frac{n^2(t-1)^2+4tq^{1-n}[n]_q^2}4\bigg) \\&=x^{n-2}(x^2-n(t+1)x+t(n^2-q^{1-n}[n]_q^2)). \end{align*} $$

Thus, the identity (1.2) holds for infinitely many values of t. Note that both sides of (1.2) are polynomials in t for any fixed $x\in \mathbb C$ . Thus, if we view both sides of (1.2) as polynomials in x and t, then the identity (1.2) still holds. This completes the proof.

3. Proof of Theorem 1.5

The following lemma is quite similar to Lemma 2.1.

Lemma 3.1. Let n be a positive integer, and let $q\neq 0$ and t be complex numbers with $[n]_{q^2}+(q^{1-n}t-q^{n-1})[n]_q-nt\neq 0$ . Suppose that

(3.1) $$ \begin{align} \gamma=\frac{nt+[n]_{q^2}\pm\sqrt{(nt-[n]_{q^2})^2+4t[n]_q^2}}2 \quad \text{and}\quad z=\frac{\gamma-q^{n-1}[n]_q-nt}{[n]_{q^2}+(q^{1-n}t-q^{n-1})[n]_q-nt}. \end{align} $$

Then, for every $j=0,1,2,\ldots ,$

(3.2) $$ \begin{align} \sum_{k=0}^{n-1}(q^{\,j+k}+t)(1+z(q^{k-n+1}-1))=\gamma(1+z(q^{\,j-n+1}-1)). \end{align} $$

Proof. Since $\gamma ^2-(nt+[n]_{q^2})\gamma +t(n[n]_{q^2}-[n]_q^2)=0,$ we have

(3.3) $$ \begin{align} (\gamma-[n]_{q^2}+q^{n-1}[n]_q)z=q^{n-1}[n]_q. \end{align} $$

For $j\in \{0,1,2,\ldots \}$ , set

$$ \begin{align*}R_{j}=\sum_{k=0}^{n-1}(q^{\,j+k}+t)(1+z(q^{k-n+1}-1))-\gamma(1+z(q^{\,j-n+1}-1)).\end{align*} $$

It is easy to see that

$$ \begin{align*} \begin{aligned} R_{j}-\sum_{k=0}^{n-1}t(1+z(q^{k-n+1}-1))+\gamma(1-z) =q^{\,j-n+1}\big({q^{n-1}[n]_q(1-z)+z[n]_{q^2}}-\gamma z\big)=0 \end{aligned}\end{align*} $$

with the aid of (3.3). So $R_0=R_1=\cdots $ . As

$$ \begin{align*} \sum_{k=0}^{n-1}(q^{n-1+k}+t)(1+z(q^{k-n+1}-1)) =\gamma=\gamma(1+z(q^{(n-1)-n+1}-1)), \end{align*} $$

we get $R_{n-1}=0$ . So the desired result follows.

Proof of Theorem 1.5.

It is easy to verify the desired result for $n=2$ . Below we assume that $n\ge 3$ .

If $[n]_{q^2}-q^{n-1}[n]_q$ and $q^{1-n}[n]_q-n$ are both zero, then $[n]_q\not =0$ and

$$ \begin{align*}(q^n+1)[n]_q=(q+1)[n]_{q^2}=(q+1)q^{n-1}[n]_q=(q^n+q^{n-1})[n]_q,\end{align*} $$

and hence $q^{n-1}=1$ and $n=[n]_q=1$ . As $n\ge 3$ , there are infinitely many $t\in \mathbb C$ such that

$$ \begin{align*}[n]_{q^2}+(q^{1-n}t-q^{n-1})[n]_q-nt\not=0 \quad\mbox{and}\quad (nt-[n]_{q^2})^2+4t[n]_q^2\not=0.\end{align*} $$

Take such a number t, and choose $\gamma $ and z as in (3.1). Then $\gamma $ given in (3.1) is an eigenvalue of the matrix $Q=[q^{\,j+k}+t]_{0\le \,j,k\le n-1}$ , and the column vector $v=(v_0,\ldots ,v_{n-1})^T$ with $v_k=1+z(q^{k-n+1}-1)$ is an eigenvector of Q associated with the eigenvalue $\gamma $ . There are two different choices for $\gamma $ since $(nt-[n]_{q^2})^2+4t[n]_q^2\not =0$ .

Let $s\in \{3,\ldots ,n\}$ . For $k\in \{0,\ldots ,n-1\}$ , define

$$ \begin{align*}v^{(s)}_k= \begin{cases} q[s-2]_q&\text{if}\ k=0, \\ -[s-1]_q&\text{if}\ k=1, \\ \delta_{s,k+1}&\text{if}\ 2\le k\le n-1. \end{cases}\end{align*} $$

It is easy to verify that

$$ \begin{align*}\sum_{k=0}^{n-1}v^{(s)}_k=0=\sum_{k=0}^{n-1}q^{\,j+k}v^{(s)}_k\quad \text{for all}\ j=1,\ldots,n.\end{align*} $$

Thus, $0$ is an eigenvalue of the matrix $Q=[q^{\,j+k}+t]_{0\le \,j,k\le {n-1}}$ , and the column vector $v^{(s)}=(v^{(s)}_0,\ldots ,v^{(s)}_{n-1})^T$ is an eigenvector of Q associated with the eigenvalue $0$ .

If $\sum _{s=3}^{n} c_sv^{(s)}$ is the zero column vector for some $c_3,\ldots ,c_{n}\in \mathbb C$ , then for each $k=2,\ldots ,n-1$ ,

$$ \begin{align*} c_{k+1}=\sum_{s=3}^{n} c_s\delta_{s,k+1}=\sum_{s=3}^{n} c_sv^{(s)}_{k}=0. \end{align*} $$

Thus, the $n-2$ column vectors $v^{(3)},\ldots ,v^{(n)}$ are linearly independent over $\mathbb C$ .

By the above, the n eigenvalues of the matrix $Q=[q^{\,j+k}+t]_{0\le \,j,k\le {n-1}}$ are the two values of $\gamma $ given by (3.2) and $\lambda _3=\cdots =\lambda _n=0$ . Thus, the characteristic polynomial of Q is

$$ \begin{align*}\det(xI_n-Q)=&\ \bigg(x-\frac{nt+[n]_{q^2}}2-\frac{\sqrt{(nt-[n]_{q^2})^2+4t[n]_q^2}}2\bigg) \\&\ \times\bigg(x-\frac{nt+[n]_{q^2}}2+\frac{\sqrt{(nt-[n]_{q^2})^2+4t[n]_q^2}}2\bigg)\prod_{s=3}^n(x-\lambda_s) \\=&\ x^{n-2}\bigg(\bigg(x-\frac{nt+[n]_{q^2}}2\bigg)^2-\frac{(nt-[n]_{q^2})^2+4t[n]_q^2}4\bigg) \\=&\ x^n-(nt+[n]_{q^2})x^{n-1}+(n[n]_{q^2}-[n]_q^2)tx^{n-2}. \end{align*} $$

Thus, the identity (1.4) holds for infinitely many values of t. Note that both sides of (1.4) are polynomials in t for any fixed $x\in \mathbb C$ . If we view both sides of (1.4) as polynomials in x and t, then the identity (1.4) still holds. This concludes the proof.

4. Proof of Theorem 1.7

Proof of Theorem 1.7.

If $w_0=w_1=0$ or $n=2$ , then the desired result can be easily verified. Below we assume that $n\ge 3$ and $\{w_0,w_1\}\not =\{0\}$ .

Let $\alpha $ and $\beta $ be the two roots of the quadratic equation $z^2-az+b=0$ . Note that $\alpha \beta =b\not =0$ . Also, $\alpha \not =\beta $ since $\Delta =a^2-4b$ is nonzero. It is well known that there are constants $c_1,c_2\in \mathbb C$ such that $w_m=c_1\alpha ^m+c_2\beta ^m$ for all $m\in \mathbb Z$ . As $c_1+c_2=w_0$ and $c_1\alpha +c_2\beta =w_1$ ,

(4.1) $$ \begin{align} c_1=\frac{w_1-\beta w_0}{\alpha-\beta}\quad \text{and}\quad c_2=\frac{\alpha w_0-w_1}{\alpha-\beta}. \end{align} $$

Since $w_0$ or $w_1$ is nonzero, one of $c_1$ and $c_2$ is nonzero. Without any loss of generality, we assume $c_1\not =0$ .

Let W denote the matrix $[w_{\,j-k}+c\delta _{jk}]_{1\le \,j,k\le n}$ . Then

$$ \begin{align*}\det(W) & = \det[c_1\alpha^{\,j-k}+c_2\beta^{\,j-k}+c\delta_{jk}]_{1\le \,j,k\le n} \\ & =c_1^n\prod_{\,j=1}^n\beta^{\,j}\times\prod_{k=1}^n\beta^{-k} \times\det\bigg[\bigg(\frac{\alpha}{\beta}\bigg)^{\,j-k}+\frac{c_2+c\delta_{jk}\beta^{k-\,j}}{c_1}\bigg]_{1\le \,j,k\le n} \\ & = c_1^n\det[q^{\,j-k}+t-x\delta_{jk}]_{1\le \,j,k\le n}=(-c_1)^n\det[x\delta_{jk}-q^{\,j-k}-t]_{1\le \,j,k\le n}, \end{align*} $$

where $q=\alpha /\beta \not =0,1$ , $t=c_2/c_1$ and $x=-c/c_1$ . By applying Theorem 1.1, we obtain

$$ \begin{align*}\det(W)&=(-c_1)^n x^{n-2}(x^2-n(t+1)x+t(n^2-q^{1-n}[n]_q^2)) \\ &=c^{n-2}\bigg(c^2+nc(c_1+c_2)+c_1c_2\bigg(n^2-\frac{\alpha^{1-n}}{\beta^{1-n}}\bigg(\frac{(\alpha/\beta)^n-1}{\alpha/\beta-1}\bigg)^2\bigg)\bigg) \\ &=c^n+nw_0c^{n-1}+c^{n-2}c_1c_2\bigg(n^2-(\alpha\beta)^{1-n}\bigg(\frac{\alpha^n-\beta^n}{\alpha-\beta}\bigg)^2\bigg) \\ &=c^n+nw_0c^{n-1}+c^{n-2}c_1c_2(n^2-b^{1-n}u_n(a,b)^2). \end{align*} $$

In view of (4.1),

$$ \begin{align*} c_1c_2=\frac{(w_1-\beta w_0)(\alpha w_0-w_1)}{(\alpha-\beta)^2}=\frac{-w_1^2+(\alpha+\beta)w_0w_1-\alpha\beta w_0^2}{\Delta}=-\frac{w_1^2-aw_0w_1+bw_0^2}{a^2-4b}. \end{align*} $$

Therefore, the desired evaluation (1.6) follows.