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In this article, we define \(q\)-generalized Fibonacci polynomials and \(q\)-generalized Lucas polynomials using \(q\)-binomial coefficient and obtain their recursive properties. In addition, we introduce generalized \(q\)-Fibonacci matrix and generalized \(q\)-Lucas matrix, then we derive their basic identities. We define \((k,q,t)\)-symmetric generalized Fibonacci matrix and \((k,q,t)\)-symmetric generalized Lucas matrix, then we give the Cholesky factorization of these matrices. Finally, we give determinantal and permanental representations of these new polynomial sequences.
MacHenry [1] defined generalized Fibonacci polynomials and generalized Lucas polynomials. The generalized Fibonacci polynomials and the generalized Lucas polynomials already have comprehensive representation properties. These polynomials are a general form of generalized bivariate Fibonacci and Lucas \(p\)-polynomials, ordinary Fibonacci, Lucas, Pell, Pell-Lucas and Perrin sequences, Chebyshev polynomials of the second kind, and the Tribonacci numbers, etc.
The generalized Fibonacci polynomials, \(F_{k,n}(t)\), and the generalized Lucas polynomials, \(G_{k,n}(t)\), are defined inductively by as follows: \[\begin{aligned} F_{k,0}(t)&=&1,\text{ }F_{k,n+1}(t)=t_{1}F_{k,n}(t)+\cdots +t_{k}F_{k,n-k+1}(t)(n>1), \\ \end{aligned}\] and \[\begin{aligned} G_{k,0}(t)&=&k,\text{ }G_{k,1}(t) =t_{1},\text{ }G_{k,n}(t)=G_{k-1,n}(t)(1% \leq n\leq k), \\ G_{k,n}(t) &=&t_{1}G_{k,n-1}(t)+\cdots +t_{k}G_{k,n-k}(t)(n>k), \end{aligned}\] where the vector \(t=(t_{1},t_{2},\ldots ,t_{k})\) and \(t_{i}\) \((1\leq i\leq k)\) are constant coefficients of the core polynomial \[P(x;t_{1},t_{2},\ldots ,t_{k})=x^{k}-t_{1}x^{k-1}-\cdots -t_{k}.\]
In [2], authors gave explicit formula for the \(F_{k,n}(t)\) and \(% G_{k,n}(t)\) as follows: \[F_{k,n}(t)=\sum\limits\limits_{a\vdash n}\binom{|a|}{a_{1,\ldots ,}a_{k}}% t_{1}^{a_{1}}\ldots t_{k}^{a_{k}}, \label{fibbin}\tag{1}\] \[G_{k,n}(t)=\sum\limits\limits_{a\vdash n}\frac{n}{|a|}\binom{|a|}{a_{1,\ldots ,}a_{k}}t_{1}^{a_{1}}\ldots t_{k}^{a_{k}}. \label{lucbin}\]
The notations \(a\vdash n\) and \(|a|\) are used instead of \(\sum\limits% \limits_{j=1}^{k}ja_{j}=n\) and \(\sum\limits\limits_{j=1}^{k}a_{j},\) respectively.
In addition, in [2, 3, 4, 5, 6], the authors studied algebraic properties of these polynomials.
On the other hand, there exists several different \(q\)-analogues of the Fibonacci-type sequences, see [7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17]. For example, Cigler [14] defined \(q\)-Fibonacci polynomials (\(F_{n,q}(x,s)\)) and \(q\)-Lucas polynomials (\(L_{n,q}(x,s)\)) as follows: \[\begin{aligned} F_{n}(x,s) &=&\sum\limits_{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\left[ \begin{array}{c} n-k-1 \\ k% \end{array}% \right] q^{\binom{k+1}{2}}x^{n-1-2k}s^{k}, \\ \text{ }L_{n}(x,s) &=&\sum\limits_{i=0}^{\left\lfloor \frac{n}{2}\right\rfloor }% \frac{\left[ n\right] }{\left[ n-i\right] }\left[ \begin{array}{c} n-i \\ i% \end{array}% \right] q^{\binom{i}{2}}x^{n-2i}s^{i}. \end{aligned}\]
The \(q\)-binomial coefficient is defined as: \[{n \brack k}_q:=\frac{(q;q)_n}{(q;q)_k(q;q_{n-k})}= \frac{ [n]_q!}{ [k]_q! [n-k]_q!},\] with \((a;q)_n:=\prod\limits_{j=0}^{n-1}(1-aq^j)\), \(\lbrack n]_{q}=1+q+\cdots +q^{n-1}\) and \(\lbrack n]_{q}!=[1]_{q}[2]_{q}\cdots \lbrack n]_{q}\).
Many researchers have studied matrices whose elements are binomial coefficients, Fibonacci-type sequences and \(q\)-binomial coefficients, see [18, 19, 20, 21, 22, 23, 24, 25]. Lee et al. [21] defined \(% n\times n\) \(k\)-Fibonacci matrix \({\mathcal{F}}% (k)_{n}=[f(k)_{i,j}]_{i,j=1,2,…,n}\) as: \[f(k)_{i,j}=\left\{ \begin{array}{ll} f_{k,i-j+k-1}, & \text{if }i-j+1\geq 0, \\ 0, & \text{otherwise},% \end{array}% \right.\] where \(f_{k,n}\) \(n\)th \(k\)-Fibonacci numbers defined by Miles in [26]. The \({\mathcal{F}}(k)_{n}^{-1}=[f_{i,j}^{\imath }]\) was given as follows: \[f_{i,j}^{\imath }=\left\{ \begin{array}{ll} 1, & \text{if }i=j \\ -1, & \text{if }i-k\leq j\leq i-1 \\ 0, & \text{otherwise.}% \end{array}% \right.\] Lee et al. [21] also defined \(n\times n\) \(k\)-symmetric Fibonacci matrix \({\mathcal{Q}}(k)_{n}=[q(k)_{i,j}]\) as follows: \[q(k)_{i,j}=q(k)_{j,i}=\left\{ \begin{array}{ll} \sum\limits\limits_{l=1}^{k}q(k)_{i,i-l}+f_{k,k-1}, & \text{if }i=j, \\ \sum\limits\limits_{l=1}^{k}q(k)_{i,j-l}, & \text{if }i+1\leq j,% \end{array}% \right.\] and obtained the Cholesky factorization of \({\mathcal{Q}}(k)_{n}\) as follows: \[{\mathcal{Q}}(k)_{n}={\mathcal{F}}(k)_{n}({\mathcal{F}}(k)_{n})^{T}.\]
In addition, many researchers have studied determinantal and permanental representations of Fibonacci-type sequences and polynomials. More examples can be found in [3, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37].
In this section, we define two families of polynomials, the \(q\)-generalized Fibonacci polynomial and the \(q\)-generalized Lucas polynomial using \(q\) -binomials and obtain properties of these polynomials. In the following two definitions, the summation takes place over all integers \(% c_{1},c_{2},…,c_{k}\) such that \(\sum\limits\limits_{j=1}^{k}jc_{j}=n\), and \(% c=\sum\limits\limits_{j=1}^{k}c_{j}.\)
Definition 1.For any integers \(n\geq 0\), the \(q\)-generalized Fibonacci polynomial, \(% F_{k,n}(t;q)\), is defined by \[F_{k,n}(t;q):=\sum\limits \frac{[c]_{q}!}{[c_{1}]_{q}![c_{2}]_{q}!…[c_{k}]_{q}!}% t_{1}^{c_{1}}\ldots t_{k}^{c_{k}}.\]
In particular, if we take \(q=1\), we obtain the \(F_{k,n}(t;1):=F_{k,n}(t)\).
The first few \(F_{k,n}(t;q)\) are \[1,\text{ }t_{1},\text{ }t_{2}+t_{1}^{2},\text{ }% t_{3}+(1+q)t_{1}t_{2}+t_{1}^{3},\text{ }% t_{4}+(1+q)t_{1}t_{3}+t_{2}^{2}+t_{1}^{4}+(1+q+q^{2})t_{1}^{2}t_{2},\ldots\,.\]
In particular, if we take \(q=1\), we obtain the \(L_{k,n}(t;1):=G_{k,n}(t)\).
The first few \(L_{k,n}(t;q)\) are \[_{q}, t_{1}, (1+q)t_{2}+t_{1}^{2}, (1+q+q^{2})t_{3}+(1+q+q^{2})t_{1}t_{2}+t_{1}^{3},\ldots\,.\] We need the following definitions and lemmas in our proofs.
Definition 3. \(\mathcal{S}_{k,n}\) is the sequence defined by \(\mathcal{S}_{k,0}=1,\) \(% \mathcal{S}_{k,1}=t_{1}\) and for \(n\geq 2\) \[\mathcal{S}_{k,n}=t_{1}\mathcal{S}_{k,n-1}+% \sum\limits_{j=1}^{n-1}(-1)^{n-j}F_{k,n-j+1}(t;q)\mathcal{S}_{k,j-1}. \label{efes}\tag{2}\]
The first few terms of \(\mathcal{S}_{k,n}\) are \[1,t_{1},\text{ }-t_{2},\text{ }t_{3}-t_{1}t_{2}+qt_{1}t_{2},\text{ }% -t_{4}+t_{1}t_{3}-qt_{1}t_{3}+qt_{1}^{2}t_{2}-q^{2}t_{1}^{2}t_{2},\dots\,.\]
Lemma 1. Let \(n\geq 1\) be an integer. Then \[F_{k,n}(t;q)=\sum\limits_{j=1}^{k}(-1)^{j+1}\mathcal{S}_{k,j}F_{k,n-j}(t;q). \label{efes2}\tag{3}\]
Proof. This is obvious from Eq. (2). ◻
Definition 4. \(\mathcal{T}_{k,n}\) is the sequence defined by \(\mathcal{T}_{k,0}=1,\) \(% \mathcal{T}_{k,1}=t_{1}\) and \(n\geq 2\) \[\mathcal{T}_{k,n}=t_{1}\mathcal{T}_{k,n-1}+% \sum\limits_{j=1}^{n-1}(-1)^{n-j}L_{k,n-j+1}(t;q)\mathcal{T}_{k,j-1}. \label{elet}\tag{4}\] The first few terms of \(\mathcal{T}_{k,n}\) are \[1,t_{1},-t_{2}-qt_{2},t_{3}+qt_{3}-t_{1}t_{2}-qt_{1}t_{2}+q^{2}t_{3}+q^{2}t_{1}t_{2},\ldots\,.\]
Lemma 2. Let \(n\geq 1\) be an integer. Then \[L_{k,n}(t;q)=\sum\limits_{j=1}^{k}(-1)^{j+1}\mathcal{T}_{k,j}L_{k,n-j}(t;q). \label{elet2}\]
Proof. This is obvious from Eq. (4). ◻
Theorem 1. Let \(F_{k,n}(t_{1},t_{2},\ldots ,t_{k})\) be the generalized Fibonacci polynomial and \(F_{k,n}(t;q)\) be the \(q\)-generalized Fibonacci polynomial, then \[F_{k,n}(t_{1},t_{2},\ldots ,t_{k};q)=F_{k,n}(\mathcal{S}_{k,1},-\mathcal{S}% _{k,2},\ldots ,(-1)^{k+1}\mathcal{S}_{k,k}).\]
Proof. We proceed by induction on \(n\). The result clearly holds for \(n=1,\) since \(F_{k,1}(t;q)=t_{1}=\mathcal{S}_{k,1}=F_{k,1}(\mathcal{S}_{k,1},-% \mathcal{S}_{k,2},\ldots ,(-1)^{k+1}\mathcal{S}_{k,k})\). Now suppose that the result is true for all positive integers less than or equal to \(n\). We prove it for (\(n+1\)). In fact, by the definition of generalized Fibonacci polynomials for the vector \(\mathcal{S}=(\mathcal{S}_{k,1},-\mathcal{S}% _{k,2},\ldots ,(-1)^{k+1}\mathcal{S}_{k,k})\), we have
\[F_{k,n+1}(\mathcal{S})=\mathcal{S}_{k,1}F_{k,n}(\mathcal{S})+\cdots +(-1)^{k+1}\mathcal{S}_{k,k}F_{k,n-k+1}(\mathcal{S}).\] From the hypothesis of induction, we obtain \[F_{k,n+1}(\mathcal{S})=\mathcal{S}_{k,1}F_{k,n}(t;q)+\cdots +(-1)^{k+1}% \mathcal{S}_{k,k}F_{k,n-k+1}(t;q).\] Thus, we obtain \[F_{k,n+1}(\mathcal{S})=F_{k,n+1}(t;q),\] using Eq. (3). ◻
Theorem 2. Let \(F_{k,n}(t_{1},t_{2},\ldots ,t_{k})\) be the generalized Fibonacci polynomial and \(L_{k,n}(t;q)\) be the \(q\)-generalized Lucas polynomial, then \[L_{k,n}(t_{1},t_{2},\ldots ,t_{k};q)=F_{k,n}(\mathcal{T}_{k,1},-\mathcal{T}% _{k,2},\ldots ,(-1)^{k+1}\mathcal{T}_{k,k}).\]
Proof. The proof runs like in Theorem 1. ◻
Corollary 1. \[F_{k,n}(t;q):=\sum\limits \frac{c!}{c_{1}!…c_{k}!}\mathcal{S}_{k,1}^{c_{1}}(-% \mathcal{S}_{k,2}^{c_{2}})\ldots ((-1)^{k}\mathcal{S}% _{k,k-1}^{c_{k-1}})((-1)^{k+1}\mathcal{S}_{k,k}^{c_{k}})\,.\]
Proof. This is obvious from Eq. (1) and Theorem 1. ◻
Corollary 2. \[L_{k,n}(t;q):=\sum\limits \frac{c!}{c_{1}!…c_{k}!}\mathcal{T}_{k,1}^{c_{1}}(-% \mathcal{T}_{k,2}^{c_{2}})\ldots ((-1)^{k}\mathcal{T}% _{k,k-1}^{c_{k-1}})((-1)^{k+1}\mathcal{T}_{k,k}^{c_{k}})\,.\]
Proof. This is obvious from Eq. (1) and Theorem 2. ◻
Corollary 3. Let \(y_{k,n}=(-1)^{n+1}\mathcal{T}_{k,n}\) for \(q=1,\) \(% F_{k,n}(t_{1},t_{2},\ldots ,t_{k})\) be the generalized Fibonacci polynomial and \(G_{k,n}(t_{1},t_{2},\ldots ,t_{k})\) be the generalized Lucas polynomial, then \[G_{k,n}(t_{1},t_{2},\ldots ,t_{k})=F_{k,n}(y_{k,1},y_{k,2},\ldots ,y_{k,k}).\]
Proof. If we rewrite Theorem 2. for \(q=1,\) we obtain \[L_{k,n}(t;1)=F_{k,n}(y_{k,1},\ldots ,y_{k,k}).\] Further, this is obvious from the definitions of \(L_{k,n}(t;q)\) and \(% G_{k,n}(t)\) that \(L_{k,n}(t;1)=G_{k,n}(t).\) Therefore, we obtain the desired result. ◻
There have been several studies on the generalized Fibonacci polynomials and the generalized Lucas polynomials and the relationship between them. Corollary 3. gives a new relation between these polynomials. Using this corollary, different results can be obtained. From Propositions \(% 1,3,4\) and Lemma \(3\) in [5], we can give the following corollaries.
Corollary 4. Let \(y_{k,n}=(-1)^{n+1}\mathcal{T}_{k,n}\) for \(q=1\) and \(% F_{k,n}(t_{1},t_{2},\ldots ,t_{k})\) be the generalized Fibonacci polynomial, then \[F_{k,n}(y_{k,1},y_{k,2},\ldots ,y_{k,k})=-t_{k-1}F_{k,n-k+1}(t)-\cdots -(k-1)t_{1}F_{k,n-1}(t)+kF_{k,n}(t).\]
Corollary 5. Let \(y_{k,n}=(-1)^{n+1}\mathcal{T}_{k,n}\) for \(q=1\) and \(% F_{k,n}(t_{1},t_{2},\ldots ,t_{k})\) be the generalized Fibonacci polynomial, then \[F_{k,n}(y_{k,1},y_{k,2},\ldots ,y_{k,k})=\sum\limits\limits_{a\vdash n}n(-1)^{a+1}% \binom{|a| -1}{a_{1,\ldots ,}a_{k}}F_{k,1}^{a_{1}}\ldots F_{k,k}^{a_{k}}.\]
Corollary 6. Let \(y_{k,n}=(-1)^{n+1}\mathcal{T}_{k,n}\) for \(q=1\) and \(% F_{k,n}(t_{1},t_{2},\ldots ,t_{k})\) be the generalized Fibonacci polynomial, then \[\frac{\partial F_{k,n}(y_{k,1},y_{k,2},\ldots ,y_{k,k})}{\partial t_{i}}% =nF_{k,n-i}(t_{1},t_{2},\ldots ,t_{k}).\]
Example 1. We give a companion matrix \(C_{(k)}\) and obtain \(C_{(k)}^{\infty }\) using the method in [5] as follows:
\[C_{(k)}=\left[ \begin{array}{ccccc} 0 & 1 & 0 & \ldots & 0 \\ 0 & 0 & 1 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \ldots & 1 \\ (-1)^{k+1}\mathcal{S}_{k} & (-1)^{k}\mathcal{S}_{k-1} & (-1)^{k-1}\mathcal{S}% _{k-2} & \ldots & \mathcal{S}_{1}% \end{array}% \right]\,.\] In particular, if we take \(q=1\), we obtain the companion matrix \(A\) in [5]. We let the companion matrix operate on its last row vector on the right and append the image vector to the companion matrix as a new last row. We repeat this process, obtaining a matrix with infinitely many rows. Note that \(C_{(k)}\) is invertible if and only if \(\mathcal{S}_{k}\neq 0\). Assuming that \(\mathcal{S}_{k}\neq 0\), we can also extend the matrix from the top row upward by operating on the top row with \(C_{(k)}^{-1}\), obtaining a doubly infinite matrix, that is, one with infinitely many rows in either direction and \(k\)-columns. We call this the infinite companion matrix \(C_{(k)}^{\infty }\). If we take \(q=1\), we obtain the companion matrix \(A^{\infty }\) in [5]. It is obvious that, the right hand column of \(% C_{(k)}^{\infty }\) in the positive direction is \(F_{k,n}(t;q).\)
In this section, we introduce the generalized \(q\)-Fibonacci matrix and generalized \(q\)-Lucas matrix, then we find their inverse matrices and give the Cholesky factorization of \((k,q,t)\)-symmetric generalized Fibonacci and Lucas matrices. These results generalize the \(k\)-Fibonacci matrix and its inverse and \(k\)-symmetric Fibonacci matrix in [21](\(q=t_{i}=1)\).
Definition 5. The generalized \(q\)-Fibonacci matrix \(\mathcal{F}% _{k,n}^{(t;q)}:=[f_{i,j}]_{0\leq i,j\leq n}\) is defined by \[f_{i,j}=% \begin{cases} F_{k,i-j}(t;q), & \text{if}\ i\geq j, \\ 0, & \mbox{otherwise.}% \end{cases}%\]
Example 2. The generalized \(q\)-Fibonacci matrix for \(n=k=4\) is \[\mathcal{F}_{4,4}^{(t;q)}=% \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ t_{1} & 1 & 0 & 0 & 0 \\ t_{2}+t_{1}^{2} & t_{1} & 1 & 0 & 0 \\ t_{3}+(1+q)t_{1}t_{2}+t_{1}^{3} & t_{2}+t_{1}^{2} & t_{1} & 1 & 0 \\ (t_{4}+(1+q)t_{1}t_{3}+t_{2}^{2}+t_{1}^{4} & & & & \\ +(1+q+q^{2})t_{1}^{2}t_{2}) & t_{3}+(1+q)t_{1}t_{2}+t_{1}^{3} & t_{2}+t_{1}^{2} & t_{1} & 1% \end{pmatrix}\,.\]
In particular, if we take \(q=t_{i}=1\), we obtain the equation \(\mathcal{F}% _{k,n}^{(1;1)}=\mathcal{F}(k)_{n}\).
Definition 6. The \(n\times n\) Hessenberg matrix \(\mathcal{D}_{k,n}:=[d_{i,j}]_{1\leq i,j\leq n}\) is defined by \[d_{i,j}=% \begin{cases} F_{k,i-j+1}(t;q), & \text{if}\ i\geq j, \\ 1, & \text{if }i+1=j, \\ 0, & \mbox{otherwise.}% \end{cases}%\]
Lemma 3. [28] Let \(A_{n}\) be an \(n\times n\) lower Hessenberg matrix for all \(n\geq 1\) and define \(\det (A_{0})=1\). Then, \(\det (A_{1})=a_{11}\) and for \(n\geq 2\) \[\det (A_{n})=a_{n,n}\det (A_{n-1})+\sum\limits\limits_{r=1}^{n-1}[(-1)^{n-r}a_{n,r}(\prod\limits% \limits_{j=r}^{n-1}a_{j,j+1})\det (A_{r-1})].\]
Lemma 4. Let \(n\geq 1\) be an integer. Then \(\det (\mathcal{D}_{k,n})=% \mathcal{S}_{k,n}.\)
Proof. We proceed by induction on \(n\). The result clearly holds for \(n=1\). Now, suppose that the result is true for all positive integers less than or equal to \(n\). We prove it for \((n+1)\). In fact, using Lemma 3 we have
\[\begin{aligned} \det (\mathcal{D}_{k,n+1}) &=&d_{n+1,n+1}\det (\mathcal{D}% _{k,n})+\sum\limits\limits_{i=1}^{n}\left[ (-1)^{n+1-i}d_{n+1,i}\prod\limits% \limits_{j=i}^{n}d_{j,j+1}\det (\mathcal{D}_{k,i-1})\right] \\ &=&t_{1}\det (\mathcal{D}_{k,n})+\sum\limits\limits_{i=1}^{n}\left[ (-1)^{n+1-i}F_{k,n-i+2}(t;q)\det (\mathcal{D}_{k,i-1})\right] \\ &=&t_{1}\mathcal{S}_{k,n}+\sum\limits\limits_{i=1}^{n}\left[ (-1)^{n+1-i}F_{k,n-i+2}(t;q)\mathcal{S}_{k,i-1}\right] =\mathcal{S}_{k,n+1}. \end{aligned}\] ◻
Theorem 3. Let \(\mathcal{F}_{k,n}^{(t;q)}\) be the \((n+1)\times (n+1)\) lower triangular generalized \(q\)-Fibonacci matrix. Then, we have \[(\mathcal{F}_{k,n}^{(t;q)})^{-1}=[b_{i,j}]=\left\{ \begin{array}{cc} (-1)^{i-j}\mathcal{S}_{k,i-j}, & \text{if}\ \ i-j\geq 0, \\ 0, & \text{otherwise.}% \end{array}% \right.\]
Proof. Note that it suffices to prove that \(\mathcal{F}_{k,n}^{(t;q)}(\mathcal{F}% _{k,n}^{(t;q)})^{-1}=I_{n+1}\). For \(i>j\geq 0\), we have \[\begin{aligned} \sum\limits_{k=0}^{n}f_{i,k}b_{k,j} &=&\sum\limits_{k=j}^{i}f_{i,k}b_{k,j} \\ &=&F_{k,i-j}(t;q)\mathcal{S}_{k,0}-F_{k,i-j-1}(t;q)\mathcal{S}_{k,1}+\cdots +F_{k,0}(t;q)(-1)^{i-j}\mathcal{S}_{k,i-j} \end{aligned}\] and we know \[F_{k,i-j}(t;q)=\sum\limits_{s=1}^{k}(-1)^{j+1}\mathcal{S}_{k,s}F_{k,i-j-s}(t;q)\] from Eq. (2). Therefore, we obtain \(% \sum\limits_{k=0}^{n}f_{i,k}b_{k,j}=0\) for \(i>j\geq 0.\) It is obvious that \(% \sum\limits_{k=0}^{n}f_{i,k}b_{k,j}=0\) for \(i-j<0\) and \(% \sum\limits_{k=0}^{n}f_{i,k}b_{k,j}=f_{i,i}b_{i,j}=\) \(\mathcal{S}% _{k,0}F_{k,0}(t;q)=1\) for \(i=j\) which implies that \(\mathcal{F}% _{k,n}^{(t;q)}(\mathcal{F}_{k,n}^{(t;q)})^{-1}=I_{n+1},\) as desired. ◻
In particular, if we take \(q=t_{i}=1\), we obtain the inverse matrix of the \(% k\)-Fibonacci matrix[21].
Definition 7. The generalized \(q\)-Lucas matrix \(\mathcal{L}_{k,n}^{(t;q)}:=[l_{i,j}]_{0% \leq i,j\leq n}\) is defined by \[l_{i,j}=% \begin{cases} L_{k,i-j}(t;q), & \text{if}\ i>j, \\ 1 & \text{if }i=j, \\ 0, & \mbox{otherwise.}% \end{cases}%\]
For example, \[\mathcal{L}_{3,3}^{(t;q)}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ t_{1} & 1 & 0 & 0 \\ (1+q)t_{2}+t_{1}^{2} & t_{1} & 1 & 0 \\ (1+q+q^{2})t_{3}+(1+q+q^{2})t_{1}t_{2}+t_{1}^{3} & (1+q)t_{2}+t_{1}^{2} & t_{1} & 1% \end{pmatrix}% .\]
Definition 8. The \(n\times n\) Hessenberg matrix \(\mathcal{E}_{k,n}:=[e_{i,j}]_{1\leq i,j\leq n}\) is defined by \[e_{i,j}=% \begin{cases} L_{k,i-j+1}(t;q), & \text{if}\ i\geq j, \\ 1, & \text{if}\ i+1=j, \\ 0, & \mbox{otherwise.}% \end{cases}%\]
Lemma 5. Let \(n\geq 1\) be an integer. Then \[\det (\mathcal{E}_{k,n})=\mathcal{T}_{k,n}.\]
Proof. The proof runs like in Lemma 4. ◻
Theorem 4. Let \(\mathcal{L}_{k,n}^{(t;q)}\) be the \((n+1)\times (n+1)\) lower triangular generalized \(q\)-Lucas matrix. Then, we have \[(\mathcal{L}_{k,n}^{(t;q)})^{-1}=[c_{i,j}]=\left\{ \begin{array}{ll} (-1)^{i-j}\mathcal{T}_{k,i-j}, & \text{if }i-j>0, \\ 0, & \text{otherwise.}% \end{array}% \right.\]
Proof. The proof runs like in Theorem 3. ◻
Definition 9. Let \(n\geq 1\) be an integer, \((k,q,t)\)-symmetric generalized Fibonacci matrix \({\mathcal{Q}}_{k,n}^{(t,q)}:=[m_{i,j}]_{0\leq i,j\leq n}\) is defined by \[m_{i,j}=\left\{ \begin{array}{ll} \sum\limits\limits_{l=0}^{i}F_{k,l}(t;q)F_{k,l}(t;q), & \text{if }i=j, \\ \sum\limits\limits_{l=0}^{i}F_{k,i-l}(t;q)F_{k,j-l}(t;q), & \text{if }i+1\leq j.% \end{array}% \right.\]
Theorem 5. The Cholesky factorization of \({\mathcal{Q}}% _{k,n}^{(t,q)}\) is given by \[{\mathcal{Q}}_{k,n}^{(t,q)}=\mathcal{F}_{k,n}^{(t;q)}(\mathcal{F}% _{k,n}^{(t;q)})^{T}\]
Proof. Note that it suffices to prove that \((\mathcal{F}_{k,n}^{(t;q)})^{-1}{% \mathcal{Q}}_{k,n}^{(t,q)}=(\mathcal{F}_{k,n}^{(t;q)})^{T}.\) We take \(% \mathcal{F}_{k,n}^{(t;q)})^{-1}=\left[ b_{i,j}\right]\), \({\mathcal{Q}}% _{k,n}^{(t,q)}=\left[ m_{i,j}\right]\) and \((\mathcal{F}_{k,n}^{(t;q)})^{T}=% \left[ \overline{f}_{i,j}\right]\) and obtain \(\sum\limits\limits_{s=1}^{k}b_{i,s}\)\(m_{s,j}\) for \(i,j=0,1,2,…,k.\) For \(i=j=n\), \[ \overline{f}_{i,i}=\sum\limits\limits_{s=1}^{k}b_{i,s}m_{s,i}\] \[ =(-1)^{n}\mathcal{S} _{k,n}F_{k,n}(t;q)+ (-1)^{n-1}\mathcal{S} _{k,n-1}[F_{k,1}(t;q)F_{k,n}(t;q)+F_{k,0}(t;q)F_{k,n-1}(t;q)]+\cdots \] \[ -\mathcal{S}_{k,1}[F_{k,n-1}(t;q)F_{k,n}(t;q)+\cdots +F_{k,0}(t;q)F_{k,1}(t;q)] \] \[ +\mathcal{S}_{k,0}[F_{k,n}(t;q)F_{k,n}(t;q)+\cdots +F_{k,0}(t;q)F_{k,0}(t;q)] \] \[ F_{k,n}(t;q)[(-1)^{n}\mathcal{S}_{k,n}+(-1)^{n-1}\mathcal{S}% _{k,n-1}F_{k,1}(t;q)+\cdots +\mathcal{S}_{k,0}F_{k,n}(t;q)] \] \[ +F_{k,n-1}(t;q)[(-1)^{n-1}\mathcal{S}_{k,n-1}+\cdots +\mathcal{S} _{k,0}F_{k,n-1}(t;q)]+\cdots \] \[ +F_{k,1}(t;q)[-\mathcal{S}_{k,1}+\mathcal{S}_{k,0}F_{k,1}(t;q)]+1. \] We know \((-1)^{n}\mathcal{S}_{k,n}+(-1)^{n-1}\mathcal{S}% _{k,n-1}F_{k,1}(t;q)+\cdots +\mathcal{S}_{k,0}F_{k,n}(t;q)=0\) from the definition of \(\mathcal{S}_{k,n},\) so \(\overline{f}_{i,i}=1\) for \(% i=0,1,2,…,k.\) For \(i>j,\) \[ \sum\limits\limits_{s=1}^{k}b_{i,s}m_{s,i} =(-1)^{i}\mathcal{S}_{k,i}F_{k,j}(t;q)+ (-1)^{i-1}\mathcal{S}% _{k,i-1}[F_{k,1}(t;q)F_{k,j}(t;q)+F_{k,0}(t;q)F_{k,j-1}(t;q)] \] \[ +(-1)^{i-2}\mathcal{S}_{k,i-2}[F_{k,2}(t;q)F_{k,j}(t;q)+\cdots +F_{k,0}(t;q)F_{k,j-2}(t;q)] \] \[ +\cdots +\mathcal{S}_{k,0}[F_{k,i}(t;q)F_{k,j}(t;q)+\cdots +F_{k,i-j}(t;q)F_{k,0}(t;q)] \] \[ F_{k,j}(t;q)[(-1)^{i}\mathcal{S}_{k,i}+(-1)^{i-1}\mathcal{S}% _{k,i-1}F_{k,1}(t;q)+\cdots +\mathcal{S}_{k,0}F_{k,i}(t;q)] \] \[ +F_{k,j-1}(t;q)[(-1)^{i-1}\mathcal{S}_{k,i-1}+(-1)^{i-2}\mathcal{S}% _{k,i-2}F_{k,1}(t;q)+\cdots +\mathcal{S}_{k,0}F_{k,i-1}(t;q)]+\cdots \] \[+F_{k,0}(t;q)[(-1)^{i-j}\mathcal{S}_{k,i-j}+(-1)^{i-j-1}\mathcal{S}% _{k,i-j-1}F_{k,1}(t;q)+\cdots +\mathcal{S}_{k,0}F_{k,i-j}(t;q)]. \] We know \((-1)^{n}\mathcal{S}_{k,n}+(-1)^{n-1}\mathcal{S}% _{k,n-1}F_{k,1}(t;q)+\cdots +\mathcal{S}_{k,0}F_{k,n}(t;q)=0\) from the definition of \(\mathcal{S}_{k,n},\) so \(\overline{f}_{i,i}=0\) for \(% i=0,1,2,…,k.\) Finally, for \(i<j,\) equation \(\overline{f}% _{i,i}=F_{k,j-i}(t;q)\) is shown in a similar way. ◻
Definition 10. Let \(n\geq 1\) be an integer. Then \((k,q,t)\)-symmetric generalized Lucas matrix \({\mathcal{P}}_{k,n}^{(t,q)}:=[n_{i,j}]_{0\leq i,j\leq n}\) is defined by \[n_{i,j}=\left\{ \begin{array}{ll} \sum\limits\limits_{l=0}^{i}L_{k,l}(t;q)L_{k,l}(t;q), & \text{if }i=j, \\ \sum\limits\limits_{l=0}^{i}L_{k,i-l}(t;q)L_{k,j-l}(t;q), & \text{if }i+1\leq j.% \end{array}% \right.\]
Theorem 6. The Cholesky factorization of \({\mathcal{P}}_{k,n}^{(t,q)}\) is given by \[{\mathcal{P}}_{k,n}^{(t,q)}=\mathcal{L}_{k,n}^{(t;q)}(\mathcal{L}% _{k,n}^{(t;q)})^{T}\]
Proof. The proof runs like in Theorem 5. ◻
In this section, we obtain any term of \(q\)-generalized Fibonacci and Lucas polynomials using determinants and permanents of Hessenberg matrices.
Theorem 7. Let \(n\geq 1\) be an integer\(,\) \(F_{k,n}(t;q)\) be the \(n\)th \(q\) -generalized Fibonacci polynomial and \(% _{-}U_{k,n}^{(q)}=[u_{i,j}]_{i,j=1,2,…,n}\) be the \(n\times n\) Hessenberg matrix defined as \[u_{ij}=% \begin{cases} (-1)^{i-j}\mathcal{S}_{k,i-j+1}, & \text{if}\ i-j\geq 0, \\ -1, & \text{if}\ i+1=j, \\ 0, & \text{otherwise.}% \end{cases} \label{haen}\] Then, \(\det (_{-}U_{k,n}^{(q)})=F_{k,n}(t;q). \label{dethes1}\)
Proof. We proceed by induction on \(m\). The result clearly holds for \(m=1,\) since \(% \det (_{-}U_{k,1}^{(q)})=\mathcal{S}_{k,1}=t_{1}=F_{k,1}(t;q)\). Now, suppose that the result is true for all positive integers less than or equal to \(m\). We prove it for \((m+1)\). In fact, using Lemma 3 we have \[ \det (_{-}U_{k,m+1}^{(q)}) =u_{m+1,m+1}\det (_{-}U_{k,m}^{(q)})+\sum\limits\limits_{i=1}^{m}\left[ (-1)^{m+1-i}u_{m+1,i}\prod\limits\limits_{j=i}^{m}u_{j,j+1}\det (_{-}U_{k,i-1}^{(q)})\right] \] \[=t_{1}\det (_{-}U_{k,m}^{(q)})+\sum\limits\limits_{i=1}^{m-k+1}\left[ (-1)^{m+1-i}u_{m+1,i}\prod\limits\limits_{j=i}^{m}u_{j,j+1}\det (_{-}U_{k,i-1}^{(q)})\right] \] \[+\sum\limits\limits_{i=m-k+2}^{m}\left[ (-1)^{m+1-i}u_{m+1,i}\prod\limits% \limits_{j=i}^{m}u_{j,j+1}\det (_{-}U_{k,i-1}^{(q)})\right] \] \[=t_{1}\det (_{-}U_{k,m}^{(q)})+\sum\limits\limits_{i=m-k+2}^{m}\left[ u_{m+1,i}\det (_{-}U_{k,i-1}^{(q)})\right] \] \[=t_{1}\det (_{-}U_{k,m}^{(q)})-\mathcal{S}_{k,2}\det (_{-}U_{k,m-1}^{(q)})+\cdots +(-1)^{k-1}\mathcal{S}_{k}\det (_{-}U_{k,m-k+1}^{(q)}). \] From the hypothesis of induction and Eq. (3), we obtain \[\det (_{-}U_{k,m+1}^{(q)})=\sum\limits_{j=1}^{k}(-1)^{j-1}\mathcal{S}% _{k,j}F_{k,m+1-j}(t;q)=F_{k,m+1}(t;q).\] Therefore, \(\det (_{-}U_{k,n}^{(q)})=F_{k,n}(t;q)\) holds for all positive integers \(n\). ◻
Theorem 8. Let \(n\geq 1\) be an integer\(,\) \(F_{k,n}(t;q)\) be the \(n\)th \(q\) -generalized Fibonacci polynomial and \(% _{+}V_{k,n}^{(q)}=[v_{s,t}]_{i,j=1,2,…,n}\) be the \(n\times n\) Hessenberg matrix defined as \[v_{s,t}=% \begin{cases} i^{3(s-t)}\mathcal{S}_{k,i-j+1}, & \text{if}\ s-t\geq 0, \\ i, & \text{if}\ s+1=t, \\ 0, & \text{otherwise.}% \end{cases}%\] Then, \(\det (_{+}V_{k,n}^{(q)})=F_{k,n}(t;q).\)
Proof. Since the proof is similar to the proof of Theorem 7 using Lemma 3, we omit the details. ◻
Theorem 9. Let \(n\geq 1\) be an integer\(,\) \(L_{k,n}(t;q)\) be the \(n\) th \(q\)-generalized Lucas polynomial, \(% _{-}W_{k,n}^{(q)}=[w_{i,j}]_{i,j=1,2,…,n}\) and \(% _{+}Y_{k,n}^{(q)}=[y_{i,j}]_{i,j=1,2,…,n}\) be the \(n\times n\) Hessenberg matrices defined as \[w_{i,j}=% \begin{cases} (-1)^{i-j}\mathcal{T}_{k,i-j+1}, & \text{if}\ i-j\geq 0, \\ -1, & \text{if}\ i+1=j, \\ 0, & \text{otherwise},% \end{cases}% \quad\text{ and }\quad y_{s,t}=% \begin{cases} i^{3(s-t)}\mathcal{T}_{k,i-j+1}, & \text{if}\ s-t\geq 0, \\ i, & \text{if}\ s+1=t, \\ 0, & \text{otherwise}.% \end{cases}%\] Then, \(\det (_{-}W_{k,n}^{(q)})=\det (_{+}Y_{k,n}^{(q)})=L_{k,n}(t;q).\)
Proof. Since the proof is similar to the proof of Theorem 7 using Lemma 3, we omit the details. ◻
The permanent of an \(n\)-square matrix is defined by per\(% A=\sum\limits_{\sigma \in S_{n}}\prod\limits_{i=1}^{n}a_{i\sigma (i)},\) where the summation extends over all permutations \(\sigma\) of the symmetric group \(% S_{n}\) (cf. [38]). There is a relation between permanent and determinant of a Hessenberg matrix (cf. [33, 39]). Then, it is clear that the following corollaries hold.
Corollary 7. Let \(n\geq 1\) be an integer\(,\) \(F_{k,n}(t;q)\) be the \(n\)th \(q\) -generalized Fibonacci polynomial, \(_{+}U_{k,n}^{(q)}=[\overline{u}% _{i,j}]_{i,j=1,2,…,n}\) and \(_{-}V_{k,n}^{(q)}=[\overline{v}% _{s,t}]_{s,t=1,2,…,n}\) be the \(n\times n\) Hessenberg matrix defined as \[\overline{u}_{ij}=% \begin{cases} (-1)^{i-j}\mathcal{S}_{k,i-j+1}, & \text{if}\ i-j\geq 0, \\ 1, & \text{if}\ i+1=j, \\ 0, & \text{otherwise}% \end{cases}% \quad \text{ and }\quad \overline{v}_{s,t}=% \begin{cases} i^{3(s-t)}\mathcal{S}_{k,i-j+1}, & \text{if}\ s-t\geq 0, \\ -i, & \text{if}\ s+1=t, \\ 0, & \text{otherwise},% \end{cases}%\] where \(i=\sqrt{-1}\). Then, \(\text{per}(_{+}U_{k,n}^{(q)})=\text{per}% (_{-}V_{k,n}^{(q)})=F_{k,n}(t;q).\)
Corollary 8. Let \(n\geq 1\) be an integer\(,\) \(L_{k,n}(t;q)\) be the \(n\)th \(q\)-generalized Lucas polynomial, \(_{+}W_{k,n}^{(q)}=[\overline{w}_{i,j}]_{i,j=1,2,…,n}\) and \(_{-}Y_{k,n}^{(q)}=[\overline{y}_{s,t}]_{i,j=1,2,…,n}\) be the \(n\times n\) Hessenberg matrices defined as \[\overline{w}_{i,j}=% \begin{cases} (-1)^{i-j}\mathcal{T}_{k,i-j+1}, & \text{if}\ i-j\geq 0, \\ 1, & \text{if}\ i+1=j, \\ 0, & \text{otherwise}% \end{cases}% \quad \text{ and }\quad \overline{y}_{s,t}=% \begin{cases} i^{3(s-t)}\mathcal{T}_{k,i-j+1}, & \text{if}\ s-t\geq 0, \\ -i, & \text{if}\ s+1=t, \\ 0, & \text{otherwise}.% \end{cases}%\] Then, \(\text{per}(_{+}W_{k,n}^{(q)})=\text{per}% (_{-}Y_{k,n}^{(q)})=L_{k,n}(t;q).\)
The author declares that they have no conflicts of interest.