1 Introduction↩︎

Let \(\mathbb{F}_q\) be the finite field with \(q\) elements, where \(q\) is a prime power. An \([n,k,d]_{q^2}\) linear code \(\mathcal{C}\) is a \(k\)-dimensional subspace of \(\mathbb{F}_{q^2}^n\) with minimum distance \(d\). The Hermitian inner product of two vectors \(\mathbf{a} = (a_1,\dots,a_n)\) and \(\mathbf{b} = (b_1,\dots,b_n)\) over \(\mathbb{F}_{q^2}^n\) is defined by \(\langle \mathbf{a},\mathbf{b} \rangle_\mathrm{H} = \sum\limits_{i=1}^n a_i b_i^q\). The Hermitian dual of \(\mathcal{C}\) is \[\mathcal{C}^{\perp_\mathrm{H}} = \{\mathbf{x} \in \mathbb{F}_{q^2}^n: \langle \mathbf{x},\mathbf{c} \rangle_\mathrm{H} = 0,\text{ for all } \mathbf{c} \in \mathcal{C}\}.\]

For a linear code \(\mathcal{C}\), the hull \(\mathrm{Hull}(\mathcal{C})\) is defined as the intersection of \(\mathcal{C}\) and its dual code. It is well-known that the value of \(\dim(\mathrm{Hull}(\mathcal{C}))\) plays a critical role in determining the computational complexity of algorithms to check the permutation equivalence of two linear codes[1], computing the automorphism group of a linear code[1], 6?, calculating the number of shared pairs that required to construct an entanglement-assisted quantum error-correcting code (in short, EAQECC)[2]. And so, it is very important to determine \(\dim(\operatorname{Hull}(\mathcal{C}))\)[1], [1], [1], [1], [1], [1], [3], [4], 1?, 2?, 3?, 4?, 5?, 2?.

In recent years, the construction of non-GRS type linear codes has attracted considerable attention due to that they can effectively resist the Sidelnikov-Shestakov attack and the Wi-eschebrink attack. So far, there are extensive study on the properties and constructions of non-GRS codesnon-GRS?, liang?, 1?, non-GRS?, 5?, non-GRS?, liang?, 2?, non-GRS?, liang?, 3?, non-GRS?, liang?, 4?, non-GRS?, 6?, non-GRS?, 9?, non-GRS?, 10?, non-GRS?, 12?, non-GRS?, 13?, non-GRS?, 14?, non-GRS?, 15?. In particular, in 2017, Beelen et al. [3] firstly introduced the twisted generalized Reed-Solomon (in short, TGRS) code. Subsequently, many scholars studied the TGRS code, including the NMDS properties, self-dual properties, self-orthogonal properties, and so on[1], [1], [3], [3], [3], [3], [3], [3], [3], [3], [3], [3], 3?, property?, 2?, 8?, property?, 4?, property?, 5?, property?, 6?, property?, 7?, property?, 8?, property?, 9?, property?, 10?. In 2025, Zhao et al.[5] generalized the definition of the TGRS code to be the arbitrary twisted generalized Reed-Solomon (in short, A-TGRS) code. And then they constructed several classes of Hermitian self-dual A-TGRS codes[5], 2?. Recently, Hu et al.L?, P-TGRS? generalized TGRS codes to be the most general form, namely, \((\mathcal{L},\mathcal{P})\)-twisted generalized Reed-Solomon (in short, \((\mathcal{L},\mathcal{P})\)-TGRS) codes, and presented an in-depth and comprehensive investigation. So far, there are many study focusing on some special \((\mathcal{L},\mathcal{P})\)-TGRS codes[6], L?, P-TGRS?, L?, P?, liang?, 1?, L?, P?, liang?, 2?, L?, P?, 3?, L?, P?, 4?.

To date, there are numerous works focused on Euclidean hulls of \((\mathcal{L},\mathcal{P})\)-TGRS codes[1], [1], [1], [1], [1], [1], [1], [1], [1], [1], [3], [3], [3], [3], [3], [3], [3], [3], [3], [3], [7], 1?, 2?, 3?, 4?, 5?, 7?, 8?, 9?, 10?, 11?. However, there exist only a few results on Hermitian hulls of \((\mathcal{L},\mathcal{P})\)-TGRS codes, as listed below.

• In 2021, Wu et al.[1], [3], 10? constructively proved that there exist \((\mathcal{L},\mathcal{P})\)-TGRS codes \(\mathcal{C}(\mathcal{L},\mathcal{P},\boldsymbol{B}_1)\) with zero-dimensional Hermitian hull, where \[\boldsymbol{B}_1 = \begin{pmatrix} \boldsymbol{0}_{h-1\times t-1}&\boldsymbol{0}_{h-1\times 1}&\boldsymbol{0}_{h-1\times n-k-t}\\ \boldsymbol{0}_{1\times t-1} &d_{h,t}& \boldsymbol{0}_{1\times n-k-t} \\ \boldsymbol{0}_{k-h\times t-1}&\boldsymbol{0}_{k-h\times 1}&\boldsymbol{0}_{k-h\times n-k-t} \end{pmatrix}_{k\times(n-k)} \left(0 \le h \le k-1, 0 \le t \le n-k-1 \right).\]

• In 2022, Lin Sok [7] constructively proved that there exist \((\mathcal{L},\mathcal{P})\)-TGRS codes \(\mathcal{C}(\mathcal{L},\mathcal{P},\boldsymbol{B}_2)\) with arbitrary Hermitian hull dimension, where \[\boldsymbol{B}_2 = \begin{pmatrix} \boldsymbol{0}_{k-1\times 1}&\boldsymbol{0}_{k-1\times n-k-1}\\ b_{k-1,0} &\boldsymbol{0}_{1\times n-k-1} \\ \end{pmatrix}_{k\times(n-k)}.\]

• In 2022, Luo et al. [8] constructively proved that there exist \((\mathcal{L},\mathcal{P})\)-TGRS codes \(\mathcal{C}(\mathcal{L},\mathcal{P},\boldsymbol{B}_3)\) with \(\mathrm{dim}(\mathrm{Hull}_{H}(\mathcal{C}_k))=k\), where \[\boldsymbol{B}_3 = \begin{pmatrix} d_{0,0}&\boldsymbol{0}_{1\times n-k-1}\\ \boldsymbol{0}_{k-1\times 1} &\boldsymbol{0}_{k-1\times n-k-1} \\ \end{pmatrix}_{k\times(n-k)}.\]

• In 2025, Gao et al. H?, Gao?, 1? determined the Hermitian hull dimension for the \((\mathcal{L},\mathcal{P})\)-TGRS code \(\mathcal{C}(\mathcal{L},\mathcal{P},\boldsymbol{B}_4)\) with \[\boldsymbol{B}_4 = \begin{pmatrix} \boldsymbol{0}_{k-1\times 1}&\boldsymbol{0}_{k-1\times n-k-1}\\ 1 &\boldsymbol{0}_{1\times n-k-1} \\ \end{pmatrix}_{k\times(n-k)}.\]

• In 2026, for a special class of \((\mathcal{L},\mathcal{P})\)-TGRS codes \(\mathcal{C}(\mathcal{L},\mathcal{P},\boldsymbol{B}_5)\), Lao et al. [9] gave a upper bound and a lower bound for \(\mathrm{dim}\left(\mathrm{Hull}_{H}\left(\mathcal{C}(\mathcal{L},\mathcal{P},\boldsymbol{B}_5)\right)\right)\) and some sufficient conditions for that the code \(\mathcal{C}(\mathcal{L},\mathcal{P},\boldsymbol{B}_5)\) has a given Hermitian hull dimensions. Here, \(\mathcal{P} \subseteq \{h_1,h_2,\ldots,h_\ell\}\subseteq \{0,1,\ldots,k-1\}\), \(\mathcal{L}\subseteq\{t_1,t_2,\ldots,t_\ell\}\subseteq \{1,2,\ldots,n-k\}\) with \(h_1,h_2,\ldots,h_\ell\) are distinct and \(t_1,t_2,\ldots,t_\ell\) are distinct, and for each integer \(s\) with \(1\le s \le \ell\), \[b_{ij}=\begin{cases} \eta_s, &\text{if }i=h_s\text{ and }j=t_s-1;\\ 0,&\text{otherwise}. \end{cases}\]

Motivated by the above works, in this paper, we focus on a class of \((\mathcal{L},\mathcal{P})\)-TGRS codes \(\mathcal{C}(\mathcal{L},\mathcal{P},\boldsymbol{B})\) with \[\boldsymbol{B} = \begin{pmatrix} \boldsymbol{0}_{k-2\times 1}&\boldsymbol{0}_{k-2\times 1}&\boldsymbol{0}_{k-2\times n-k-2}\\ b_{k-2,0}&b_{k-2,1} &\boldsymbol{0}_{1\times n-k-2} \\ b_{k-1,0}&b_{k-1,1} &\boldsymbol{0}_{1\times n-k-2} \\ \end{pmatrix}_{k\times(n-k)},\] where \(\left\{b_{k-2,0}, b_{k-2,1}, b_{k-1,0},b_{k-1,1}\right\} \subseteq \mathbb{F}_{q^2}\). By taking a special class of the vector \(\boldsymbol{\alpha}\), we divide three cases to completely determine the corresponding Hermitian Hull dimensions. The paper is organized as follows. In Section 2, we give the definition of the \((\mathcal{L},\mathcal{P})\)-TGRS code and some necessary lemmas. In Section 3, we determine the Hermitian hull dimension for a class of \((\mathcal{L},\mathcal{P})\)-TGRS codes, and then obtain two classes of EAQECCs. In Section 4, we give some corresponding examples. In Section 5, we conclude the whole paper.

2 Preliminaries↩︎

Throughout this paper, for convenience, we fix some notations as follows.

• \(q=p^m\) and \(j=\frac{q-3}{2}\), where \(p\) is an odd prime and \(m\) is a positive integer.

• \(\mathbb{F}_{q^2}\) is the finite field with \(q^2\) elements and \(\mathbb{F}_{q^2}^{*}=\mathbb{F}_{q^2}\setminus \{0\}=\left< \gamma \right>\).

• \(\mathbb{Z}\) denotes the set of all integers.

• For any matrix \(\boldsymbol{G}\), \(\boldsymbol{G}^\dagger\) denotes the transpose of \(\boldsymbol{G}\) with respect to the Hermitian inner product.

• For the linear code \(\mathcal{C}\), \(\dim(\mathrm{Hull}_{H}(\mathcal{C}))\) denotes the Hermitian hull dimension of \(\mathcal{C}\).

• For any integer \(r\) with \(1\le r\le q\), \(c_{r}=\left( q-1 \right) \beta ^{r\left( q-1 \right) }\).

• For any integers \(r\) and \(i\) with \(2\le i\le q\), \(1\le r\le q\), \(e_{r}=\left( q-1 \right) \sum\limits_{t=0}^{i-1}{\gamma ^{rt\left( q-1 \right)}}\).

• \(\varGamma=\Delta\left[1+\left(\Delta\cdot\gamma^{(i-1)}\right)^{q-1}\right]\left(1+\gamma ^{i\left( q-1 \right)}\right)\left(1+\gamma ^{2\left( q-1 \right)}\right)-2\left(\frac{b_{k-2,1}}{b_{k-2,0}}\right)^{q+1}\gamma^{i(q-1)}\cdot\left(1+\gamma ^{\left( q-1 \right)}\right)\), where

  \(\Delta=b_{k-1,1}-\frac{b_{k-2,1}b_{k-1,0}}{b_{k-2,0}}\).

• \(\varGamma_{1}=b_{k-1,1}\left(\gamma^{i(q-1)}+1\right)\left(1+\gamma^{q-1}\right)\left(1+b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}\right)-b_{k-1,0}^{q+1}\left(\gamma^{(i-1)(q-1)}-1\right)\left(\gamma^{(i+1)(q-1)}-1\right)\).

• \(\varGamma_{2}=2b_{k-1,1}\left(1+\gamma^{1-q}\right)\left(1+b_{k-1,1}^{q-1}\gamma^{(q-1)(i-1)}\right)+b_{k-1,0}^{q+1}\left(1+\gamma^{i(q-1)}\right)\left(1+\gamma^{2(1-q)}\right)\).

In this section, we recall the definition of the \((\mathcal{L},\mathcal{P})\)-twisted generalized Reed-Solomon code, and give some necessary lemmas.

The definition of the \((\mathcal{L},\mathcal{P})\)-twisted generalized Reed-Solomon code is given in the following

In this paper, we consider a special class of \((\mathcal{L},\mathcal{P})\)-TRS codes with \[\boldsymbol{B} = \begin{pmatrix} 0&0&0&\cdots&0\\ \vdots&\vdots&\vdots& &\vdots\\ 0&0&0&\cdots&0\\ b_{k-2,0} & b_{k-2,1}&0&\cdots&0 \\ b_{k-1,0} & b_{k-1,1}&0&\cdots&0 \\ \end{pmatrix}_{k\times(n-k)},\] where \(\left\{b_{k-2,0}, b_{k-2,1}, b_{k-1,0},b_{k-1,1}\right\} \subseteq \mathbb{F}_{q^2}\), and briefly denote it as \(\mathcal{C}_{k}(\boldsymbol{\alpha})\).

The following Lemma 1 is crucial for calculating the matrix \(\boldsymbol{GG}^\dagger\).

For an \([n,k,d]_{q^2}\) linear code \(\mathcal{C}\), the following Lemmas 2-3 provide a method for calculating \(\dim(\mathrm{Hull}_H(\mathcal{C}))\) and constructing EAQECCs.

The following Lemmas 4-18 are crucial for proving our main results. And the proofs of Lemmas 8-18 are given in the Appendix 6.

Proof. For any \(2\le i\le q\) and \(1 \le r \le q\), it is easy to know that \(\gamma^{r(q-1)}-1\in \mathbb{F}_{q^2}^{*}\), then \(e_r=0\) if and only if \(\left(\gamma^{r(q-1)}-1\right)\left( q-1 \right)\sum\limits_{t=0}^{i-1}{ \gamma ^{rt\left( q-1 \right)}}=0\), i.e., \((q-1)\left(\gamma^{ri(q-1)}-1\right)=0\). Now by \(p\nmid q-1\). we know that \(e_{r} = 0\) if and only if \(\gamma^{ri(q-1)}=1\), i.e., \(ri \equiv 0 \pmod{q+1}\).

Proof. By Lemma 4 and \(\gcd\left( i,q+1 \right) = h\), we know that \(e_r = 0\) if and only if \(r\equiv 0\left(\bmod \frac{q+1}{h}\right)\), i.e., there exists some \(k_r\in \mathbb{Z}\) such that \(r=\frac{k_r(q+1)}{h}\). Namely, \(e_r=0\) if and only if \(k_r=\dfrac{rh}{q+1}\in\mathbb{Z}\).

Proof. (1) From \(2\le i\le q\), we have \(h=\gcd\left( i,q+1 \right)\le q<q+1\), then \(\frac{h}{q+1}\notin \mathbb{Z}\). Note that \(k_{r+1}=\frac{(r+1)h}{q+1}=\frac{h}{q+1}+k_r\), thus \(k_r\) and \(k_{r+1}\) are not integers simultaneously. Furthermore, by Lemma 5, \(e_r\) and \(e_{r+1}\) are not zeros simultaneously.

(2) Note that \(k_{q+1-r}=\frac{(q+1-r)h}{q+1}=1-k_r\), and so, \(k_r\) and \(k_{q+1-r}\) are both integers or not simultaneously. Furthermore, by Lemma 5, we know that \(e_r\) and \(e_{q+1-r}\) are both zeros or not simultaneously.

Proof. By \(j=\frac{q-3}{2}\), we have \(j+2 =\frac{q+1}{2}\), then \(\gamma^{(j+2)(q-1)} = \gamma^{\frac{q^2-1}{2}} = -1\). Furthermore, \[e_{j+2}=\left( q-1 \right) \sum\limits_{t=0}^{i-1}{\gamma ^{(j+2)t\left( q-1 \right)}}= \left( q-1 \right) \sum\limits_{t=0}^{i-1}{(-1)^t}= \begin{array}{c} \begin{cases} 0, & \text{if } 2\mid i;\\ q-1,& \text{if } 2\nmid i. \\ \end{cases} \end{array}\]

By taking \(b_{k-1,0}=0\) in Lemma 14, it is easy to obtain the following

By taking \(b_{k-1,0}=0\) or \(b_{k-1,1}=0\) in Lemma 15, it is easy to obtain the following Corollaries 2-3.

3 The Hermitian Hull dimensions of a class of \((\mathcal{L},\mathcal{P})\)-TRS codes↩︎

Throughout this section, we fix \(q\ge7\), \(\mathbb{F}_{q^2}^{*}=\left< \gamma \right>\), \(2\le i\le q\) with \(i\equiv 0\pmod p\) or \(i^2\equiv 1\pmod p\), \(j=\frac{q-3}{2}\), \(k=q+j\) and \[\boldsymbol{\alpha}=\left(\alpha_1,\alpha_2,\ldots \alpha_{q-1},\alpha_1\gamma,\alpha_2\gamma,\ldots,\alpha_{q-1}\gamma,\ldots \alpha_1\gamma^{i-1},\alpha_2\gamma^{i-1},\ldots ,\alpha_{q-1}\gamma^{i-1}\right),\] where \(\alpha_s=\gamma^{s(q+1)}\) \(\left(1\le s\le q-1\right)\).

In this section, by taking a special class of the vector \(\boldsymbol{\alpha}\), we completely determine the corresponding \(\mathrm{dim}\left(\mathrm{Hull}_H(\mathcal{C}_{q+j}(\boldsymbol{\alpha}))\right)\).

3.1 Main results↩︎

In this subsection, we present the value of \(\mathrm{dim}\left(\mathrm{Hull}_H(\mathcal{C}_{q+j}(\boldsymbol{\alpha}))\right)\) by \(3\) cases, and then obtain two classes of EAQECCs.

Then by combining with Lemma 3 and Theorems 4-6, we can immediately obtain two classes of EAQECCs as follows.

3.2 Three Crucial Propositions↩︎

In this subsection, we present three important propositions and their proofs.

Proof. By Remark 1, we have \[\boldsymbol{G}=\left(\boldsymbol{G}_1:\boldsymbol{G}_{\gamma}:\cdots:\boldsymbol{G}_{\gamma^{i-1}}\right),\] where \[\label{M3461} \boldsymbol{G}_{\beta}=\left( \begin{matrix} 1& \cdots& 1\\ \alpha _1\beta& \cdots& \alpha _{q-1}\beta\\ \vdots& & \vdots\\ \left( \alpha _1\beta \right) ^{k-3}& \cdots& \left( \alpha _{q-1}\beta \right) ^{k-3}\\ \left( \alpha _1\beta \right) ^{k-2}+\sum\limits_{\ell=0}^1{b_{k-2,\ell}\left( \alpha _1\beta \right) ^{k+\ell}}& \cdots& \left( \alpha _{q-1}\beta \right) ^{k-2}+\sum\limits_{\ell=0}^1{b_{k-2,\ell}\left( \alpha _{q-1}\beta \right) ^{k+\ell}}\\ \left( \alpha _1\beta \right) ^{k-1}+\sum\limits_{\ell=0}^1{b_{k-1,\ell}\left( \alpha _1\beta \right) ^{k+\ell}}& \cdots& \left( \alpha _{q-1}\beta \right) ^{k-1}+\sum\limits_{\ell=0}^1{b_{k-1,\ell}\left( \alpha _{q-1}\beta \right) ^{k+\ell}}\\ \end{matrix} \right)\tag{1}\] with \(\beta \in \left\{1,\gamma,\gamma^2,\ldots ,\gamma^{i-1}\right\}\). Hence, \[\label{M3462} \boldsymbol{G}\boldsymbol{G}^\dagger=\left(\boldsymbol{G}_1:\boldsymbol{G}_{\gamma}:\cdots:\boldsymbol{G}_{\gamma^{i-1}}\right)\left( \begin{array}{c} \boldsymbol{G}_1^\dagger\\ \boldsymbol{G}_{\gamma}^\dagger\\ \vdots\\ \boldsymbol{G}_{\gamma ^{i-1}}^\dagger\\ \end{array} \right) =\sum\limits_{t=0}^{i-1}\boldsymbol{G}_{\gamma^t}\boldsymbol{G}_{\gamma^t}^\dagger.\tag{2}\]

Now for any \(\beta \in \left\{1,\gamma,\gamma^2,\ldots ,\gamma^{i-1}\right\}\), by directly calculating, we have \[\label{M3463} \boldsymbol{G}_{\beta}\boldsymbol{G}_{\beta}^\dagger=\left(a_{uv}\right)_{k\times k},\tag{3}\] where \[\label{a32uv} a_{uv}=\begin{array}{c} \begin{cases} \sum\limits_{s=1}^{q-1}\left(\alpha_s\beta\right)^{(u-1)+q(v-1)}, & \text{if } u,v\in \left\{1, 2, \ldots , k-2 \right\};\\ \sum\limits_{s=1}^{q-1}\left[\left( \alpha _s\beta \right) ^{k-2+\left( v-1 \right) q}+b_{k-2,0}\left( \alpha _s\beta \right) ^{k+\left( v-1 \right) q}+b_{k-2,1}\left( \alpha _s\beta \right) ^{k+1+\left( v-1 \right) q}\right],& \text{if } u=k-1,v\in \left\{1, 2, \ldots , k-2 \right\};\\ \sum\limits_{s=1}^{q-1}\left[\left( \alpha _s\beta \right) ^{k-1+\left( v-1 \right) q}+b_{k-1,0}\left( \alpha _s\beta \right) ^{k+\left( v-1 \right) q}+b_{k-1,1}\left( \alpha _s\beta \right) ^{k+1+\left( v-1 \right) q}\right], & \text{if } u=k, v\in \left\{1, 2, \ldots , k-2 \right\};\\ \sum\limits_{s=1}^{q-1}\left[\left( \alpha _s\beta \right) ^{q\left( k-2 \right) +u-1}+b_{k-2,0}^{q}\left( \alpha _s\beta \right) ^{qk+u-1}+b_{k-2,1}^{q}\left( \alpha _s\beta \right) ^{q\left( k+1 \right) +u-1}\right], & \text{if } v=k-1,u\in \left\{1, 2, \ldots , k-2 \right\};\\ \sum\limits_{s=1}^{q-1}\left[\left( \alpha _s\beta \right) ^{q\left( k-1 \right) +u-1}+b_{k-1,0}^{q}\left( \alpha _s\beta \right) ^{qk+u-1}+b_{k-1,1}^{q}\left( \alpha _s\beta \right) ^{q\left( k+1 \right) +u-1}\right], & \text{if } v=k-1,u\in \left\{1, 2, \ldots , k-2 \right\};\\ \sum\limits_{s=1}^{q-1}A_{11}A_{12},&\text{if } u=v=k-1;\\ \sum\limits_{s=1}^{q-1}A_{11}A_{22},&\text{if } u=k-1,v=k;\\ \sum\limits_{s=1}^{q-1}A_{21}A_{12},&\text{if } u=k,v=k-1;\\ \sum\limits_{s=1}^{q-1}A_{21}A_{22},&\text{if } u=v=k, \end{cases} \end{array}\tag{4}\] with \(A_{11}=\left( \alpha _s\beta \right) ^{k-2}+b_{k-2,0}\left( \alpha _s\beta \right) ^k+b_{k-2,1}\left( \alpha _s\beta \right) ^{k+1}\), \(A_{12}=\left( \alpha _s\beta \right) ^{q\left( k-2 \right)}+b_{k-2,0}^{q}\left( \alpha _s\beta \right) ^{qk}+b_{k-2,1}^{q}\left( \alpha _s\beta \right) ^{q\left( k+1 \right)}\),
\(A_{21}=\left( \alpha _s\beta \right) ^{k-1}+b_{k-1,0}\left( \alpha _s\beta \right) ^k+b_{k-1,1}\left( \alpha _s\beta \right) ^{k+1}\) and \(A_{22}=\left( \alpha _s\beta \right) ^{q\left( k-1 \right)}+b_{k-1,0}^{q}\left( \alpha _s\beta \right) ^{qk}+b_{k-1,1}^{q}\left( \alpha _s\beta \right) ^{q\left( k+1 \right)}.\)

Next, according to the ranges of \(u\) and \(v\), we divide the following \(6\) cases to determine the values of \(a_{uv}\). By combining with Remark 2 and the range of \(u+v-2\), we deduce that only the following \(a_{uv}\) are non-zero, and all other entries are zero. The detailed proofs are given in the Appendix 7.

case 1 For \(u,v\in \{1,2,\dots,k-2\}\), we have \(a_{11}=a_{qq}=q-1\) and \[\begin{align} &a_{u(q-u+1)}=(q-1)\beta^{(q-1)(q-u+1)}=c_{q-u+1},\quad(1\le u \le q);\\ &a_{u(2q-u)}=(q-1)\beta^{(q-1)(q-u)}=c_{q-u},\quad(j+5\le u\le q-1);\\ &a_{u(2q-u)}=(q-1)\beta^{(q-1)(2q-u+1)}=c_{2q-u+1},\quad(q+1\le u \le k-2).\\ \end{align}\]

case 2 For \(u=k-1, v\in \left\{1, 2, \ldots , k-2 \right\}\), we have \(a_{( k-1 ) (j+1)}= b_{k-2,1}(q-1)\beta^{(q-1)(j+2)}=b_{k-2,1}c_{j+2}\) and \[a_{( k-1 )(j+2)}= b_{k-2,0}(q-1)\beta^{(q-1)(j+3)}=b_{k-2,0}c_{j+3}, a_{\left( k-1 \right) (j+4)}= (q-1)\beta^{(q-1)(j+5)}=c_{j+5}.\]

case 3 For \(u=k, v\in \left\{1, 2, \ldots , k-2 \right\}\), we have \(a_{k (j+1)}=b_{k-1,1}(q-1)\beta^{(q-1)(j+2)}=b_{k-1,1}c_{j+2}\) and \[a_{k (j+2)}=b_{k-1,0}(q-1)\beta^{(q-1)(j+3)}=b_{k-1,0}c_{j+3}, a_{k (j+3)}=(q-1)\beta^{(q-1)(j+4)}=c_{j+4}.\]

case 4 For \(u\in {1, 2, \ldots , k-2}, v=k-1\), we have \(a_{(j+1)\left( k-1 \right) }= b_{k-2,1}^q(q-1)\beta^{(q-1)(j+2)}=b_{k-2,1}^qc_{j+2}\) and \[a_{(j+2)\left( k-1 \right) }= b_{k-2,0}^q(q-1)\beta^{(q-1)(j+1)}=b_{k-2,0}^qc_{j+1}, a_{(j+4)\left( k-1 \right) }= (q-1)\beta^{(q-1)(j-1)}=c_{j-1}.\]

case 5 For \(u\in \left\{1, 2, \ldots , k-2 \right\}, v=k\), we have \(a_{(j+1) k}= b_{k-1,1}^q(q-1)\beta^{(q-1)(j+2)}=b_{k-1,1}^qc_{j+2}\) and \[a_{(j+2) k}= b_{k-1,0}^q(q-1)\beta^{(q-1)(j+1)}=b_{k-1,0}^qc_{j+1}, a_{(j+3)k}=(q-1)\beta^{(q-1)j}=c_{j}.\]

case 6 For \(u,v \in \{k-1,k\}\), we have \(a_{(k-1)(k-1)}=\left( q-1 \right) b_{k-2,0}^{q+1}\beta ^{(q-1)(j+2)}=b_{k-2,0}^{q+1}c_{j+2}\) and \[\begin{align} &a_{(k-1)k}=\left( q-1 \right) \left[ b_{k-2,0}b_{k-1,0}^{q}\beta ^{(q-1)(j+2)}+b_{k-2,1}\beta ^{(q-1)(j+1)} \right]= b_{k-2,0}b_{k-1,0}^{q}c_{j+2}+b_{k-2,1}c_{j+1};\\ &a_{k(k-1)}=\left( q-1 \right) \left[ b_{k-2,1}^{q}\beta ^{(q-1)(j+3)}+b_{k-2,0}^{q}b_{k-1,0}\beta ^{(q-1)(j+2)} \right]=b_{k-2,1}^{q}c_{j+3}+b_{k-2,0}^{q}b_{k-1,0}c_{j+2};\\ &a_{kk}=\left( q-1 \right) \left[ b_{k-1,1}^{q}\beta ^{(q-1)(j+3)}+b_{k-1,0}^{q+1}\beta ^{(q-1)(j+2)}+b_{k-1,1}\beta ^{(q-1)(j+1)} \right]=b_{k-1,1}^{q}c_{j+3}+b_{k-1,0}^{q+1}c_{j+2}+b_{k-1,1}c_{j+1}. \end{align}\]

Thus \[\label{GG32beta} \boldsymbol{G}_{\beta}\boldsymbol{G}_{\beta}^\dagger =\left( \begin{matrix} q-1& & & & & & & & & & c_q& & & & \\ 0& & & & & & & & & c_{q-1}& & & & & \\ \vdots& & & & & & & & \iddots & & & & & & \\ 0& & & & & & & c_{j+4}& & & & & & & \\ 0& & & & & & c_{j+3}& \cdot& \cdots& \cdot& \cdot& \cdot& \cdots& D_5& D_6\\ 0& & & & & c_{j+2}& \cdot& \cdot& \cdots& \cdot& \cdot& \cdot& \cdots& D_7& D_8\\ 0& & & & c_{j+1}& \cdot& \cdot& \cdot& \cdots& \cdot& \cdot& \cdot& \cdots& 0& c_j\\ 0& & & c_j& \cdot& \cdot& \cdot& \cdot& \cdots& \cdot& \cdot& \cdot& \cdots& c_{j-1}& \cdot\\ \vdots& & \iddots & & \vdots& \vdots& \vdots& \vdots& & & & & \iddots & \vdots& \vdots\\ 0& c_2& \cdots& \cdot& \cdot& \cdot& \cdot& \cdot& \cdots& \cdot& \cdot& c_1& \cdots& \cdot& \cdot\\ c_1& \cdot& \cdots& \cdot& \cdot& \cdot& \cdot& \cdot& \cdots& \cdot& q-1& \cdot& \cdots& \cdot& \cdot\\ 0& \cdot& \cdots& \cdot& \cdot& \cdot& \cdot& \cdot& \cdots& c_q& \cdot& \cdot& \cdots& \cdot& \cdot\\ \vdots& & & & \vdots& \vdots& \vdots& \vdots& \iddots & & & & & \vdots& \vdots\\ 0& \cdot& \cdots& \cdot& D_1& D_2& 0& c_{j+5}& \cdots& \cdot& \cdot& \cdot& \cdots& A_1& A_2\\ 0& \cdot& \cdots& \cdot& D_3& D_4& c_{j+4}& \cdot& \cdots& \cdot& \cdot& \cdot& \cdots& A_3& A_4\\ \end{matrix} \right),\tag{5}\] where \(D_1=b_{k-2,1}c_{j+2}\), \(D_2=b_{k-2,0}c_{j+3}\), \(D_3=b_{k-1,1}c_{j+2}\), \(D_4=b_{k-1,0}c_{j+3}\), \(D_5=b_{k-2,1}^{q}c_{j+2}\), \(D_6=b_{k-1,1}^{q}c_{j+2}\),
\(D_7=b_{k-2,0}^{q}c_{j+1}\), \(D_8=b_{k-1,0}^{q}c_{j+1}\), \(A_1=b_{k-2,0}^{q+1}c_{j+2}\), \(A_2=b_{k-2,0}b_{k-1,0}^{q}c_{j+2}+b_{k-2,1}c_{j+1}\), \(A_3=b_{k-2,1}^{q}c_{j+3}+b_{k-2,0}^{q}b_{k-1,0}c_{j+2}\) and \(A_4=b_{k-1,1}^{q}c_{j+3}+b_{k-1,0}^{q+1}c_{j+2}+b_{k-1,1}c_{j+1}\).

Furthermore, by substituting the Equation (5 ) into the Equation (2 ) and directly calculating, Proposition 10 is immediately.

From the following Proposition 11, to determine \(\operatorname{rank}(\boldsymbol{GG}^\dagger)\) for \(\gcd(i,q+1)=1\), it’s enough to compute \(\operatorname{rank}\left(\boldsymbol{A}_{2\times 2}\right)\).

Proof. By \(\gcd (i, q+1)=1\) and Corollary 3, we have \(e_r \neq 0\) for \(1 \le r \le q\). For convenience, we denote \(\boldsymbol{a}_{u,1}\) be the \(u\)-th row of the matrix \(\boldsymbol{GG}^{\dagger}\) given by (?? ), where \(1 \leq u\leq k\). Now, for the matrix \(\boldsymbol{GG}^{\dagger}\) given by (?? ), we perform the following \(4\) steps of elementary transformations.

Step 1. Replace \(\boldsymbol{a}_{u+q-1,1}\) with \(\boldsymbol{a}_{u+q-1,1}-\frac{e_{q+2-u}}{e_{q+1-u}}\boldsymbol{a}_{u,1}\) for each \(u=2,3, \ldots , j+1\), we obtain the matrix \(\boldsymbol{A}^{(1)}\) and denote the vector \(\boldsymbol{a}_{u,1}^{(1)}\) be the \(u\)-th row of \(\boldsymbol{A}^{(1)}\), where \(1 \leq u\leq k\);

Step 2. Replace \(\boldsymbol{a}_{k-1,1}^{(1)}\) with \(\boldsymbol{a}_{k-1,1}^{(1)}-\frac{b_{k-2,0}e_{j+3}}{e_{j+2}}\boldsymbol{a}_{j+2,1}^{(1)}-\frac{b_{k-2,1}e_{j+2}}{e_{j+1}}\boldsymbol{a}_{j+3,1}^{(1)}\), and replace \(\boldsymbol{a}_{k,1}^{(1)}\) with \(\boldsymbol{a}_{k,1}^{(1)}-\frac{b_{k-1,0}e_{j+3}}{e_{j+2}}\boldsymbol{a}_{j+2,1}^{(1)}-\frac{b_{k-1,1}e_{j+2}}{e_{j+1}}\boldsymbol{a}_{j+3,1}^{(1)}\), we obtain the matrix \(\boldsymbol{A}^{(2)}\) and denote the vector \(\boldsymbol{b}_{1,v}^{(2)}\) be the \(v\)-th column of \(\boldsymbol{A}^{(2)}\), where \(1 \leq v\leq k\);

Step 3. Replace \(\boldsymbol{b}_{1,v+q-1}^{(2)}\) with \(\boldsymbol{b}_{1,v+q-1}^{(2)}-\frac{e_{v-1}}{e_{v}}\boldsymbol{b}_{1,v}^{(2)}\) for each \(v=2,3, \ldots , j+1\), we obtain the matrix \(\boldsymbol{A}^{(3)}\) and denote the vector \(\boldsymbol{b}_{1,v}^{(3)}\) be the \(v\)-th column of \(\boldsymbol{A}^{(3)}\), where \(1 \leq v\leq k\);

Step 4. Replace \(\boldsymbol{b}_{1,k-1}^{(3)}\) with \(\boldsymbol{b}_{1,k-1}^{(3)}-\frac{b_{k-2,0}^qe_{j+1}}{e_{j+2}}\boldsymbol{b}_{1,j+2}^{(3)}-\frac{b_{k-2,1}^qe_{j+2}}{e_{j+3}}\boldsymbol{b}_{1,j+3}^{(3)}\), and replace \(\boldsymbol{b}_{1,k}^{(3)}\) with \(\boldsymbol{b}_{1,k}^{(3)}-\frac{b_{k-1,0}^qe_{j+1}}{e_{j+2}}\boldsymbol{b}_{1,j+2}^{(3)}-\frac{b_{k-1,1}^qe_{j+2}}{e_{j+3}}\boldsymbol{b}_{1,j+3}^{(3)}\), we obtain the matrix

\[\boldsymbol{A}^{(4)}=\left( \begin{matrix} i(q-1)& & & & & & & & e_q& & & \\ \vdots& & & & & & & \iddots & & & & \\ 0& & & & & & e_{j+4}& & & & & \\ 0& & & & & e_{j+3}& \cdots& \cdots& \cdots& \cdots& 0& 0\\ 0& & & & e_{j+2}& \cdots& \cdots& \cdots& \cdots& \cdots& 0& 0\\ 0& & & e_{j+1}& \vdots& & & & & & & 0\\ 0& & e_j& \vdots& \vdots& & & & & & 0& \\ \vdots& \iddots & & \vdots& \vdots& & & & & \iddots & & \\ e_1& & & \vdots& \vdots& & & & i(q-1)& & & \\ \vdots& & & \vdots& \vdots& & & \iddots& & & & & \\ 0& & & 0& 0& & 0& & & & C_1 & C_2 \\ 0& & & 0& 0& 0& & & & & C_3 & C_4 \\ \end{matrix} \right)\] with \[\begin{align} &C_1=b_{k-2,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right),\\ &C_2=b_{k-2,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right) +b_{k-2,0}b_{k-1,0}^{q}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right), \\ &C_3=b_{k-2,0}^{q}b_{k-1,0}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)+b_{k-2,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)\\ \end{align}\] and \[C_4=b_{k-1,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)+b_{k-1,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)+b_{k-1,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right).\] Now set \[\boldsymbol{A}_{2\times 2}=\left(\begin{matrix} C_1 & C_2\\ C_3 & C_4\\ \end{matrix}\right),\] and by combining with Lemma 12, we can get \[\mathrm{rank}(\boldsymbol{G}\boldsymbol{G}^\dagger)=\mathrm{rank}(\boldsymbol{A}^{(4)})=q+\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right).\]

From the following Proposition 12, to determine \(\operatorname{rank}(\boldsymbol{GG}^\dagger)\) for \(\gcd(i,q+1)=h>1\), it’s enough to compute \(\operatorname{rank}\left(\boldsymbol{A}_{7\times 7}\right)\).

Proof. Firstly, we denote the matrix \(\boldsymbol{GG}^\dagger\) given in the Equation (?? ) by \(\left(a_{uv}^{1}\right)_{k\times k}\), then by Lemma 6, we know that \(e_r\) and \(e_{r+1}\) are not zeros simultaneously, where \(1\le r\le q-1\). Furthermore, \(a_{u(q+1-u)}^{1}\) and \(a_{(q-1+u)(q+1-u)}^{1}\) are not zeros simultaneously, \(a_{(q+1-v)v}^{1}\) and \(a_{(q+1-v)(q-1+v)}^{1}\) are also not zeros simultaneously, where \(2\le u, v\le j-1\).

Secondly, for convenience, we denote the vectors \(\boldsymbol{a}_{u,2}\) and \(\boldsymbol{b}_{2,v}\) be the \(u\)-th row and the \(v\)-th column of the matrix \(\boldsymbol{GG}^\dagger\) given by the Equation (?? ), respectively, where \(1 \le u,v\le k\). Then for the matrix \(\boldsymbol{GG}^\dagger\) given by (?? ), we perform the following \(2\) steps of elementary transformations.

Step 1. For each \(u=2,3,\ldots,j-1\), if \(a_{u(q+1-u)}^{1} = 0\), then swap \(\boldsymbol{a}_{u,2}\) and \(\boldsymbol{a}_{q-1+u,2}\); if \(a_{u(q+1-u)}^{1} \neq 0\), then replace \(\boldsymbol{a}_{q-1+u,2}\) with \(\boldsymbol{a}_{q-1+u,2}-\frac{e_{q+2-u}}{e_{q+1-u}}\boldsymbol{a}_{u,2}\);

Step 2. For each \(v=2,3,\ldots,j-1\), if \(a_{(q+1-v)v}^{1} = 0\), then swap \(\boldsymbol{b}_{2,v}\) and \(\boldsymbol{b}_{2,q-1+v}\); if \(a_{(q+1-v)v}^{1} \neq 0\), then replace \(\boldsymbol{b}_{2,q-1+v}\) with \(\boldsymbol{b}_{2,q-1+v}-\frac{e_{v-1}}{e_{v}}\boldsymbol{b}_{2,v}\), and so, we obtain the matrix \(\boldsymbol{A}_{1}^{(1)}\), and denote \(\boldsymbol{A}_{1}^{(1)}=\left(a_{uv}^{(1)}\right)_{k\times k}\) and \(e_{u_{1}}^{1}=a_{q+1-u_{1},u_{1}}^{(1)}\in \mathbb{F}_{q^2}^{*}\), where \(u_{1}\in \left\{2,3,\ldots,j-1,j+5,\ldots,q-1\right\}\). Thus, \[\begin{align} &\boldsymbol{A}_{1}^{(1)}\\ &=\left( \begin{matrix} i(q-1)& & & & & & & & & & & & e_q& & & & & \\ & & & & & & & & & & & e_{q-1}^{1} & & & & & & \\ & & & & & & & & & & \iddots & & & & & & & \\ & & & & & & & & & e_{j+5}^{1}& & & & & & & & \\ & & & & & & & & e_{j+4} & & & & & & & & & \\ & & & & & & &e_{j+3}& \cdots & \cdots &\cdots &\cdots &\cdots & \cdots &\cdots &\cdots & D_5^{1}& D_6^{1}\\ & & & & & &e_{j+2}&\cdots &\cdots &\cdots &\cdots & \cdots & \cdots &\cdots &\cdots &\cdots &D_7^{1}& D_8^{1}\\ & & & & &e_{j+1}& \vdots& & & & & & & & & & & e_j\\ & & & &e_j& \vdots& \vdots& & & & & & & & & & e_{j-1}& \\ & & &e_{j-1}^{1}& &\vdots&\vdots& & & & & & & & & 0& & \\ & & \iddots & & &\vdots&\vdots& & & & & & & & \iddots& & & & \\ &e_2^{1}& & & &\vdots & \vdots & & & & & & & 0 & & & & \\ e_1& & && & \vdots & \vdots & & & & & & i(q-1) & & & & & \\ & & & & &\vdots& \vdots & & & & & 0 & & & & & & \\ & & & & &\vdots&\vdots & & & & \iddots & & & & & & & \\ & & & & &\vdots&\vdots & & & 0 & & & & & & & & \\ & & & & &D_1^{1}& D_2^{1}& &e_{j+5} & & & & & & & & B_1& B_2\\ & & & & &D_3^{1}& D_4^{1} &e_{j+4} & & & & & & & & & B_3& B_4\\ \end{matrix} \right), \end{align}\] where \(D_1^{1}=b_{k-2,1}e_{j+2}\), \(D_2^{1}=b_{k-2,0}e_{j+3}\), \(D_3^{1}=b_{k-1,1}e_{j+2}\), \(D_4^{1}=b_{k-1,0}e_{j+3}\), \(D_5^{1}=b_{k-2,1}^{q}e_{j+2}\), \(D_6^{1}=b_{k-1,1}^{q}e_{j+2}\), \(D_7^{1}=b_{k-2,0}^{q}e_{j+1}\), \(D_8^{1}=b_{k-1,0}^{q}e_{j+1}\), \(B_1=b_{k-2,0}^{q+1}e_{j+2}\), \(B_2=b_{k-2,1}e_{j+1}+b_{k-2,0}b_{k-1,0}^{q}e_{j+2}\), \(B_3=b_{k-2,0}^{q}b_{k-1,0}e_{j+2}+b_{k-2,1}^{q}e_{j+3}\) and \(B_4=b_{k-1,1}e_{j+1}+b_{k-1,0}^{q+1}e_{j+2}+b_{k-1,1}^{q}e_{j+3}\).

Next, let \[\boldsymbol{g}_i =\left(0,\dots,0,\underset{\text{the i-th component}}{1},0,\dots,0\right)^\mathrm{T}\in\mathbb{F}_{q^2}^{k},\]

\[\boldsymbol{P}=\big(\boldsymbol{g}_{1}\;\ldots\;\boldsymbol{g}_{j-1}\quad\boldsymbol{g}_{k-6}\quad \boldsymbol{g}_{k-5}\quad\boldsymbol{g}_{k-4}\quad \boldsymbol{g}_{k-3}\quad\boldsymbol{g}_{k-2}\quad \boldsymbol{g}_{j}\;\ldots\;\boldsymbol{g}_{k-7}\quad \boldsymbol{g}_{k-1}\quad\boldsymbol{g}_{k}\big),\] and \[\boldsymbol{Q}=\big(\boldsymbol{g}_{1}\;\ldots\;\boldsymbol{g}_{j-1}\quad\boldsymbol{g}_{j+5}\;\ldots\;\boldsymbol{g}_{k-2}\quad\boldsymbol{g}_{j}\quad \boldsymbol{g}_{j+1}\quad\boldsymbol{g}_{j+2}\quad \boldsymbol{g}_{j+3}\quad\boldsymbol{g}_{j+4}\quad \boldsymbol{g}_{k-1}\quad\boldsymbol{g}_{k}\big).\]

It’s easy to know that the matrices \(\boldsymbol{P}\) and \(\boldsymbol{Q}\) are both non-singular, and then by directly calculating, we have

\[\boldsymbol{P}\boldsymbol{A}_{1}^{(1)}\boldsymbol{Q}=\begin{pmatrix} \boldsymbol{B}_{q-5\times q-5} &\boldsymbol{0}_{q-5\times j-2} &\boldsymbol{0}_{q-5\times 7}\\ \boldsymbol{0}_{j-2\times q-5} &\boldsymbol{0}_{j-2\times j-2} &\boldsymbol{0}_{j-2\times 7}\\ \boldsymbol{0}_{7\times q-5} &\boldsymbol{0}_{7\times j-2} & \boldsymbol{A}_{7\times 7} \end{pmatrix},\] where \[\boldsymbol{B}_{q-5\times q-5}=\begin{pmatrix} i(q-1) & & & & & & & e_q\\ & & & & & & e_{q-1}^{1}& \\ & & & & & \iddots& & \\ & & & & e_{j+5}^{1}& & & \\ & & & e_{j-1}^{1}& & & & \\ & & \iddots& & & & & \\ & e_{2}^{1}& & & & & & \\ e_{1}& & & & & & & i(q-1)\\ \end{pmatrix}\] and \[\boldsymbol{A}_{7\times 7}=\begin{pmatrix} 0 & 0 & 0 & 0 & e_{j+4}& 0& 0 \\ 0 & 0 & 0 & e_{j+3}& 0& b_{k-2,1}^{q}e_{j+2}& b_{k-1,1}^{q}e_{j+2} \\ 0 & 0 & e_{j+2}& 0& 0& b_{k-2,0}^{q}e_{j+1}& b_{k-1,0}^{q}e_{j+1} \\ 0 & e_{j+1}& 0& 0& 0& 0 & e_j \\ e_j& 0& 0& 0& 0& e_{j-1} & 0 \\ 0 & b_{k-2,1}e_{j+2}& b_{k-2,0}e_{j+3}& 0 & e_{j+5} & B_1 & B_2 \\ 0 & b_{k-1,1}e_{j+2}& b_{k-1,0}e_{j+3}& e_{j+4}&0 & B_3 & B_4 \\ \end{pmatrix}\] with \(B_1=b_{k-2,0}^{q+1}e_{j+2}\), \(B_2=b_{k-2,1}e_{j+1}+b_{k-2,0}b_{k-1,0}^{q}e_{j+2}\), \(B_3=b_{k-2,0}^{q}b_{k-1,0}e_{j+2}+b_{k-2,1}^{q}e_{j+3}\) and
\(B_4=b_{k-1,1}e_{j+1}+b_{k-1,0}^{q+1}e_{j+2}+b_{k-1,1}^{q}e_{j+3}\).

By combining with Lemma 12, we have \(\mathrm{rank}\left(\boldsymbol{B}_{q-5\times q-5}\right)=q-5\), furthermore, \[\mathrm{rank}\left(\boldsymbol{GG}^\dagger\right)=\mathrm{rank}\left(\boldsymbol{P}\boldsymbol{A}_{1}^{(1)}\boldsymbol{Q}\right)=q-5+\mathrm{rank}\left(\boldsymbol{A}_{7\times 7}\right).\]

3.3 The proofs of Theorems 4-6↩︎

By Lemma 2.5, we only need to focus on computing \(\operatorname{rank}\left(\boldsymbol{GG}^\dagger\right)\) to determine \(\dim\left(\operatorname{Hull}_H\left(\mathcal{\mathcal{C}}\right)\right)\).

The proof of Theorem 4

By \(\gcd(i,q+1)=1\) and Proposition 11, we only need to compute \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)\). Next, depending on \(i=q\) or not, we divide the following 2 cases to determine the value of \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)\).

case 1. If \(i=q\), then by Lemma 13, we know that \(e_{j+1}^2=e_{j+2}e_{j}\), \(e_{j+2}^2=e_{j+3}e_{j+1}\) and \(e_{j+3}^2=e_{j+4}e_{j+2}\), i.e., \[e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}=e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}=e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}=0.\] Furthermore, we can get \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=0\), i.e., \(\mathrm{rank}\left(\boldsymbol{G}\boldsymbol{G}^\dagger\right)=q\).

case 2. If \(i\neq q\), then by Lemma 13, we know that \(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\), \(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\) and \(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\) are all non-zero. Next, according to the number of zero entries among \(b_{k-2,0},b_{k-2,1},b_{k-1,0},b_{k-1,1}\), we divide the following \(5\) cases to determine the value of \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)\).

case 2.1. If \(b_{k-2,0}\), \(b_{k-2,1}\), \(b_{k-1,0}\) and \(b_{k-1,1}\) are all zero, it is easy to get \(\mathrm{rank}\left(\boldsymbol{A}_{2\times2}\right)=0\), i.e., \(\mathrm{rank}\left( \boldsymbol{G}\boldsymbol{G}^\dagger \right)=q\).

case 2.2. If only one of \(b_{k-2,0},b_{k-2,1},b_{k-1,0},b_{k-1,1}\) is non-zero, then we have the following \(4\) cases.

\((1)\) If \(b_{k-2,0}\in \mathbb{F}_{q^2}^{*}\), then we have \[\boldsymbol{A}_{2\times 2}=\left(\begin{matrix} b_{k-2,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right) & 0\\ 0 & 0\\ \end{matrix}\right).\] It is easy to get \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=1\), i.e., \(\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right)=q+1\).

\((2)\) If \(b_{k-2,1}\in \mathbb{F}_{q^2}^{*}\), then we have

\[\boldsymbol{A}_{2\times 2}=\left(\begin{matrix} 0 & b_{k-2,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)\\ b_{k-2,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)& 0\\ \end{matrix}\right).\] It is easy to get \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=2\), i.e., \(\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right)=q+2\).

\((3)\) If \(b_{k-1,0}\in \mathbb{F}_{q^2}^{*}\), then we have

\[\boldsymbol{A}_{2\times 2}=\left(\begin{matrix} 0 & 0\\ 0 & b_{k-1,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)\\ \end{matrix}\right).\] It is easy to get \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=1\), i.e., \(\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right)=q+1\).

\((4)\) If \(b_{k-1,1}\in \mathbb{F}_{q^2}^{*}\), then we have

\[\boldsymbol{A}_{2\times 2}=\left(\begin{matrix} 0 & 0\\ 0 & b_{k-1,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)+b_{k-1,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)\\ \end{matrix}\right).\] Thus, \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=0\) if and only if \(b_{k-1,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)+b_{k-1,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)=0\). And by Corollary 1, we know that \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=0\) if and only if \(b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}=-1\). Thus \[\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right) = \begin{array}{c} \begin{cases} q, &\text{if } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}=-1 ;\\ q+1, &\text{if }b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}\neq -1. \end{cases} \end{array}\]

case 2.3. If two of \(b_{k-2,0},b_{k-2,1},b_{k-1,0},b_{k-1,1}\) are exactly zero, then we have the following \(6\) cases.

(1) If \(b_{k-2,0}=b_{k-2,1}=0\), then we have

\[\boldsymbol{A}_{2\times 2}=\left(\begin{matrix} 0 & 0\\ 0 & b_{k-1,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)+b_{k-1,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)+b_{k-1,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)\\ \end{matrix}\right).\] Thus, \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=0\) if and only if \(b_{k-1,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)+b_{k-1,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)+b_{k-1,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)=0\). And by Lemma 14, we know that \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=0\) if and only if \(\varGamma_{2}=0\). Thus \[\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right) = \begin{array}{c} \begin{cases} q, &\text{if } \varGamma_{2}=0;\\ q+1, &\text{if }\varGamma_{2}\neq 0. \end{cases} \end{array}\]

\((2)\) If \(b_{k-2,0}=b_{k-1,0}=0\), then we have

\[\boldsymbol{A}_{2\times 2}=\left(\begin{matrix} 0 & b_{k-2,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)\\ b_{k-2,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right) & b_{k-1,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)+b_{k-1,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)\\ \end{matrix}\right).\] And so, \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=2\), i.e., \(\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right)=q+2\).

\((3)\) If \(b_{k-2,0}=b_{k-1,1}=0\), then we have

\[\boldsymbol{A}_{2\times 2}=\left(\begin{matrix} 0 & b_{k-2,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)\\ b_{k-2,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right) & b_{k-1,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)\\ \end{matrix}\right).\] And so, \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=2\), i.e., \(\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right)=q+2\).

\((4)\) If \(b_{k-1,0}=b_{k-2,1}=0\), then we have

\[\boldsymbol{A}_{2\times 2}=\left(\begin{matrix} b_{k-2,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right) & 0\\ 0 & b_{k-1,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)+b_{k-1,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)\\ \end{matrix}\right).\] Thus, \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=1\) if and only if \(b_{k-1,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)+b_{k-1,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)=0\). Now by Corollary 1, we know that \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=1\) if and only if \(b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}=-1\). Thus \[\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right) = \begin{array}{c} \begin{cases} q+1,&\text{if } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}=-1;\\ q+2,&\text{if } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}\neq -1. \end{cases} \end{array}\]

\((5)\) If \(b_{k-2,1}=b_{k-1,1}=0\), then we have

\[\boldsymbol{A}_{2\times 2}=\left(\begin{matrix} b_{k-2,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right) &b_{k-2,0}b_{k-1,0}^{q}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)\\ b_{k-2,0}^{q}b_{k-1,0}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right) &b_{k-1,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)\\ \end{matrix}\right).\] Note that \[\frac{ b_{k-2,0}^{q}b_{k-1,0}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)}{b_{k-2,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)}=\frac{b_{k-1,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)}{b_{k-2,0}b_{k-1,0}^{q}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)}=\frac{b_{k-1,0}}{b_{k-2,0}},\] and so, \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=1\), i.e., \(\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right)=q+1\).

\((6)\) If \(b_{k-1,0}=b_{k-1,1}=0\), then we have \[\boldsymbol{A}_{2\times 2}=\left(\begin{matrix} b_{k-2,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right) & b_{k-2,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)\\ b_{k-2,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right) & 0\\ \end{matrix}\right).\] And so, \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=2\), i.e., \(\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right)=q+2\).

case 2.4. If only one of \(b_{k-2,0},b_{k-2,1},b_{k-1,0},b_{k-1,1}\) is zero, then we have the following \(4\) cases.

\((1)\) If \(b_{k-2,0} = 0\), then we have

\[\boldsymbol{A}_{2\times 2}=\left(\begin{matrix} 0 & b_{k-2,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)\\ b_{k-2,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right) & b_{k-1,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)+b_{k-1,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)+b_{k-1,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)\\ \end{matrix}\right).\] And so, \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=2\), i.e., \(\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right)=q+2\).

\((2)\) If \(b_{k-2,1} = 0\), then we have

\[\label{M34617} \boldsymbol{A}_{2\times 2}=\left(\begin{matrix} b_{k-2,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)& b_{k-2,0}b_{k-1,0}^{q}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)\\ b_{k-2,0}^{q}b_{k-1,0}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right) & b_{k-1,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)+b_{k-1,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)+b_{k-1,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)\\ \end{matrix}\right).\tag{6}\]

Next, for convenience, we denote \(\boldsymbol{a}_{u_1,3}\) be the \(u_1\)-th row of the matrix \(\boldsymbol{A}_{2\times 2}\) given by the Equation (6 ), where \(u_1=1,2\). Then for the matrix \(\boldsymbol{A}_{2\times 2}\) given by (6 ), we perform the following \(2\) steps of elementary transformations.

Step 1. Replace \(\boldsymbol{a}_{2,3}\) with \(\boldsymbol{a}_{2,3}-\frac{b_{k-1,0}}{b_{k-2,0}}\boldsymbol{a}_{1,3}\), we obtain the matrix \(\boldsymbol{A}_{2\times 2}^{(1)}\) and denote the vector \(\boldsymbol{b}_{3,v_1}^{(1)}\) be the \(v_1\)-th column of \(\boldsymbol{A}_{2\times 2}^{(1)}\), where \(v_1=1,2\);

Step 2. Replace \(\boldsymbol{b}_{3,2}^{(1)}\) with \(\boldsymbol{b}_{3,2}^{(1)}-\left(\frac{b_{k-1,0}}{b_{k-2,0}}\right)^q\boldsymbol{b}_{3,1}^{(1)}\), we obtain the matrix \[\boldsymbol{A}_{2\times 2}^1=\left(\begin{matrix} b_{k-2,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)& 0\\ 0 & b_{k-1,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)+b_{k-1,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)\\ \end{matrix}\right).\] Hence, \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=\mathrm{rank}(\boldsymbol{A}_{2\times 2}^1)=1\) if and only if \[b_{k-1,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)+b_{k-1,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)=0 .\] Now by Corollary 1, we know that \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=1\) if and only if \(b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}=-1\). Thus \[\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right) =\begin{array}{c} \begin{cases} q+1,&\text{if } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}=-1;\\ q+2,& \text{if } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}\neq -1. \end{cases} \end{array}\]

\((3)\) If \(b_{k-1,0} = 0\), then we have

\[\boldsymbol{A}_{2\times 2}=\left(\begin{matrix} b_{k-2,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)& b_{k-2,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)\\ b_{k-2,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right) & b_{k-1,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)+b_{k-1,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)\\ \end{matrix}\right).\] Hence, \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=1\) if and only if \[\frac{b_{k-2,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)}{b_{k-2,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)}=\frac{b_{k-1,1}}{b_{k-2,1}}+\frac{b_{k-1,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)}{b_{k-2,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)} .\] Now by Corollary 2, we know that \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=1\) if and only if \[b_{k-1,1}\left(1+b_{k-1,1}^{q-1}\cdot\gamma^{(i-1)(q-1)} \right)\left(1+\gamma ^{i\left( q-1 \right)}\right)\left(1+\gamma ^{2\left( q-1 \right)}\right) =2\left(\frac{b_{k-2,1}}{b_{k-2,0}}\right)^{q+1}\gamma^{i(q-1)}\cdot\left(1+\gamma ^{\left( q-1 \right)}\right).\] Thus \[\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right) =\begin{array}{c} \begin{cases} q+1, &\text{if } b_{k-1,1}\left(1+b_{k-1,1}^{q-1}\cdot\gamma^{(i-1)(q-1)} \right)\left(1+\gamma ^{i\left( q-1 \right)}\right)\left(1+\gamma ^{2\left( q-1 \right)}\right)\\ &\quad=2\left(\frac{b_{k-2,1}}{b_{k-2,0}}\right)^{q+1}\gamma^{i(q-1)}\cdot\left(1+\gamma ^{\left( q-1 \right)}\right);\\ q+2,&\text{if } b_{k-1,1}\left(1+b_{k-1,1}^{q-1}\cdot\gamma^{(i-1)(q-1)} \right)\left(1+\gamma ^{i\left( q-1 \right)}\right)\left(1+\gamma ^{2\left( q-1 \right)}\right)\\ &\quad\neq 2\left(\frac{b_{k-2,1}}{b_{k-2,0}}\right)^{q+1}\gamma^{i(q-1)}\cdot\left(1+\gamma ^{\left( q-1 \right)}\right). \end{cases} \end{array}\]

\((4)\) If \(b_{k-1,1} = 0\), then we have

\[\label{M34618} \begin{align} &\boldsymbol{A}_{2\times 2}\\ &=\left(\begin{matrix} b_{k-2,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)& b_{k-2,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right) +b_{k-2,0}b_{k-1,0}^{q}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)\\ b_{k-2,0}^{q}b_{k-1,0}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)+b_{k-2,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right) & b_{k-1,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)\\ \end{matrix}\right). \end{align}\tag{7}\] Next, for convenience, we denote \(\boldsymbol{b}_{4,v_1}\) be the \(v_1\)-th column of the matrix \(\boldsymbol{A}_{2\times 2}\) given by the Equation (7 ), where \(v_1=1,2,\) then replace \(\boldsymbol{b}_{4,2}\) with \(\boldsymbol{b}_{4,2}-\left(\frac{b_{k-1,0}}{b_{k-2,0}}\right)^q\boldsymbol{b}_{4,1}\), we obtain the matrix \[\boldsymbol{A}_{2\times 2}^2=\left(\begin{matrix} b_{k-2,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)& b_{k-2,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right) \\ b_{k-2,0}^{q}b_{k-1,0}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)+b_{k-2,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right) & -\left(\frac{b _{10}b_{k-2,1}}{b_{k-2,0}}\right)^q\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)\\ \end{matrix}\right).\] Hence, \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=\mathrm{rank}(\boldsymbol{A}_{2\times 2}^2)=1\) if and only if

\[\frac{b_{k-2,1}^q\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)}{b_{k-2,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)}+\frac{b_{k-1,0}}{b_{k-2,0}}=\frac{-b_{k-1,0}^qb_{k-2,1}^q\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)}{b_{k-2,0}^qb_{k-2,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right) }.\] Now by Corollary 3, we know that \(\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right)=1\) if and only if

\[\Delta_{1}\left[1+\left(\Delta_{1}\cdot\gamma^{(i-1)}\right)^{q-1} \right]\left(1+\gamma ^{i\left( q-1 \right)}\right)\left(1+\gamma ^{2\left( q-1 \right)}\right)=2\left(\frac{b_{k-2,1}}{b_{k-2,0}}\right)^{q+1}\gamma^{i(q-1)}\cdot\left(1+\gamma ^{\left( q-1 \right)}\right),\] where \(\Delta_{1}=-\frac{b_{k-2,1}b_{k-1,0}}{b_{k-2,0}}\). Thus

\[\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right) =\begin{array}{c} \begin{cases} q+1, &\text{if } \Delta_{1}\left[1+\left(\Delta_{1}\cdot\gamma^{(i-1)}\right)^{q-1} \right]\left(1+\gamma ^{i\left( q-1 \right)}\right)\left(1+\gamma ^{2\left( q-1 \right)}\right)\\ &\quad=2\left(\frac{b_{k-2,1}}{b_{k-2,0}}\right)^{q+1}\gamma^{i(q-1)}\cdot\left(1+\gamma ^{\left( q-1 \right)}\right);\\ q+2,&\text{if } \Delta_{1}\left[1+\left(\Delta_{1}\cdot\gamma^{(i-1)}\right)^{q-1} \right]\left(1+\gamma ^{i\left( q-1 \right)}\right)\left(1+\gamma ^{2\left( q-1 \right)}\right)\\ &\quad\neq 2\left(\frac{b_{k-2,1}}{b_{k-2,0}}\right)^{q+1}\gamma^{i(q-1)}\cdot\left(1+\gamma ^{\left( q-1 \right)}\right), \end{cases} \end{array}\] where \(\Delta_{1}=-\frac{b_{k-2,1}b_{k-1,0}}{b_{k-2,0}}\).

case 2.5. If \(b_{k-2,0}\), \(b_{k-2,1}\), \(b_{k-1,0}\) and \(b_{k-1,1}\) are all non-zero, then we have \[\label{M34619} \boldsymbol{A}_{2\times 2}=\left(\begin{matrix} C_1 & C_2\\ C_3 & C_4\\ \end{matrix}\right),\tag{8}\] where \[\begin{align} &C_1=b_{k-2,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right),\\ &C_2=b_{k-2,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right) +b_{k-2,0}b_{k-1,0}^{q}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right), \\ &C_3=b_{k-2,0}^{q}b_{k-1,0}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)+b_{k-2,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)\\ \end{align}\] and \[C_4=b_{k-1,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)+b_{k-1,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)+b_{k-1,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right).\]

Next, for convenience, we denote \(\boldsymbol{a}_{u_1,5}\) be the \(u_1\)-th row of the matrix \(\boldsymbol{A}_{2\times 2}\) given by the Equation (8 ), where \(u_1=1,2\). Then for the matrix \(\boldsymbol{A}_{2\times 2}\) given by (8 ), we perform the following \(2\) steps of elementary transformations.

Step 1. Replace \(\boldsymbol{a}_{2,5}\) with \(\boldsymbol{a}_{2,5}-\frac{b_{k-1,0}}{b_{k-2,0}}\boldsymbol{a}_{1,5}\), we obtain the matrix \(\boldsymbol{A}_{2\times 2}^{(2)}\) and denote the vector \(\boldsymbol{b}_{5,v_1}^{(2)}\) be the \(v_1\)-th column of \(\boldsymbol{A}_{2\times 2}^{(2)}\), where \(v_1=1,2\);

Step 2. Replace \(\boldsymbol{b}_{5,2}^{(2)}\) with \(\boldsymbol{b}_{5,2}^{(2)}-\left(\frac{b_{k-1,0}}{b_{k-2,0}}\right)^q\boldsymbol{b}_{5,1}^{(2)}\), we obtain the matrix \[\boldsymbol{A}_{2\times 2}^{3}=\left(\begin{matrix} b_{k-2,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right) & b_{k-2,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)\\ b_{k-2,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right) & C_4^{1}\\ \end{matrix}\right),\] where \[C_4^{1}=\left(b_{k-1,1}-\frac{b_{k-2,1}b_{k-1,0}}{b_{k-2,0}}\right)\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)+\left(b_{k-1,1}^{q}-\frac{b_{k-1,0}^qb_{k-2,1}^q}{b_{k-2,0}^q}\right)\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right).\] Hence, \(\mathrm{rank}(A_{2\times 2})=\mathrm{rank}(A_{2\times 2}^3)=1\) if and only if

\[\frac{b_{k-1,1}}{b_{k-2,1}}-\frac{b_{k-1,0}}{b_{k-2,0}}+\frac{\left(b_{k-1,1}^{q}-\frac{b_{k-1,0}^qb_{k-2,1}^q}{b_{k-2,0}^q}\right)\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)}{b_{k-2,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)}=\frac{b_{k-2,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)}{b_{k-2,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)}.\] And by Lemma 15, we know that \(\mathrm{rank}(A_{2\times 2})=1\) if and only if \(\varGamma=0\). Thus \[\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right) =\begin{array}{c} \begin{cases} q+1,& \text{if } \varGamma=0; \\ q+2,&\text{if } \varGamma\neq 0. \end{cases} \end{array}\]

Now by combining with the discussions of the above \(4\) cases, we conclude that \[\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right) =\begin{array}{c} \begin{cases} q, & \text{if } i=q\\ &\quad \text{or } i\neq q ,b_{k-2,1}=b_{k-2,0}=0 , b_{k-1,1}\neq 0 \text{ and }\\ &\quad 2b_{k-1,1}\left(1+\gamma^{1-q}\right)\left(1+b_{k-1,1}^{q-1}\gamma^{(q-1)(i-1)}\right)=-b_{k-1,0}^{q+1}\left(1+\gamma^{i(q-1)}\right)\left(1+\gamma^{2(1-q)}\right);\\ q+2,&\text{if } i\neq q \text{ and } b_{k-2,0}=0, b_{k-2,1}\neq 0\\ &\quad \text{or } i\neq q \text{ and } b_{k-1,0}=b_{k-1,1}=0, b_{k-2,0}, b_{k-2,1}\neq 0\\ &\quad \text{or } i\neq q, b_{k-2,1}=0, b_{k-2,0},b_{k-1,1}\neq 0 \text{ and } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}\neq -1\\ &\quad \text{or } i\neq q , b_{k-2,1},b_{k-2,0}\neq 0,\left(b_{k-1,0},b_{k-1,1}\right)\neq (0,0) \text{ and } \varGamma\neq 0;\\ q+1, &\text{otherwise}. \end{cases} \end{array}\]

And so, by Lemma 2, we complete the proof of Theorem 4.

The proofs of Theorems 5-6

By \(\gcd(i,q+1)=h>1\) and Proposition 12, we only need to compute \(\mathrm{rank}\left(\boldsymbol{A}_{7\times 7}\right)\).

The proof of Theorem 5

By \(2\mid i\) and Lemma 7, we have \(e_{j+2}=0\). Then by Lemma 6, we have \(e_{j+1}\in \mathbb{F}_{q^2}^{*}\) and \(e_{j+3}\in \mathbb{F}_{q^2}^{*}\). Thus,

\[\label{M34621} \boldsymbol{A}_{7\times 7}=\left(\begin{matrix} & & & & e_{j+4}& 0 & 0 \\ & & & e_{j+3}& & 0& 0 \\ & & 0& & & b_{k-2,0}^{q}e_{j+1}& b_{k-1,0}^{q}e_{j+1} \\ & e_{j+1}& & & & 0 & e_j \\ e_j& & & & & e_{j-1} & 0 \\ 0 & 0& b_{k-2,0}e_{j+3}& 0 & e_{j+5} & 0 & b_{k-2,1}e_{j+1} \\ 0 & 0& b_{k-1,0}e_{j+3}& e_{j+4}&0 & b_{k-2,1}^{q}e_{j+3} & b_{k-1,1}e_{j+1}+b_{k-1,1}^{q}e_{j+3} \\ \end{matrix}\right)\;.\tag{9}\] Next, depending on \(i=\frac{q+1}{2}\) or not, we divide the following 2 cases to determine the value of \(\mathrm{rank}\left(\boldsymbol{A}_{7\times 7}\right)\).

case 1. If \(i=\frac{q+1}{2}\), then by Lemma 8 and Lemma 6, we have \(e_{j+4}=e_{j}=0\), \(e_{j-1}\in \mathbb{F}_{q^2}^{*}\) and \(e_{j+5}\in \mathbb{F}_{q^2}^{*}\). Thus,

\[\label{A11} \boldsymbol{A}_{7\times 7}=\left(\begin{matrix} & & & & 0& 0 & 0 \\ & & & e_{j+3}& & 0& 0 \\ & & 0& & & b_{k-2,0}^{q}e_{j+1}& b_{k-1,0}^{q}e_{j+1} \\ & e_{j+1}& & & & 0 & 0\\ 0& & & & & e_{j-1} & 0 \\ 0 & 0& b_{k-2,0}e_{j+3}& 0 & e_{j+5} & 0& b_{k-2,1}e_{j+1} \\ 0 & 0& b_{k-1,0}e_{j+3}& 0&0 & b_{k-2,1}^{q}e_{j+3} & b_{k-1,1}e_{j+1}+b_{k-1,1}^{q}e_{j+3}\\ \end{matrix}\right)\;.\tag{10}\] Now, for convenience, we denote the vector \(\boldsymbol{a}_{u_2,6}\) be the \(u_2\)-th row of the matrix \(\boldsymbol{A}_{7\times 7}\) given by the Equation (10 ), where \(1 \le u_2\le 7\). Next, for the matrix \(\boldsymbol{A}_{7\times 7}\) given by (10 ), we perform the following \(2\) steps of elementary transformations.

Step 1. Replace \(\boldsymbol{a}_{3,6}\) with \(\boldsymbol{a}_{3,6}-\frac{b_{k-2,0}^{q}e_{j+1}}{e_{j-1}}\boldsymbol{a}_{5,6}\), and replace \(\boldsymbol{a}_{7,6}\) with \(\boldsymbol{a}_{7,6}-\frac{b_{k-2,1}^{q}e_{j+3}}{e_{j-1}}\boldsymbol{a}_{5,6}\), we obtain the matrix \(\boldsymbol{A}_{7\times 7}^{(1)}\) and denote the vector \(\boldsymbol{b}_{6,v_{2}}^{(1)}\) be the \(v_{2}\)-th column of \(\boldsymbol{A}_{7\times 7}^{(1)}\), where \(1\le v_{2}\le 7\);

Step 2. Replace \(\boldsymbol{b}_{6,3}^{(1)}\) with \(\boldsymbol{b}_{6,3}^{(1)}-\frac{b_{k-2,0}e_{j+3}}{e_{j+5}}\boldsymbol{b}_{6,5}^{(1)}\), and replace \(\boldsymbol{b}_{6,7}^{(1)}\) with \(\boldsymbol{b}_{6,7}-\frac{b_{k-2,1}e_{j+1}}{e_{j+5}}\boldsymbol{b}_{6,5}^{(1)}\), we obtain the matrix \[\boldsymbol{A}_{7\times 7}^1=\left(\begin{matrix} & & & & 0& 0 & 0 \\ & & & e_{j+3}& & 0& 0 \\ & & 0& & & 0& b_{k-1,0}^{q}e_{j+1} \\ & e_{j+1}& & & & 0 & 0\\ 0& & & & & e_{j-1} & 0 \\ 0 & 0& 0& 0 & e_{j+5} & 0& 0 \\ 0 & 0& b_{k-1,0}e_{j+3}& 0&0 & 0 & b_{k-1,1}e_{j+1}+b_{k-1,1}^{q}e_{j+3}\\ \end{matrix}\right)\;,\] and \(\mathrm{rank}\left(\boldsymbol{A}_{7\times 7}\right)=\mathrm{rank}\left(\boldsymbol{A}_{7\times 7}^{1}\right)\). Next, we divide the following \(2\) cases basing on \(b_{k-1,0}=0\) or not.

(1) If \(b_{k-1,0}=0\), then we have

\[\boldsymbol{A}_{7\times 7}^{1}=\left(\begin{matrix} & & & & 0& 0 & 0 \\ & & & e_{j+3}& & 0& 0 \\ & & 0& & & 0& 0 \\ & e_{j+1}& & & & 0 & 0\\ 0& & & & & e_{j-1} & 0 \\ 0 & 0& 0& 0 & e_{j+5} & 0& 0 \\ 0 & 0& 0& 0&0 & 0 & b_{k-1,1}e_{j+1}+b_{k-1,1}^{q}e_{j+3}\\ \end{matrix}\right)\;.\] Thus, \(\mathrm{rank}\left(\boldsymbol{A}_{7\times 7}\right)=\mathrm{rank}\left(\boldsymbol{A}_{7\times 7}^{1}\right)=4\) if and only if \(b_{k-1,1}e_{j+1}+b_{k-1,1}^qe_{j+3}=0\). And by Lemma 16, we know that \(\mathrm{rank}\left(\boldsymbol{A}_{7\times 7}\right)=4\) if and only if \(b_{k-1,1}=0\) or \(b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}=1\). Thus

\[\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right)= \begin{array}{c} \begin{cases} q-1, &\text{if } b_{k-1,1}=0 \text{ or } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}=1;\\ q,& \text{if } b_{k-1,1}\neq 0 \text{ and } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}\neq 1.\\ \end{cases} \end{array}\]

(2) If \(b_{k-1,0}\neq 0\), then \(\mathrm{rank}(\boldsymbol{A}_{7\times 7}^{1})=6\), i.e. \(\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right)=q+1\).

And so, by combining with (1) and (2), we know that for \(i=\frac{q+1}{2}\),

\[\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right)= \begin{array}{c} \begin{cases} q-1, &\text{if } b_{k-1,0}=b_{k-1,1}= 0\\ &\quad \text{or } b_{k-1,0}= 0 \text{ and } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}=1;\\ q,&\text{if } b_{k-1,0}=0 , b_{k-1,1}\neq 0 \text{ and } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}\neq 1;\\ q+1,& \text{if } b_{k-1,0}\neq 0. \end{cases} \end{array}\]

case 2. If \(i\neq\frac{q+1}{2}\), then by Lemma 8, we have \(e_{j}\in \mathbb{F}_{q^2}^{*}\) and \(e_{j+4}\in \mathbb{F}_{q^2}^{*}\).

For convenience, we denote the vectors \(\boldsymbol{a}_{u_2,7}\) and \(\boldsymbol{b}_{7,v_2}\) be the \(u_2\)-th row and the \(v_2\)-th column of the matrix \(\boldsymbol{A}_{7\times 7}\) given by the Equation (9 ), respectively, where \(1 \le u_2,v_2\le 7\). Next, for the matrix \(\boldsymbol{A}_{7\times 7}\) given by (9 ), we perform the following \(2\) steps of elementary transformations.

Step 1. Replace \(\boldsymbol{a}_{6,7}\) with \(\boldsymbol{a}_{6,7}-\frac{e_{j+5}}{e_{j+4}}\boldsymbol{a}_{1,7}\), and replace \(\boldsymbol{a}_{7,7}\) with \(\boldsymbol{a}_{7,7}-\frac{e_{j+4}}{e_{j+3}}\boldsymbol{a}_{2,7}\);

Step 2. Replace \(\boldsymbol{b}_{7,6}\) with \(\boldsymbol{b}_{7,6}-\frac{e_{j-1}}{e_{j}}\boldsymbol{b}_{7,1}\), and replace \(\boldsymbol{a}_{7,7}\) with \(\boldsymbol{b}_{7,7}-\frac{e_{j}}{e_{j+1}}\boldsymbol{b}_{7,2}\), we obtain the matrix \[\label{A2} \boldsymbol{A}_{7\times 7}^2=\left(\begin{matrix} & & & & e_{j+4}& 0 & 0 \\ & & & e_{j+3}& & 0& 0 \\ & & 0& & & b_{k-2,0}^{q}e_{j+1}& b_{k-1,0}^{q}e_{j+1} \\ & e_{j+1}& & & & 0 & 0\\ e_j& & & & & 0 & 0 \\ 0 & 0& b_{k-2,0}e_{j+3}& 0 & 0 & 0 & b_{k-2,1}e_{j+1} \\ 0 & 0& b_{k-1,0}e_{j+3}& 0&0 & b_{k-2,1}^{q}e_{j+3} &b_{k-1,1}e_{j+1}+b_{k-1,1}^{q}e_{j+3} \\ \end{matrix}\right)\;,\tag{11}\] and \(\mathrm{rank}\left(\boldsymbol{A}_{7\times 7}\right)=\mathrm{rank}\left(\boldsymbol{A}_{7\times 7}^{2}\right)\).

Next, we divide the following \(5\) cases basing on the values of \(b_{k-2,0}\), \(b_{k-2,1}\), \(b_{k-1,0}\) and \(b_{k-1,1}\) are zero or not.

(1) If \(b_{k-2,0}=b_{k-1,0}=b_{k-2,1}=0\), then we have \[\boldsymbol{A}_{7\times 7}^{2}=\left(\begin{matrix} & & & & e_{j+4}& 0 & 0 \\ & & & e_{j+3}& & 0& 0 \\ & & 0& & & 0& 0 \\ & e_{j+1}& & & & 0 & 0\\ e_j& & & & & 0 & 0 \\ 0 & 0& 0& 0 & 0 & 0 & 0 \\ 0 & 0& 0& 0&0 & 0 &b_{k-1,1}e_{j+1}+b_{k-1,1}^{q}e_{j+3} \\ \end{matrix}\right)\;.\] And so, \(\mathrm{rank}\left(\boldsymbol{A}_{7\times 7}\right)=\mathrm{rank}\left(\boldsymbol{A}_{7\times 7}^{2}\right)=4\) if and only if \(b_{k-1,1}e_{j+1}+b_{k-1,1}^qe_{j+3}=0\). Furthermore, by Lemma 16, we know that \(\mathrm{rank}\left(\boldsymbol{A}_{7\times 7}\right)=4\) if and only if \(b_{k-1,1}=0\) or \(b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}=1\). Thus

\[\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right) =\begin{array}{c} \begin{cases} q-1 , &\text{if } b_{k-1,1}=0 \text{ or } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}=1;\\ q ,& \text{if } b_{k-1,1}\neq 0 \text{ and } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}\neq 1. \end{cases} \end{array}\]

(2) If \(b_{k-2,0}=b_{k-1,0}=0\) and \(b_{k-2,1}\neq 0\), then we have \[\boldsymbol{A}_{7\times 7}^{2}=\left(\begin{matrix} & & & & e_{j+4}& 0 & 0 \\ & & & e_{j+3}& & 0& 0 \\ & & 0& & & 0& 0 \\ & e_{j+1}& & & & 0 & 0\\ e_j& & & & & 0 & 0 \\ 0 & 0& 0& 0 & 0 & 0 & b_{k-2,1}e_{j+1} \\ 0 & 0& 0& 0&0 & b_{k-2,1}^{q}e_{j+3} &b_{k-1,1}e_{j+1}+b_{k-1,1}^{q}e_{j+3} \\ \end{matrix}\right)\;.\] And so, \(\mathrm{rank}(\boldsymbol{A}_{7\times 7}^{2})=6\), i.e., \(\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right)=q+1\).

(3) If \(b_{k-2,0}=b_{k-2,1}=0\) and \(b_{k-1,0}\neq 0\), then we have \[\boldsymbol{A}_{7\times 7}^{2}=\left(\begin{matrix} & & & & e_{j+4}& 0 & 0 \\ & & & e_{j+3}& & 0& 0 \\ & & 0& & & 0& b_{k-1,0}^{q}e_{j+1} \\ & e_{j+1}& & & & 0 & 0\\ e_j& & & & & 0 & 0 \\ 0 & 0& 0& 0 & 0 & 0 & 0 \\ 0 & 0& b_{k-1,0}e_{j+3}& 0&0 & 0 &b_{k-1,1}e_{j+1}+b_{k-1,1}^{q}e_{j+3} \\ \end{matrix}\right)\;.\] And so, \(\mathrm{rank}(\boldsymbol{A}_{7\times 7}^{2})=6\), i.e., \(\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right)=q+1\).

(4) If \(b_{k-2,0}=0\), \(b_{k-2,1}\neq 0\) and \(b_{k-1,0}\neq 0\), then we have \[\boldsymbol{A}_{7\times 7}^{2}=\left(\begin{matrix} & & & & e_{j+4}& 0 & 0 \\ & & & e_{j+3}& & 0& 0 \\ & & 0& & & 0& b_{k-1,0}^{q}e_{j+1} \\ & e_{j+1}& & & & 0 & 0\\ e_j& & & & & 0 & 0 \\ 0 & 0& 0& 0 & 0 & 0 & b_{k-2,1}e_{j+1} \\ 0 & 0& b_{k-1,0}e_{j+3}& 0&0 & b_{k-2,1}^{q}e_{j+3} &b_{k-1,1}e_{j+1}+b_{k-1,1}^{q}e_{j+3} \\ \end{matrix}\right)\;.\] And so, \(\mathrm{rank}(\boldsymbol{A}_{7\times 7}^{2})=6\), i.e., \(\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right)=q+1\).

(5) If \(b_{k-2,0}\neq 0\), then for convenience, we denote the vector \(\boldsymbol{a}_{u_2,8}\) be the \(u\)-th row of the matrix \(\boldsymbol{A}_{7\times 7}^{2}\) given by the Equation (11 ), where \(1 \le u_2\le 7\). Next, for the matrix \(\boldsymbol{A}_{7\times 7}^{2}\) given by (11 ), we perform the following \(2\) steps of elementary transformations.

Step 1. Replace \(\boldsymbol{a}_{7,8}\) with \(\boldsymbol{a}_{7,8}-\frac{b_{k-1,0}}{b_{k-2,0}}\boldsymbol{a}_{6,8}\), and replace \(\boldsymbol{a}_{7,8}\) with \(\boldsymbol{a}_{7,8}-\left(\frac{b_{k-2,1}}{b_{k-2,0}}\right)^q\boldsymbol{a}_{3,8}\), we obtain the matrix \(\boldsymbol{A}_2^{(1)}\), and denote the vector \(\boldsymbol{b}_{8,v_2}^{(1)}\) be the \(v_2\)-th column of \(\boldsymbol{A}_2^{(1)}\), where \(1 \le v_2\le 7\).

Step 2. Replace \(\boldsymbol{b}_{8,7}^{(1)}\) with \(\boldsymbol{b}_{8,7}^{(1)}-\frac{b_{k-2,1}}{b_{k-2,0}}\boldsymbol{b}_{8,3}^{(1)}-\left(\frac{b_{k-1,0}}{b_{k-2,0}}\right)^q\boldsymbol{b}_{8,6}^{(1)}\), we obtain the matrix \[\boldsymbol{A}_{2}^{(2)}=\left(\begin{matrix} & & & & e_{j+4}& 0 & 0 \\ & & & e_{j+3}& & 0& 0 \\ & & 0& & & b_{k-2,0}^{q}e_{j+1}& 0 \\ & e_{j+1}& & & & 0 & 0\\ e_j& & & & & 0 & 0 \\ 0 & 0& b_{k-2,0}e_{j+3}& 0 & 0 & 0 & 0 \\ 0 & 0& 0& 0&0 & 0 &\left(b_{k-1,1}-\frac{b_{k-1,0}b_{k-2,1}}{b_{k-2,0}}\right)e_{j+1}+\left(b_{k-1,1}^q-\frac{b_{k-1,0}^qb_{k-2,1}^q}{b_{k-2,0}^q}\right)e_{j+3} \\ \end{matrix}\right)\;.\] Hence, \(\mathrm{rank}\left(\boldsymbol{A}_{7\times 7}^{2}\right) =\mathrm{rank}\left(\boldsymbol{A}_{2}^{(2)}\right)=6\) if and only if \[\left(b_{k-1,1}-\frac{b_{k-1,0}b_{k-2,1}}{b_{k-2,0}}\right)e_{j+1}+\left(b_{k-1,1}^q-\frac{b_{k-1,0}^qb_{k-2,1}^q}{b_{k-2,0}^q}\right)e_{j+3}=0.\] Now by Lemma 17, we know that \(\mathrm{rank}\left(\boldsymbol{A}_{7\times 7}^{2}\right)=\mathrm{rank}\left(\boldsymbol{A}_{2}^{(2)}\right)=6\) if and only if \(b_{k-2,1}b_{k-1,0}=b_{k-1,1}b_{k-2,0}\) or \[\left(b_{k-1,1}-\frac{b_{k-2,1}b_{k-1,0}}{b_{k-2,0}}\right)^{q-1}\gamma^{(i-1)(q-1)}=1.\] Thus \[\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right) =\begin{array}{c} \begin{cases} q+1 , &\text{if } b_{k-2,1}b_{k-1,0}=b_{k-1,1}b_{k-2,0} \text{ or } \left(b_{k-1,1}-\frac{b_{k-2,1}b_{k-1,0}}{b_{k-2,0}}\right)^{q-1}\gamma^{(i-1)(q-1)}=1;\\ q +2, & \text{if } b_{k-2,1}b_{k-1,0}\neq b_{k-1,1}b_{k-2,0} \text{ and } \left(b_{k-1,1}-\frac{b_{k-2,1}b_{k-1,0}}{b_{k-2,0}}\right)^{q-1}\gamma^{(i-1)(q-1)}\neq 1. \end{cases} \end{array}\] And so, for \(i\neq\frac{q+1}{2}\), we have

\[\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right) =\begin{array}{c} \begin{cases} q-1, &\text{if } b_{k-2,0}=b_{k-2,1}=b_{k-1,0}=b_{k-1,1}=0\\ &\quad \text{or } b_{k-2,0}=b_{k-2,1}=b_{k-1,0}=0 \text{ and } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}=1;\\ q, & \text{if } b_{k-2,0}=b_{k-2,1}=b_{k-1,0}=0, b_{k-1,1}\neq 0 \text{ and } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}\neq 1;\\ q+2, &\text{if } b_{k-2,0}\neq 0 , b_{k-2,1}b_{k-1,0}\neq b_{k-1,1}b_{k-2,0} \text{ and } \left(b_{k-1,1}-\frac{b_{k-2,1}b_{k-1,0}}{b_{k-2,0}}\right)^{q-1}\gamma^{(i-1)(q-1)}\neq 1;\\ q+1,& \text{otherwise}. \end{cases} \end{array}\]

Now by combining with the discussions of the above \(2\) cases, we conclude that

\[\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right) = \begin{array}{c} \begin{cases} q-1, & \text{if } i=\frac{q+1}{2}, b_{k-1,0}=b_{k-1,1}=0 \\ &\quad \text{or } i=\frac{q+1}{2}, b_{k-1,0}= 0 \text{ and } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}=1\\ &\quad \text{or } i\neq \frac{q+1}{2}, b_{k-2,0}=b_{k-2,1}=b_{k-1,0}=b_{k-1,1}=0\\ &\quad \text{or } i\neq \frac{q+1}{2} , b_{k-2,0}=b_{k-2,1}=b_{k-1,0}=0 \text{ and } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}=1;\\ q, & \text{if } i=\frac{q+1}{2}, b_{k-1,0}=0 , b_{k-1,1}\neq 0 \text{ and } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}\neq 1\\ &\quad \text{or } i\neq \frac{q+1}{2} , b_{k-2,0}=b_{k-2,1}=b_{k-1,0}=0 , b_{k-1,1}\neq 0 \text{ and } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}\neq 1;\\ q+2, &\text{if } i\neq \frac{q+1}{2} , b_{k-2,0}\neq 0, b_{k-2,1}b_{k-1,0}\neq b_{k-1,1}b_{k-2,0} \text{ and } \left(b_{k-1,1}-\frac{b_{k-2,1}b_{k-1,0}}{b_{k-2,0}}\right)^{q-1}\gamma^{(i-1)(q-1)}\neq 1;\\ q+1,&\text{otherwise}. \end{cases} \end{array}\]

And so, by Lemma 2, we complete the proof of Theorem 5.

The proof of Theorem 6

By \(2\nmid i\) and Lemma 7, we have \(e_{j+2}\neq 0\). Next, depending on the relation between \(i\) and \(q+1\), we divide the following \(3\) cases to determine the value of \(\mathrm{rank}\left(\boldsymbol{A}_{7\times 7}\right)\).

case 1. If \(i=\frac{q+1}{2}\), then by Lemma 9, we have \(e_{j+1}=e_{j+3}=0\), and by Lemmas 5-6, we can get \(e_{j-1}=e_{j+5}=0\), \(e_{j}\in \mathbb{F}_{q^2}^{*}\) and \(e_{j+4}\in \mathbb{F}_{q^2}^{*}\). And so, we have

\[\label{M34626} \boldsymbol{A}_{7\times 7}=\left(\begin{matrix} & & & & e_{j+4}& 0 & 0 \\ & & & 0& & b_{k-2,1}^{q}e_{j+2}& b_{k-1,1}^{q}e_{j+2} \\ & & e_{j+2}& & & 0& 0 \\ & 0& & & & 0 & e_j \\ e_j& & & & & 0 & 0 \\ 0 & b_{k-2,1}e_{j+2}& & 0 & 0 & b_{k-2,0}^{q+1}e_{j+2}& b_{k-2,0}b_{k-1,0}^{q}e_{j+2} \\ 0 & b_{k-1,1}e_{j+2}& 0& e_{j+4}&0 &b_{k-2,0}^{q}b_{k-1,0}e_{j+2}& b_{k-1,0}^{q+1}e_{j+2} \\ \end{matrix}\right)\;.\tag{12}\] Now, for convenience, we denote the vectors \(\boldsymbol{a}_{u_2,9}\) and \(\boldsymbol{b}_{9,v_2}\) be the \(u_2\)-th row and the \(v_2\)-th column of the matrix \(\boldsymbol{A}_{7\times 7}\) given by the Equation (12 ), respectively, where \(1 \le u_2,v_2\le 7\). Next, for the matrix \(\boldsymbol{A}_{7\times 7}\) given by (12 ), we perform the following 2 steps of elementary transformations.

Step 1. Replace \(\boldsymbol{a}_{2,9}\) with \(\boldsymbol{a}_{2,9}-b_{k-1,1}^{q}\frac{e_{j+2}}{e_{j}}\boldsymbol{a}_{4,9}\), replace \(\boldsymbol{a}_{6,9}\) with \(\boldsymbol{a}_{6,9}-b_{k-2,0}b_{k-1,0}^{q}\frac{e_{j+2}}{e_{j}}\boldsymbol{a}_{4,9}\), and replace \(\boldsymbol{a}_{7,9}\) with \(\boldsymbol{a}_{7,9}-b_{k-1,0}^{q+1}\frac{e_{j+2}}{e_{j}}\boldsymbol{a}_{4,9}\);

Step 2. Replace \(\boldsymbol{b}_{9,2}\) with \(\boldsymbol{b}_{9,2}-b_{k-1,1}\frac{e_{j+2}}{e_{j+4}}\boldsymbol{b}_{9,4}\), and replace \(\boldsymbol{b}_{9,6}\) with \(\boldsymbol{b}_{9,6}-b_{k-2,0}^{q}b_{k-1,0}\frac{e_{j+2}}{e_{j+4}}\boldsymbol{b}_{9,4}\), we obtain the matrix \[\boldsymbol{A}_{7\times 7}^3=\left(\begin{matrix} & & & & e_{j+4}& 0 & 0 \\ & & & 0& & b_{k-2,1}^{q}e_{j+2}& 0 \\ & & e_{j+2}& & & 0& 0 \\ & 0& & & & 0 & e_j \\ e_j& & & & & 0 & 0 \\ 0 & b_{k-2,1}e_{j+2}& & 0 & 0 & b_{k-2,0}^{q+1}e_{j+2}& 0\\ 0 & 0& 0& e_{j+4}&0 &0 & 0 \\ \end{matrix}\right)\;.\] Thus, it is easy to get \[\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right)=q-5+\mathrm{rank}(\boldsymbol{A}_{7\times 7})=q-5+\mathrm{rank}(\boldsymbol{A}_{7\times 7}^3)=\begin{array}{c} \begin{cases} q,& \text{if } b_{k-2,1}=b_{k-2,0}=0;\\ q+1,& \text{if } b_{k-2,1}=0 \text{ and } b_{k-2,0}\neq 0;\\ q+2,& \text{if } b_{k-2,1}\neq 0. \end{cases} \end{array}\]

case 2. If \(i=\frac{q+1}{4}\) or \(\frac{3(q+1)}{4}\), then by Lemma 10, we have \(e_{j}=e_{j+4}=0\), and by Lemma 6, \(e_{j-1}\), \(e_{j+1}\), \(e_{j+3}\) and \(e_{j+5}\) are all non-zero. Hence, \[\label{M34627} \boldsymbol{A}_{7\times 7}=\left(\begin{matrix} & & & & 0& 0 & 0 \\ & & & e_{j+3}& & b_{k-2,1}^{q}e_{j+2}& b_{k-1,1}^{q}e_{j+2} \\ & & e_{j+2}& & & b_{k-2,0}^{q}e_{j+1}& b_{k-1,0}^{q}e_{j+1} \\ & e_{j+1}& & & & 0 & 0 \\ 0& & & & & e_{j-1} & 0 \\ 0 & b_{k-2,1}e_{j+2}& b_{k-2,0}e_{j+3}& 0 & e_{j+5} & b_{k-2,0}^{q+1}e_{j+2} & b_{k-2,1}e_{j+1}+b_{k-2,0}b_{k-1,0}^{q}e_{j+2} \\ 0 & b_{k-1,1}e_{j+2}& b_{k-1,0}e_{j+3}& 0 &0 & b_{k-2,0}^{q}b_{k-1,0}e_{j+2}+b_{k-2,1}^{q}e_{j+3} & b_{k-1,1}e_{j+1}+b_{k-1,0}^{q+1}e_{j+2}+b_{k-1,1}^{q}e_{j+3} \\ \end{matrix}\right)\;.\tag{13}\] Now, for convenience, we denote the vectors \(\boldsymbol{a}_{u_2,10}\) and \(\boldsymbol{b}_{10,v_2}\) be the \(u_2\)-th row and the \(v_2\)-th column of the matrix \(\boldsymbol{A}_{7\times 7}\) given by the Equation (13 ), respectively, where \(1 \le u_2,v_2\le 7\). Next, for the matrix \(\boldsymbol{A}_{7\times 7}\) given by (13 ), we perform the following \(4\) steps of elementary transformations.

Step 1. Replace \(\boldsymbol{a}_{2,10}\) with \(\boldsymbol{a}_{2,10}-b_{k-2,1}^{q}\frac{e_{j+2}}{e_{j-1}}\boldsymbol{a}_{5,10}\), and replace \(\boldsymbol{a}_{3,10}\) with \(\boldsymbol{a}_{3,10}-b_{k-2,0}^{q}\frac{e_{j+1}}{e_{j-1}}\boldsymbol{a}_{5,10}\);

Step 2. Replace \(\boldsymbol{a}_{6,10}\) with \(\boldsymbol{a}_{6,10}-b_{k-2,1}\frac{e_{j+2}}{e_{j+1}}\boldsymbol{a}_{4,10}-b_{k-2,0}^{q+1}\frac{e_{j+2}}{e_{j-1}}\boldsymbol{a}_{5,10}\), and replace \(\boldsymbol{a}_{7,10}\) with \(\boldsymbol{a}_{7,10}-b_{k-1,1}\frac{e_{j+2}}{e_{j+1}}\boldsymbol{a}_{4,10}-\left(b_{k-2,0}^{q}b_{k-1,0}\frac{e_{j+2}}{e_{j-1}}+b_{k-2,1}^q\frac{e_{j+3}}{e_{j-1}}\right)\boldsymbol{a}_{5,10}\);

Step 3. Replace \(\boldsymbol{b}_{10,7}\) with \(\boldsymbol{b}_{10,7}-b_{k-1,1}^{q}\frac{e_{j+2}}{e_{j+3}}\boldsymbol{b}_{10,4}-\left(b_{k-2,1}\frac{e_{j+1}}{e_{j+5}}+b_{k-2,0}b_{k-1,0}^q\frac{e_{j+2}}{e_{j+5}}\right)\boldsymbol{b}_{10,5}\), and replace \(\boldsymbol{b}_{10,3}\) with \(\boldsymbol{b}_{10,3}-b_{k-2,0}\frac{e_{j+3}}{e_{j+5}}\boldsymbol{b}_{10,5}\), we obtain the matrix \(\boldsymbol{A}_{3}^{(1)}\), and denote the vector \(\boldsymbol{b}_{10,v_2}^{(1)}\) be the \(v_2\)-th column of \(\boldsymbol{A}_{3}^{(1)}\), where \(1 \le v_2\le 7\);

Step 4. Replace \(\boldsymbol{b}_{10,7}^{(1)}\) with \(\boldsymbol{b}_{10,7}^{(1)}-b_{k-1,0}^{q+1}\frac{e_{j+1}}{e_{j+2}}\boldsymbol{b}_{10,3}^{(1)}\), we obtain the matrix \[\boldsymbol{A}_{7\times 7}^4=\left(\begin{matrix} & & & & 0& 0 & 0 \\ & & & e_{j+3}& & 0& 0 \\ & & e_{j+2}& & & 0& 0 \\ & e_{j+1}& & & & 0 & 0 \\ 0& & & & & e_{j-1} & 0 \\ 0 & 0& 0& 0 & e_{j+5} & 0 & 0 \\ 0 & 0& b_{k-1,0}e_{j+3}& 0&0 & 0 & b_{k-1,1}e_{j+1}+b_{k-1,0}^{q+1}\left(e_{j+2}-\frac{e_{j+1}}{e_{j+2}}e_{j+3}\right)+b_{k-1,1}^{q}e_{j+3} \\ \end{matrix}\right)\;.\] Hence, \(\mathrm{rank}\left( \boldsymbol{A}_{7\times 7}\right)=\mathrm{rank}\left( \boldsymbol{A}_{7\times 7}^{4}\right)=5\) if and only if \[b_{k-1,1}e_{j+1}+b_{k-1,0}^{q+1}\left(e_{j+2}-\frac{e_{j+1}}{e_{j+2}}e_{j+3}\right)+b_{k-1,1}^{q}e_{j+3}=0.\] Now by Lemma 18, we know that \(\mathrm{rank}\left( \boldsymbol{A}_{7\times 7}\right)=\mathrm{rank}\left( \boldsymbol{A}_{7\times 7}^{4}\right)=5\) if and only if \(\varGamma_{1}=0\). Thus \[\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right)= \begin{array}{c} \begin{cases} q, & \text{if } \varGamma_{1}=0;\\ q+1,& \text{if } \varGamma_{1}\neq 0. \end{cases} \end{array}\]

case 3. If \(\frac{q+1}{2}\nmid 2i\), then by Lemma 11, we know that \(e_{j}\), \(e_{j+1}\), \(e_{j+3}\) and \(e_{j+4}\) are all non-zero.

Now, for convenience, we denote the vector \(\boldsymbol{a}_{u_2,11}\) be the \(u_2\)-th row of the matrix \(\boldsymbol{A}_{7\times 7}\) given by the Equation (?? ), where \(1 \le u_2\le 7\). Next, for the matrix \(\boldsymbol{A}_{7\times 7}\) given by (?? ), we perform the following \(2\) steps of elementary transformations.

Step 1. Replace \(\boldsymbol{a}_{6,11}\) with \(\boldsymbol{a}_{6,11}-\frac{e_{j+5}}{e_{j+4}}\boldsymbol{a}_{1,11}-b_{k-2,0}\frac{e_{j+3}}{e_{j+2}}\boldsymbol{a}_{3,11}-b_{k-2,0}\frac{e_{j+2}}{e_{j+1}}\boldsymbol{a}_{4,11}\), and replace \(\boldsymbol{a}_{7,11}\) with \(\boldsymbol{a}_{7,11}-\frac{e_{j+4}}{e_{j+3}}\boldsymbol{a}_{2,11}-b_{k-1,0}\frac{e_{j+3}}{e_{j+2}}\boldsymbol{a}_{3,11}-b_{k-1,1}\frac{e_{j+2}}{e_{j+1}}\boldsymbol{a}_{4,11}\), we obtain the matrix \(\boldsymbol{A}_{4}^{(1)}\), and denote the vector \(\boldsymbol{b}_{11,v_2}^{(1)}\) be the \(v_2\)-th column of \(\boldsymbol{A}_{4}^{(1)}\), where \(1 \le v_2\le 7\);

Step 2. Replace \(\boldsymbol{b}_{11,6}^{(1)}\) with \(\boldsymbol{b}_{11,6}^{(1)}-\frac{e_{j-1}}{e_{j}}\boldsymbol{b}_{11,1}^{(1)}-b_{k-2,0}^q\frac{e_{j+1}}{e_{j+2}}\boldsymbol{b}_{11,3}^{(1)}-b_{k-2,1}^q\frac{e_{j+2}}{e_{j+3}}\boldsymbol{b}_{11,4}^{(1)}\), and replace \(\boldsymbol{b}_{11,7}^{(1)}\) with \(\boldsymbol{b}_{11,7}^{(1)}-\frac{e_{j}}{e_{j+1}}\boldsymbol{b}_{11,2}^{(1)}-b_{k-1,0}^q\frac{e_{j+1}}{e_{j+2}}\boldsymbol{b}_{11,3}^{(1)}-b_{k-1,1}^q\frac{e_{j+2}}{e_{j+3}}\boldsymbol{b}_{11,4}^{(1)}\), we obtain the matrix \[\boldsymbol{A}_{7\times 7}^5= \left(\begin{matrix} & & & & e_{j+4}& 0 & 0 \\ & & & e_{j+3}& &0& 0 \\ & & e_{j+2}& & & 0& 0 \\ & e_{j+1}& & & & 0 & 0\\ e_j& & & & & 0 & 0 \\ 0 & 0& 0& 0 &0 & C_1& C_2 \\ 0 & 0& 0& 0&0 & C_3& C_4 \\ \end{matrix}\right)\;,\] where \[\begin{align} &C_1=b_{k-2,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right),\\ &C_2=b_{k-2,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right) +b_{k-2,0}b_{k-1,0}^{q}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right), \\ &C_3=b_{k-2,0}^{q}b_{k-1,0}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)+b_{k-2,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)\\ \end{align}\] and \[C_4=b_{k-1,1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)+b_{k-1,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)+b_{k-1,1}^{q}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right).\] Thus, we have

\[\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right)=q-5+\mathrm{rank}\left(\boldsymbol{A}_{7\times 7}\right)=q-5+\mathrm{rank}(\boldsymbol{A}_{7\times 7}^5)=q+\mathrm{rank}\left(\boldsymbol{A}_{2\times 2}\right).\] Now by the proof of Theorem 4, we know that

\[mathrm{rank}\left( \boldsymbol{GG}^\dagger \right) = \begin{array}{c} \begin{cases} q, & \text{if } b_{k-2,0}=b_{k-2,1}=b_{k-1,0}=b_{k-1,1}=0\\ &\quad \text{or } b_{k-2,1}=b_{k-2,0}=0 , b_{k-1,1}\neq 0 \text{ and } \varGamma_{2}=0;\\ q+2,& \text{if } b_{k-2,0}=0 \text{ and } b_{k-2,1}\neq 0\\ &\quad \text{or } b_{k-1,0}=b_{k-1,1}=0 \text{ and } b_{k-2,0}, b_{k-2,1}\neq 0\\ &\quad \text{or } b_{k-2,1}=0, b_{k-2,0},b_{k-1,1}\neq 0 \text{ and } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}\neq -1\\ &\quad \text{or } b_{k-2,1},b_{k-2,0}\neq 0,\left(b_{k-1,0},b_{k-1,1}\right)\neq (0,0) \text{ and } \varGamma\neq 0; \\ q+1, &\text{otherwise}. \end{cases} \end{array}\]

Thus, by combining with the discussions of the above \(3\) cases, we conclude that

\[\mathrm{rank}\left( \boldsymbol{GG}^\dagger \right) =\begin{array}{c} \begin{cases} q, &\text{if } i=\frac{q+1}{2}\quad and \quad b_{k-2,1}=b_{k-2,0}=0\\ &\quad \text{or } i=\frac{q+1}{4} \text{ or } \frac{3(q+1)}{4} , b_{k-1,1}=b_{k-1,0}=0 \text{ and } \left(b_{k-2,0},b_{k-2,1}\right)\neq (0,0)\\ &\quad \text{or } i=\frac{q+1}{4} \text{ or } \frac{3(q+1)}{4} , \left(b_{k-1,0},b_{k-1,1}\right)\neq (0,0) \text{ and } \varGamma_{1}=0 \\ &\quad \text{or }\frac{q+1}{2}\nmid 2i \text{ and } b_{k-2,0}=b_{k-2,1}=b_{k-1,0}=b_{k-1,1}=0\\ &\quad \text{or } \frac{q+1}{2}\nmid 2i , b_{k-2,1}=b_{k-2,0}=0, b_{k-1,1}\neq 0 \text{ and } \varGamma_{2}=0;\\ q+2, &\text{if } i = \frac{q+1}{2} \text{ and } b_{k-2,1}\neq 0\\ &\quad \text{or } \frac{q+1}{2}\nmid 2i , b_{k-2,0}=0 \text{ and } b_{k-2,1}\neq 0\\ &\quad \text{or } \frac{q+1}{2}\nmid 2i, b_{k-1,0}=b_{k-1,1}=0 \text{ and } b_{k-2,0}, b_{k-2,1}\neq 0\\ &\quad \text{or } \frac{q+1}{2}\nmid 2i, b_{k-2,1}=0, b_{k-2,0},b_{k-1,1}\neq 0 \text{ and } b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}\neq -1\\ &\quad \text{or }\frac{q+1}{2}\nmid 2i, b_{k-2,1},b_{k-2,0}\neq 0,\left(b_{k-1,0},b_{k-1,1}\right)\neq (0,0) \text{ and } \varGamma\neq 0; \\ q+1,& \text{otherwise}. \end{cases} \end{array}\]

And so, by Lemma 2, we complete the proof of Theorem 6

4 Some examples↩︎

In this section, we present the corresponding illustrative examples for each case of Theorems 4-6 given in Section 3I, which are also checked by the Magma program. For convenience, we adopt the same case-wise classifications in Table I as those used in the proofs of Theorems 4-6.

5 Conclusions↩︎

In this paper, by taking a special class of the vector \(\boldsymbol{\alpha}\), we obtain the following three main results about a class of \(\left(\mathcal{L},\mathcal{P}\right)\)-TGRS codes \(\mathcal{C}_{q+j}\left(\boldsymbol{\alpha}\right)\).

• Determine the value of the matrix \(\boldsymbol{GG}^\dagger\) (Proposition 10).

• Completely determine the Hermitian hull dimension of \(\mathcal{C}_{q+j}\left(\boldsymbol{\alpha}\right)\) for three cases(Theorems 4-6).

• Construct two classes of EAQECCs.

6 Proofs of Several Lemmas↩︎

The proof of Lemma 8

By \(2\mid i\) and Lemma 7, we have \(e_{j+2}=0\).

If \(e_{j}=e_{j+2}=0\), then by Lemma 5, we have \(k_{j+2}\in \mathbb{Z}\) and \(k_{j}\in \mathbb{Z}\), i.e., \(\frac{(j+2)h}{q+1}\in\mathbb{Z}\) and \(\frac{jh}{q+1}\in\mathbb{Z}\), it means \(\frac{2h}{q+1}\in\mathbb{Z}\), i.e., \(q+1\mid 2h\). Note that \(2\mid q+1\), thus, \(\frac{q+1}{2}\mid h\). Now by \(h\mid i\), we have \(\frac{q+1}{2}\mid i\). And so, by \(2\leq i\leq q\), we have \(i=\frac{q+1}{2}\).

Conversely, if \(i=\frac{q+1}{2}\), then \(h=\frac{q+1}{2}\), thus, from \(1\le r \le q-2\), we can get \[k_{r+2}=\frac{(r+2)h}{q+1}=\frac{2h}{q+1}+k_r=1+k_r,\] and so, \(k_{r+2}\) and \(k_r\) are both integers or not simultaneously. Furthermore, by Lemma 5, we know that \(e_{r}\) and \(e_{r+2}\) are both zeros or not simultaneously. Thus, by \(e_{j+2}=0\), we have \(e_j=e_{j+4}=0\).

\(\quad\)

The proof of Lemma 9

\((1)\Longrightarrow(2)\) On the one hand, by \(e_{j+1}=e_{j+3}=0\) and Lemma 6, we have \(e_{j+2}\neq 0\), thus, by Lemma 7, we have \(2\nmid i\). On the other hand, by \(e_{j+1}=e_{j+3}=0\) and Lemma 5, we have \(\frac{(j+1)h}{q+1}\in \mathbb{Z}\) and \(\frac{(j+3)h}{q+1}\in\mathbb{Z}\), it means \(\frac{2h}{q+1}\in\mathbb{Z}\), i.e., \(q+1\mid2h\). Note that \(2\mid q+1\), thus, \(\frac{q+1}{2}\mid h\). Now by \(h\mid i\), we have \(\frac{q+1}{2}\mid i\). And so, by \(2\leq i\leq q\), we can get \(i=\frac{q+1}{2}\).

\((2)\Longrightarrow(3)\) On the one hand, by \(2 \nmid i\) and \(i = \frac{q+1}{2}\), we have \(2 \nmid \frac{q+1}{2}\), it means \(q \equiv 1 \pmod 4\). On the other hand, by \(i = \frac{q+1}{2}\), it’s easy to get \(\frac{q+1}{2}\mid i\).

\((3)\Longrightarrow(1)\) On the one hand, by \(\frac{q+1}{2}\mid i\) and \(2\le i\le q\), we have \(i=\frac{q+1}{2}\), thus, \(h=\frac{q+1}{2}\). By Lemma 5, we know that \(e_{r}=0\) if and only if \(k_{r}=\frac{r}{2}\in \mathbb{Z}\), i.e., \(2\mid r\), it means that exactly one of \(e_{j+1}\) and \(e_{j+2}\) equals zero. On the other hand, by \(q \equiv 1 \pmod 4\), we have \(4\nmid q+1\). Note that \(2\mid q+1\), thus, \(2\nmid \frac{q+1}{2}\). By combining with \(i=\frac{q+1}{2}\), it follows that \(2\nmid i\). Now by Lemma 7, we have \(e_{j+2}\neq 0\), thus, \(e_{j+1}=0\). Furthermore, by Lemma 6, we can get \(e_{j+1}=e_{j+3}=0\).

The proof of Lemma 10

By \(e_{j+2}\neq 0\), Lemma 5 and Lemma 7, we have \(2\nmid i\) and \(k_{j+2}\notin \mathbb{Z}\), i.e., \(2\nmid i\) and \(\frac{(j+2)h}{q+1}\notin\mathbb{Z}\).

\((1)\Longrightarrow(2)\) By \(e_{j+4}=e_{j}=0\) and Lemma 5, we have \(\frac{(j+4)h}{q+1}\in\mathbb{Z}\) and \(\frac{jh}{q+1}\in\mathbb{Z}\), then by combining with \(\frac{(j+2)h}{q+1}\notin\mathbb{Z}\), it follows that \(\frac{4h}{q+1}\in\mathbb{Z}\) and \(\frac{2h}{q+1}\notin\mathbb{Z}\), i.e., \(q+1\mid 4h\) and \(q+1\nmid 2h\). Note that \(2\mid q+1\), thus, \(\frac{q+1}{2}\mid 2h\) and \(\frac{q+1}{2}\nmid h\), namely, \(2\mid\frac{q+1}{2}\), i.e., \(4\mid q+1\), and so, \(q\equiv-1\pmod 4\). Hence, \(\frac{q+1}{4}\mid h\) and \(\frac{q+1}{2}\nmid h\). Now by \(h\mid i\), we have \(\frac{q+1}{4}\mid i\) and \(\frac{q+1}{2}\nmid i\). And so, by \(2\le i\le q\), we can get \(i=\frac{q+1}{4}\) or \(\frac{3(q+1)}{4}\).

\((2)\Longrightarrow(3)\) By \(i=\frac{q+1}{4}\) or \(\frac{3(q+1)}{4}\), it is easy to get \(\frac{q+1}{4}\mid i\).

\((3)\Longrightarrow(1)\) By \(q\equiv-1\pmod 4\), we have \(4\mid q+1\), i.e, \(2\mid \frac{q+1}{2}\), then by combining with \(2\nmid i\), it follows that \(i\neq \frac{q+1}{2}\). While by \(\frac{q+1}{4}\mid i\) and \(2\le i\le q\), we have \(i=\frac{q+1}{4}\) or \(\frac{3(q+1)}{4}\), thus, \(h=\frac{q+1}{4}\). By Lemma 5, we know that \(e_{r}=0\) if and only if \(k_{r}=\frac{r}{4}\in \mathbb{Z}\), i.e., \(4\mid r\), it means that only one of \(e_{j}\), \(e_{j+1}\), \(e_{j+2}\), \(e_{j+3}\) equals zero. And by Lemma 6, we know that \(e_{j+1}\) and \(e_{j+3}\) are both zeros or not simultaneously, thus, \(e_{j+1}\) and \(e_{j+3}\) are both non-zeros. By combining with \(e_{j+2}\neq 0\), we have \(e_{j}=0\). Furthermore, by Lemma 6, it’s easy to get \(e_{j}=e_{j+4}=0\).

The proof of Lemma 11

For any odd prime \(p\) and \(q=p^m\), it’s easy to know that \(q\equiv1\pmod 4\) or \(q\equiv-1\pmod 4\).

Then, by Lemma 7 and Lemmas 9-10, the following three statements are true,

\((1)\) \(e_{j+2}\neq 0\) if and only if \(2\nmid i\);

\((2)\) Both \(e_{j+1}\neq 0\) and \(e_{j+3}\neq 0\) if and only if \(q\equiv -1\pmod 4\) or \(q\equiv 1\pmod 4\) and \(\frac{q+1}{2}\nmid i\).

\((3)\) If \(e_{j+2}\neq 0\), then both \(e_{j}\neq 0\) and \(e_{j+4}\neq 0\) if and only if \(q\equiv 1\pmod 4\) or \(q\equiv -1\pmod 4\) and \(\frac{q+1}{4}\nmid i\).

Thus, \(e_{r}\neq 0\) \(\left(r=j, j+1, j+2, j+3, j+4\right)\) if and only if both \(2\nmid i\) and \(q\equiv 1\pmod 4\) with \(\frac{q+1}{2}\nmid i\) or \(q\equiv -1\pmod 4\) with \(\frac{q+1}{4}\nmid i\). If \(q\equiv 1\pmod 4\), then \(2\nmid \frac{q+1}{2}\), i.e., \(\gcd\left(2,\frac{q+1}{2}\right)=1\), furthermore, \(\frac{q+1}{2}\nmid 2i\) if and only if \(\frac{q+1}{2}\nmid i\). If \(q\equiv -1\pmod 4\), then \(4\mid q+1\), i.e., \(2\mid \frac{q+1}{2}\), furthermore, \(\frac{q+1}{2}\nmid 2i\) if and only if \(\frac{q+1}{4}\nmid i\).

The proof of Lemma 12

By Corollary 3, we have \(e_1\), \(e_q \neq 0\). If \(i\equiv 0\pmod p\), then \(i(q-1)=0\), furthermore, \(\mathrm{rank}(A)=2\). If \(i^2 \equiv 1\pmod p\), then \(p\nmid i\), furthermore, \(2\le i\le q-1\) and \(i(q-1)\neq 0\), thus, \(\mathrm{rank}(A)=1\) or \(2\). Note that \(\mathrm{rank}(A)=1\) if and only if \(\left[i(q-1)\right]^2=e_1e_q\), i.e., \[\label{E2461} i^2(q-1)^2=(q-1)^2\left(\sum\limits_{t=0}^{i-1}{ \gamma ^{t\left( q-1 \right)}}\right)\left(\sum\limits_{t=0}^{i-1}{ \gamma ^{qt\left( q-1 \right)}}\right).\tag{14}\] By \(i^2 \equiv 1\pmod p\) and \(q-1\not\equiv 0\pmod p\), we know that the Equation (14 ) is equivalent to \[\label{E2462} \left(\sum\limits_{t=0}^{i-1}{ \gamma ^{t\left( q-1 \right)}}\right)\left(\sum\limits_{t=0}^{i-1}{ \gamma ^{qt\left( q-1 \right)}}\right)=1.\tag{15}\] Note that \(\gamma^{q-1}-1\) and \(\gamma^{q(q-1)}-1\) are both non-zero, and so, by multiplying \(\left(\gamma^{q-1}-1\right)\left(\gamma^{q(q-1)}-1\right)\) on both sides of the Equation (15 ), we have \[\label{E2463} \left(\gamma^{i(q-1)}-1\right)\left(\gamma^{iq(q-1)}-1\right)=\left(\gamma^{q-1}-1\right)\left(\gamma^{q(q-1)}-1\right).\tag{16}\] Next, by expanding and simplifying the Equation (16 ), we have \[\label{E2464} \gamma^{q-1}-\gamma^{i(q-1)}=\gamma^{iq(q-1)}-\gamma^{q(q-1)},\tag{17}\] i.e., \[\label{E2465} \gamma^{q-1}-\gamma^{i(q-1)}=\gamma^{i(1-q)}-\gamma^{(1-q)}.\tag{18}\] And so, the Equation (18 ) is equivalent to \[\label{E2466} \gamma^{q-1}\left(\gamma^{(i+1)(1-q)}-1\right)\left(1-\gamma^{(i-1)(q-1)}\right)=0.\tag{19}\] Note that \(\gamma^{q-1}\in \mathbb{F}_{q^2}^{*}\), thus, \(\gamma^{(i+1)(1-q)}=1\) or \(\gamma^{(i-1)(q-1)}=1\), i.e., \[(i+1)(1-q)\equiv 0\pmod {q^2-1} \text{ or } (i-1)(q-1)\equiv 0\pmod {q^2-1},\] namely, \[\label{E246632} i+1\equiv 0\pmod {q+1} \text{ or } i-1\equiv 0\pmod {q+1},\tag{20}\] which contradicts the assumption \(2 \le i\le q-1\). Thus \(\mathrm{rank}(A)\neq 1\), it means \(\mathrm{rank}(A)=2\).

The proof of Lemma 13

(1) By \(2\nmid i\) and Lemma 7, we have \(e_{j+2}=q-1\) and \(\gamma^{(j+2)(q-1)} = -1\), thus, \(e_{j+2}^2=e_{j+3}e_{j+1}\) if and only if \((q-1)^2=e_{j+3}e_{j+1}\), i.e. \[\label{E24611} ( q-1 )^2 =( q-1 )^2 \left( \sum_{t=0}^{i-1}{ \gamma ^{\left( j+3 \right) t\left( q-1 \right)}} \right) \left( \sum\limits_{t=0}^{i-1}{\gamma ^{\left( j+1 \right) t\left( q-1 \right)}} \right) .\tag{21}\] Now by \(\gamma^{(j+2)(q-1)} = -1\), we have \(\gamma^{(j+1)(q-1)}=-\gamma^{1-q} \neq 1\) and \(\gamma^{(j+3)(q-1)}=-\gamma^{q-1}\neq 1\), i.e., \(\gamma^{(j+1)(q-1)}-1\) and \(\gamma^{(j+3)(q-1)}-1\) are both non-zero, thus, by multiplying \(\left(\gamma^{(j+1)(q-1)} - 1 \right) \left(\gamma^{(j+3)(q-1)} - 1\right)\) on both sides of the Equation (21 ), we have \[\left(\gamma^{(j+1)(q-1)} - 1 \right) \left(\gamma^{(j+3)(q-1)} - 1\right)=\left(\gamma^{i(j+1)(q-1)} - 1 \right) \left(\gamma^{i(j+3)(q-1)} - 1\right),\] i.e., \[\label{E24613} \left(1+\gamma^{(1-q)} \right) \left(1+\gamma^{(q-1)} \right)=\left(1+\gamma^{i(1-q)} \right) \left(1+\gamma^{i(q-1)} \right).\tag{22}\] Next, by expanding and simplifying the Equation (22 ), we have \(\gamma ^{\left( 1-q \right)}+\gamma ^{\left( q-1 \right)}=\gamma ^{i\left( 1-q \right)}+\gamma ^{i\left( q-1 \right)}\), i.e., \[\label{E24614} \gamma^{(1-q)}\left(\gamma^{(i+1)(q-1)}-1\right) \left( \gamma ^{\left( i-1 \right) \left( 1-q \right)}-1 \right) =0.\tag{23}\] Note that \(\gamma^{1-q}\in \mathbb{F}_{q^2}^{*}\), and so, \(\gamma^{(i+1)(q-1)}=1\) or \(\gamma^{(i-1)(1-q)}=1\), i.e., \[(i+1)(q-1)\equiv 0\pmod {q^2-1} \text{ or } (i-1)(1-q)\equiv 0\pmod {q^2-1},\] namely, \[\label{E2461432} i+1\equiv 0\pmod {q+1} \text{ or } i-1\equiv 0\pmod {q+1}.\tag{24}\] Thus, by \(2\leq i\leq q\), we know that the Equation (24 ) holds if and only if \(i=q\).

(2) By \(2\nmid i\), \(\frac{q+1}{2}\nmid 2i\), Lemma 7 and Lemma 11, we have \(e_{j+2}=q-1\) and \(e_{r}\in \mathbb{F}_{q^2}^{*}\) for \(r\in \left\{j,j+1,j+3,j+4\right\}\), thus, \(e_{j+1}^2=e_{j+2}e_{j}\) if and only if \(e_{j+1}^2=(q-1)e_{j}\), i.e. \[\label{E2467} \left( q-1 \right)^2\left( \sum_{t=0}^{i-1}{ \gamma ^{\left( j+1 \right) t\left( q-1 \right)}} \right) ^2=(q-1)^2 \left( \sum_{t=0}^{i-1}{\gamma ^{jt\left( q-1 \right)}} \right) .\tag{25}\] Note that \(\gamma^{j(q-1)}-1\in \mathbb{F}_{q^2}^{*}\) and \(\gamma^{(j+1)(q-1)}-1\in \mathbb{F}_{q^2}^{*}\), and so, by multiplying \(\left(\gamma^{j(q-1)} - 1 \right) \left(\gamma^{(j+1)(q-1)} - 1\right)^2\) on both sides of the Equation (25 ), we have \[\left( \gamma ^{i(j+1)\left( q-1 \right)}-1 \right) ^2\left( \gamma ^{j\left( q-1 \right)} -1\right) =\left( \gamma ^{ (j+1)(q-1) }-1 \right) ^2\left( \gamma ^{ij\left( q-1 \right)}-1 \right) ,\] i.e., \[\label{E2469} \left( 1+\gamma ^{i\left( 1-q \right)} \right) ^2\left( 1+\gamma ^{2\left( 1-q \right)} \right) =\left( 1+\gamma ^{\left( 1-q \right)} \right) ^2\left( 1+\gamma ^{2i\left( 1-q \right)} \right) .\tag{26}\] Next, by expanding and simplifying the Equation (26 ), we have \(\gamma ^{i\left( 1-q \right)}+\gamma ^{\left( i+2 \right) \left( 1-q \right)}=\gamma ^{\left( 1-q \right)}+\gamma ^{\left( 2i+1 \right) \left( 1-q \right)}\), i.e., \[\label{E2461632} \gamma ^{\left( 1-q \right)}\left( \gamma ^{\left( i-1 \right) \left( 1-q \right)}-1 \right) \left( \gamma ^{\left( i+1 \right) \left( 1-q \right)}-1 \right)=0.\tag{27}\] Now by \(\gamma^{1-q}\in \mathbb{F}_{q^2}^{*}\), we have \(\gamma^{(i-1)(1-q)}=1\) or \(\gamma^{(i+1)(1-q)}=1\), i.e., \[(i-1)(1-q)\equiv 0\pmod {q^2-1} \text{ or } (i+1)(1-q)\equiv 0\pmod {q^2-1},\] namely, \[\label{E2461732} i-1\equiv 0\pmod {q+1} \text{ or } i+1\equiv 0\pmod {q+1}.\tag{28}\] Hence, by \(2\leq i\leq q\), we know that the Equation (28 ) holds if and only if \(i=q\).

In the similar proof as that of \((2)\), Lemma 13 (3) holds too.

The proof of Lemma 14

By \(2\nmid i\), \(\frac{q+1}{2}\nmid 2i\), Lemma 7 and Lemma 11, we have \(e_{j+2}=q-1\) and \(e_{r}\in \mathbb{F}_{q^2}^{*}\) for \(r\in \left\{j,j+1,j+3,j+4\right\}\), thus, the Equation (?? ) holds if and only if \[\label{E24616} b_{k-1,1}\left(e_{j+1}^2-e_{j+2}e_{j}\right)e_{j+2}e_{j+3}+b_{k-1,1}^{q}\left(e_{j+3}^2-e_{j+4}e_{j+2}\right)e_{j+1}e_{j+2}+b_{k-1,0}^{q+1}\left(e_{j+2}^2-e_{j+1}e_{j+3}\right)e_{j+1}e_{j+3}=0.\tag{29}\] Now by the proof of Lemma 13, we know that for \(m\in \left\{0,1,3,4\right\}\), \(\gamma^{(j+m)(q-1)}=-\gamma^{(m-2)(q-1)}\neq 1\) and \(\gamma^{i(j+m)(q-1)}=-\gamma^{i(m-2)(q-1)}\), thus \[\label{A3219} e_{j+m}=\sum\limits_{t=0}^{i-1}{\left( q-1 \right) \gamma ^{t(j+m)\left( q-1 \right)}}=\left( q-1 \right) \frac{1-\gamma ^{i(j+m)\left( q-1 \right)}}{1-\gamma ^{(j+m)\left( q-1 \right)}}=(q-1)\frac{1+\gamma^{i(m-2)(q-1)}}{1+\gamma^{(m-2)(q-1)}}.\tag{30}\] Hence, the Equation (29 ) is equivalent to \[\label{E24617} \begin{align} &b_{k-1,1}\left[\left(\frac{1+\gamma ^{i\left( 1-q \right)}}{1+\gamma ^{1-q}}\right)^2-\frac{1+\gamma ^{2i\left( 1-q \right)}}{1+\gamma ^{2\left( 1-q \right)}}\right]\cdot\frac{1+\gamma ^{i\left( q-1 \right)}}{1+\gamma ^{ q-1 }} +b_{k-1,1}^{q}\left[\left(\frac{1+\gamma ^{i\left( q-1 \right)}}{1+\gamma ^{ q-1}}\right)^2-\frac{1+\gamma ^{2i\left( q-1 \right)}}{1+\gamma ^{2\left( q-1 \right)}}\right]\cdot\frac{1+\gamma ^{i\left( 1-q \right)}}{1+\gamma ^{1-q }}\\ =&-b_{k-1,0}^{q+1}\left[1-\frac{1+\gamma^{i(1-q)}}{1+\gamma^{1-q}}\cdot \frac{1+\gamma^{i(q-1)}}{1+\gamma^{q-1}}\right]\cdot \frac{1+\gamma^{i(1-q)}}{1+\gamma^{1-q}}\cdot \frac{1+\gamma^{i(q-1)}}{1+\gamma^{q-1}}. \end{align}\tag{31}\] Now by multiplying \(\left(1+\gamma ^{1-q}\right)^2\left(1+\gamma ^{2\left(1-q\right)}\right)\left(1+\gamma ^{\left(q-1\right)}\right)^2\left(1+\gamma ^{2\left(q-1\right)}\right)\) on both sides of the Equation (31 ), we have \[\label{E24618} \begin{align} &b_{k-1,1}\left[\left(1+\gamma ^{i\left(1-q\right)}\right)^2\left(1+\gamma ^{2\left(1-q \right)}\right)-\left(1+\gamma ^{2i\left(1-q \right)}\right)\left(1+\gamma ^{1-q}\right)^2\right]\cdot\left(1+\gamma ^{i\left(q-1\right)}\right)\left(1+\gamma ^{q-1}\right)\left(1+\gamma ^{2\left(q-1 \right)}\right)\\ +&b_{k-1,1}^{q}\left[\left(1+\gamma ^{i\left(q-1\right)}\right)^2\left(1+\gamma ^{2\left(q-1 \right)}\right)-\left(1+\gamma ^{2i\left(q-1 \right)}\right)\left(1+\gamma ^{q-1}\right)^2\right]\cdot\left(1+\gamma ^{i\left(1-q\right)}\right)\left(1+\gamma ^{1-q}\right)\left(1+\gamma ^{2\left(1-q \right)}\right)\\ =&b_{k-1,0}^{q+1}\left[\left(1+\gamma^{i(q-1)}\right)\left(1+\gamma^{i(1-q)}\right)-\left(1+\gamma^{1-q}\right)\left(1+\gamma^{q-1}\right)\right] \left(1+\gamma^{i(1-q)}\right)\left(1+\gamma^{i(q-1)}\right)\left(1+\gamma ^{2\left(1-q \right)}\right)\left(1+\gamma ^{2\left(q-1 \right)}\right). \end{align}\tag{32}\] Next, by expanding and simplifying the Equation (32 ), we have \[\label{E24620} \begin{align} &2b_{k-1,1}\left[-\gamma^{1-q}\left(\gamma^{(i-1)(1-q)}-1\right)\left(\gamma^{(i+1)(1-q)}-1\right)\right]\cdot\left(1+\gamma ^{i\left(q-1\right)}\right)\left(1+\gamma ^{q-1}\right)\left(1+\gamma ^{2\left(q-1 \right)}\right)\\ +&2b_{k-1,1}^{q}\left[-\gamma^{q-1}\left(\gamma^{(i-1)(q-1)}-1\right)\left(\gamma^{(i+1)(q-1)}-1\right)\right]\cdot\left(1+\gamma ^{i\left(1-q\right)}\right)\left(1+\gamma ^{1-q}\right)\left(1+\gamma ^{2\left(1-q \right)}\right)\\ =&b_{k-1,0}^{q+1}\left[\gamma^{q-1}\left(\gamma^{(i-1)(q-1)}-1\right)\left(1-\gamma^{(i+1)(1-q)}\right)\right]\cdot \left(1+\gamma^{i(1-q)}\right)\left(1+\gamma^{i(q-1)}\right)\left(1+\gamma ^{2\left(1-q \right)}\right)\left(1+\gamma ^{2\left(q-1 \right)}\right). \end{align}\tag{33}\] Now by multiplying \(\gamma ^{(2i+1)\left(q-1\right)}\) on both sides of the Equation (33 ), we have \[\label{E24621} \begin{align} &2b_{k-1,1}\left[\left(1-\gamma^{(i-1)(q-1)}\right)\left(1-\gamma^{(i+1)(q-1)}\right)\right]\cdot\left(1+\gamma ^{i\left(q-1\right)}\right)\left(1+\gamma ^{q-1}\right)\left(1+\gamma ^{2\left(q-1 \right)}\right)\\ +&2b_{k-1,1}^{q}\left[\gamma^{(i-1)(q-1)}\left(\gamma^{(i-1)(q-1)}-1\right)\left(\gamma^{(i+1)(q-1)}-1\right)\right]\cdot\left(1+\gamma ^{i\left(q-1\right)}\right)\left(1+\gamma ^{q-1}\right)\left(1+\gamma ^{2\left(q-1 \right)}\right)\\ =&-b_{k-1,0}^{q+1}\left[\left(\gamma^{(i-1)(q-1)}-1\right)\left(\gamma^{(i+1)(q-1)}-1\right)\right]\cdot \left(1+\gamma^{i(q-1)}\right)\left(1+\gamma^{i(q-1)}\right)\left(1+\gamma ^{2\left(q-1 \right)}\right)\left(1+\gamma ^{2\left(q-1 \right)}\right)\gamma^{1-q}. \end{align}\tag{34}\] Note that \(\frac{q+1}{2}\nmid 2i\), then we have \(\gamma^{i(q-1)}+1\in \mathbb{F}_{q^2}^{*}\). Thus, the Equation (34 ) holds if and only if \[2b_{k-1,1}\left(1+\gamma^{q-1}\right)\left(1+b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}\right)=-b_{k-1,0}^{q+1}\left(1+\gamma^{2(q-1)}\right)\left(1+\gamma^{i(q-1)}\right)\gamma^{1-q},\] namely, \[2b_{k-1,1}\left(1+\gamma^{1-q}\right)\left(1+b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}\right)=-b_{k-1,0}^{q+1}\left(1+\gamma^{2(1-q)}\right)\left(1+\gamma^{i(q-1)}\right),\] i.e., \(\varGamma_{2}=0\).

The proof of Lemma 15

By multiplying \(b_{k-2,1}b_{k-2,0}^{q+1}\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right)\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)\) on both sides of the Equation (?? ), we have \[\label{E24623} \begin{align} &\left[\Delta\cdot\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right) +\Delta^q\cdot \left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)\right] b_{k-2,0}^{q+1}\left(e_{j+2}-\frac{e_{j+3}}{e_{j+2}}e_{j+1}\right)\\ =&b_{k-2,1}^{q+1}\left(e_{j+3}-\frac{e_{j+4}}{e_{j+3}}e_{j+2}\right)\left(e_{j+1}-\frac{e_{j+2}}{e_{j+1}}e_{j}\right), \end{align}\tag{35}\] where \(\Delta=b_{k-1,1}-\frac{b_{k-2,1}b_{k-1,0}}{b_{k-2,0}}\). Note that \(2\nmid i\), \(\frac{q+1}{2}\nmid 2i\), Lemma 7 and Lemma 11, we have \(e_{j+2}=q-1\) and \(e_{r}\in \mathbb{F}_{q^2}^{*}\) for \(r\in \left\{j,j+1,j+3,j+4\right\}\), and then the Equation (35 ) holds if and only if \[\label{E24624} \begin{align} &\left[\Delta\cdot\left(e_{j+1}^2-e_{j}e_{j+2}\right)e_{j+3} +\Delta^q\cdot \left(e_{j+3}^{2}-e_{j+2}e_{j+4}\right)e_{j+1}\right] b_{k-2,0}^{q+1}\left(e_{j+2}^{2}-e_{j+1}e_{j+3}\right)\\ =&b_{k-2,1}^{q+1}\left(e_{j+3}^2-e_{j+2}e_{j+4}\right)\left(e_{j+1}^2-e_{j}e_{j+2}\right)e_{j+2}. \end{align}\tag{36}\] where \(\Delta=b_{k-1,1}-\frac{b_{k-2,1}b_{k-1,0}}{b_{k-2,0}}\).

Now by the proof of Lemma 14, we know that for \(m\in \left\{0,1,3,4\right\}\), \(\gamma^{(j+m)(q-1)}=-\gamma^{(m-2)(q-1)}\neq 1\) and \(\gamma^{i(j+m)(q-1)}=-\gamma^{i(m-2)(q-1)}\), thus \[e_{j+m}=\sum\limits_{t=0}^{i-1}{\left( q-1 \right) \gamma ^{t(j+m)\left( q-1 \right)}}=\left( q-1 \right) \frac{1-\gamma ^{i(j+m)\left( q-1 \right)}}{1-\gamma ^{(j+m)\left( q-1 \right)}}=(q-1)\frac{1+\gamma^{i(m-2)(q-1)}}{1+\gamma^{(m-2)(q-1)}}.\] Hence, the Equation (36 ) is equivalent to \[\label{E24625} \begin{align} &\left\{\Delta\cdot\left[\left(\frac{1+\gamma ^{i\left( 1-q \right)}}{1+\gamma ^{\left( 1-q \right)}}\right)^2-\frac{1+\gamma ^{2i\left( 1-q \right)}}{1+\gamma ^{2\left( 1-q \right)}}\right]\frac{1+\gamma ^{i\left( q-1 \right)}}{1+\gamma ^{\left( q-1 \right)}}\right.\\ &\left.+\Delta^q\cdot\left[\left(\frac{1+\gamma ^{i\left( q-1 \right)}}{1+\gamma ^{\left( q-1 \right)}}\right)^2-\frac{1+\gamma ^{2i\left( q-1 \right)}}{1+\gamma ^{2\left( q-1 \right)}}\right]\frac{1+\gamma ^{i\left( 1-q \right)}}{1+\gamma ^{\left( 1-q \right)}}\right\}\cdot b_{k-2,0}^{q+1}\left[1-\frac{1+\gamma ^{i\left( 1-q \right)}}{1+\gamma ^{\left( 1-q \right)}}\cdot \frac{1+\gamma ^{i\left( q-1 \right)}}{1+\gamma ^{\left( q-1 \right)}}\right]\\ =&b_{k-2,1}^{q+1}\left[\left(\frac{1+\gamma ^{i\left( q-1 \right)}}{1+\gamma ^{\left( q-1 \right)}}\right)^2-\frac{1+\gamma ^{2i\left( q-1 \right)}}{1+\gamma ^{2\left( q-1 \right)}}\right]\cdot\left[\left(\frac{1+\gamma ^{i\left( 1-q \right)}}{1+\gamma ^{\left(1-q \right)}}\right)^2-\frac{1+\gamma ^{2i\left( 1-q \right)}}{1+\gamma ^{2\left( 1-q \right)}}\right] \end{align}\tag{37}\] Now by multiplying \(\left(1+\gamma ^{ 1-q }\right)^3\left(1+\gamma ^{ q-1 }\right)^3\left(1+\gamma ^{2\left( 1-q \right)}\right)\left(1+\gamma ^{2\left( q-1 \right)}\right)\) on both sides of the Equation (37 ), we have \[\label{E24626} \begin{align} &\left\{\Delta\cdot\varPsi\cdot\left(1+\gamma ^{i\left( q-1 \right)}\right)\left(1+\gamma ^{\left( q-1 \right)}\right)\left(1+\gamma ^{2\left( q-1 \right)}\right)\right.\\ +&\left.\Delta^q\cdot\varUpsilon\cdot\left(1+\gamma ^{i\left( 1-q \right)}\right)\left(1+\gamma ^{\left( 1-q \right)}\right)\left(1+\gamma ^{2\left( 1-q \right)}\right)\right\}\cdot b_{k-2,0}^{q+1}\Omega\\ =&b_{k-2,1}^{q+1}\cdot\varPsi\cdot\varUpsilon\cdot\left(1+\gamma ^{ 1-q }\right)\left(1+\gamma ^{ q-1 }\right) , \end{align}\tag{38}\] where \[\varPsi=\left(1+\gamma ^{i\left( 1-q \right)}\right)^2\left(1+\gamma ^{2\left( 1-q \right)}\right)-\left(1+\gamma ^{2i\left( 1-q \right)}\right)\left(1+\gamma ^{\left( 1-q \right)}\right)^2,\] \[\varUpsilon=\left(1+\gamma ^{i\left( q-1 \right)}\right)^2\left(1+\gamma ^{2\left( q-1 \right)}\right)-\left(1+\gamma ^{2i\left( q-1 \right)}\right)\left(1+\gamma ^{\left( q-1 \right)}\right)^2\] and \[\varOmega=\left(1+\gamma ^{\left( q-1 \right)}\right)\left(1+\gamma ^{\left( 1-q \right)}\right)-\left(1+\gamma ^{i\left( q-1 \right)}\right)\left(1+\gamma ^{i\left( 1-q \right)}\right).\] Note that \[\varPsi=-2\gamma^{1-q}\left(\gamma ^{(i-1)\left( 1-q \right)}-1\right)\left(\gamma ^{(i+1)\left( 1-q \right)}-1\right),\varUpsilon=-2\gamma^{(q-1)}\left(\gamma ^{(i-1)\left( q-1 \right)}-1\right)\left(\gamma ^{(i+1)\left( q-1 \right)}-1\right)\] and \[\Omega=\gamma^{(q-1)}\left(\gamma ^{(i-1)\left( q-1 \right)}-1\right)\left(\gamma ^{(i+1)\left( 1-q \right)}-1\right),\] thus, the Equation (38 ) is equivalent to \[\label{E24627} \begin{align} &\left\{-2\Delta\cdot\gamma^{1-q}\left(\gamma ^{(i-1)\left( 1-q \right)}-1\right)\left(\gamma ^{(i+1)\left( 1-q \right)}-1\right)\cdot\left(1+\gamma ^{i\left( q-1 \right)}\right)\left(1+\gamma ^{\left( q-1 \right)}\right)\left(1+\gamma ^{2\left( q-1 \right)}\right)\right.\\ &-\left.2\Delta^q\cdot\gamma^{(q-1)}\left(\gamma ^{(i-1)\left( q-1 \right)}-1\right)\left(\gamma ^{(i+1)\left( q-1 \right)}-1\right)\cdot\left(1+\gamma ^{i\left( 1-q \right)}\right)\left(1+\gamma ^{\left( 1-q \right)}\right)\left(1+\gamma ^{2\left( 1-q \right)}\right)\right\}\\ &\cdot \left[b_{k-2,0}^{q+1}\gamma^{(q-1)}\left(\gamma ^{(i-1)\left( q-1 \right)}-1\right)\left(\gamma ^{(i+1)\left( 1-q \right)}-1\right)\right]\\ =&4b_{k-2,1}^{q+1}\cdot\gamma^{1-q}\left(\gamma ^{(i-1)\left( 1-q \right)}-1\right)\left(\gamma ^{(i+1)\left( 1-q \right)}-1\right)\cdot\gamma^{(q-1)}\left(\gamma ^{(i-1)\left( q-1 \right)}-1\right)\left(\gamma ^{(i+1)\left( q-1 \right)}-1\right)\\ &\cdot \left(1+\gamma ^{ 1-q }\right)\left(1+\gamma ^{ q-1 }\right), \end{align}\tag{39}\] i.e., \[\label{E24628} \begin{align} &\Delta\cdot\gamma^{1-q}\left(\gamma ^{(i-1)\left( 1-q \right)}-1\right)\left(\gamma ^{(i+1)\left( 1-q \right)}-1\right)\cdot\left(1+\gamma ^{i\left( q-1 \right)}\right)\left(1+\gamma ^{\left( q-1 \right)}\right)\left(1+\gamma ^{2\left( q-1 \right)}\right)\\ +&\Delta^q\cdot\gamma^{(q-1)}\left(\gamma ^{(i-1)\left( q-1 \right)}-1\right)\left(\gamma ^{(i+1)\left( q-1 \right)}-1\right)\cdot\left(1+\gamma ^{i\left( 1-q \right)}\right)\left(1+\gamma ^{\left( 1-q \right)}\right)\left(1+\gamma ^{2\left( 1-q \right)}\right)\\ =&-2\left(\frac{b_{k-2,1}}{b_{k-2,0}}\right)^{q+1}\cdot\gamma^{1-q}\left(\gamma ^{(i-1)\left( 1-q \right)}-1\right)\left(\gamma ^{(i+1)\left( q-1 \right)}-1\right)\left(1+\gamma ^{ 1-q }\right)\left(1+\gamma ^{ q-1 }\right), \end{align}\tag{40}\] Now by multiplying \(\gamma^{(2i+1)(q-1)}\) on both sides of the Equation (40 ), we have \[\label{E24629} \begin{align} &\Delta\left(1-\gamma ^{(i-1)\left( q-1 \right)}\right)\left(1-\gamma ^{(i+1)\left( q-1 \right)}\right)\cdot\left(1+\gamma ^{i\left( q-1 \right)}\right)\left(1+\gamma ^{\left( q-1 \right)}\right)\left(1+\gamma ^{2\left( q-1 \right)}\right)\\ +&\Delta^q\cdot\gamma^{(i-1)(q-1)}\left(\gamma ^{(i-1)\left( q-1 \right)}-1\right)\left(\gamma ^{(i+1)\left( q-1 \right)}-1\right)\cdot\left(1+\gamma ^{i\left( q-1 \right)}\right)\left(1+\gamma ^{\left( q-1 \right)}\right)\left(1+\gamma ^{2\left( q-1 \right)}\right)\\ =&-2\left(\frac{b_{k-2,1}}{b_{k-2,0}}\right)^{q+1}\cdot\gamma^{i(q-1)}\left(1-\gamma ^{(i-1)\left( q-1 \right)}\right)\left(\gamma ^{(i+1)\left( q-1 \right)}-1\right)\left(1+\gamma ^{ q-1 }\right)\left(1+\gamma ^{ q-1 }\right) . \end{align}\tag{41}\] Note that \(1-\gamma ^{(i-1)\left( q-1 \right)}\in \mathbb{F}_{q^2}^{*}\) and \(\gamma ^{(i+1)\left( q-1 \right)}-1\in \mathbb{F}_{q^2}^{*}\), and so, the Equation (41 ) is equivalent to \[\label{E24630} \Delta\left(1+\Delta^{q-1}\cdot\gamma^{(i-1)(q-1)} \right)\left(1+\gamma ^{i\left( q-1 \right)}\right)\left(1+\gamma ^{2\left( q-1 \right)}\right) =2\left(\frac{b_{k-2,1}}{b_{k-2,0}}\right)^{q+1}\gamma^{i(q-1)}\cdot\left(1+\gamma ^{\left( q-1 \right)}\right),\tag{42}\] i.e., \(\varGamma =0\).

The proof of Lemma 16

By Lemma 7, we have \(\gamma^{(j+2)(q-1)}=-1\), and so, \(\gamma^{(j+3)(q-1)}=-\gamma^{q-1}\neq 1\) and \(\gamma^{(j+1)(q-1)}=-\gamma^{1-q}\neq 1\). Note that \(2\mid i\), thus, \(\gamma^{i(j+3)(q-1)}=\gamma^{i(q-1)}\) and \(\gamma^{i(j+1)(q-1)}=\gamma^{i(1-q)}\). Furthermore, it is easy to know that \[e_{j+1}=\left( q-1 \right)\sum\limits_{t=0}^{i-1}{ \gamma ^{t(j+1)\left( q-1 \right)}}=\left( q-1 \right) \frac{1-\gamma ^{i(j+1)\left( q-1 \right)}}{1-\gamma ^{(j+1)\left( q-1 \right)}}=(q-1)\frac{1-\gamma^{i(1-q)}}{1+\gamma^{(1-q)}}\] and \[e_{j+3}=\left( q-1 \right)\sum\limits_{t=0}^{i-1}{ \gamma ^{t(j+3)\left( q-1 \right)}}=\left( q-1 \right) \frac{1-\gamma ^{i(j+3)\left( q-1 \right)}}{1-\gamma ^{(j+1)\left( q-1 \right)}}=(q-1)\frac{1-\gamma^{i(q-1)}}{1+\gamma^{(q-1)}}.\] Hence, by multiplying \(\left(1+\gamma^{q-1}\right)\left(1+\gamma^{1-q}\right)\) on both sides of the Equation (?? ), we have \[\label{E24633} b_{k-1,1}\left(\gamma^{i(1-q)}-1\right)\left(\gamma^{q-1}+1\right)+b_{k-1,1}^{q}\left(\gamma^{i(q-1)}-1\right)\left(\gamma^{1-q}+1\right)=0.\tag{43}\] Now by multiplying \(\gamma^{i(q-1)}\) on both sides of Equation (43 ), we have \[b_{k-1,1}\left(1-\gamma^{i(q-1)}\right)\left(\gamma^{q-1}+1\right)+b_{k-1,1}^{q}\gamma^{(i-1)(q-1)}\left(\gamma^{i(q-1)}-1\right)\left(1+\gamma^{q-1}\right)=0.\] i.e., \[\label{E24634} b_{k-1,1}\left(1-\gamma^{i(q-1)}\right)\left(\gamma^{q-1}+1\right)\left(1-b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}\right)=0.\tag{44}\] Note that \(1-\gamma^{i(q-1)}\in \mathbb{F}_{q^2}^{*}\) and \(\gamma^{q-1}+1\in \mathbb{F}_{q^2}^{*}\), thus the Equation (44 ) is equivalent to \(b_{k-1,1}=0\) or \(b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}=1\).

The proof of Lemma 17

In the similar proof as those of Lemmas 16-17 is immediately.

The proof of Lemma 18

By \(2\nmid i\) and Lemma 7, we have \(e_{j+2}=q-1\neq 0\). Then by multiplying \(e_{j+2}\) on both sides of the Equation (?? ), \[\label{E24636} b_{k-1,1}e_{j+1}e_{j+2}+b_{k-1,0}^{q+1}\left(e_{j+2}^2-e_{j+1}e_{j+3}\right)+b_{k-1,1}^{q}e_{j+2}e_{j+3}=0.\tag{45}\] Now by the proof of Lemma 14, we know that \(\gamma^{(j+1)(q-1)}=-\gamma^{1-q}\neq 1\), \(\gamma^{(j+3)(q-1)}=-\gamma^{q-1}\neq 1\) and \[e_{j+1}=\left( q-1 \right)\sum\limits_{t=0}^{i-1}{ \gamma ^{t(j+1)\left( q-1 \right)}}=\left( q-1 \right) \frac{1-\gamma ^{i(j+1)\left( q-1 \right)}}{1-\gamma ^{(j+1)\left( q-1 \right)}}=(q-1)\frac{1+\gamma^{i(1-q)}}{1+\gamma^{(1-q)}},\] \[e_{j+3}=\left( q-1 \right)\sum\limits_{t=0}^{i-1}{ \gamma ^{t(j+3)\left( q-1 \right)}}=\left( q-1 \right) \frac{1-\gamma ^{i(j+3)\left( q-1 \right)}}{1-\gamma ^{(j+1)\left( q-1 \right)}}=(q-1)\frac{1+\gamma^{i(q-1)}}{1+\gamma^{(q-1)}}.\] Hence, the Equation (45 ) is equivalent to \[\label{E24637} b_{k-1,1}\left(\frac{1+\gamma^{i(1-q)}}{1+\gamma^{1-q}}\right)+b_{k-1,0}^{q+1}\left(1-\frac{1+\gamma^{i(1-q)}}{1+\gamma^{1-q}}\cdot \frac{1+\gamma^{i(q-1)}}{1+\gamma^{(q-1)}}\right)+b_{k-1,1}^{q}\left(\frac{1+\gamma^{i(q-1)}}{1+\gamma^{(q-1)}}\right)=0.\tag{46}\] Note that \(\left(1+\gamma^{1-q}\right)\in \mathbb{F}_{q^2}^{*}\) and \(\left(1+\gamma^{q-1}\right)\in \mathbb{F}_{q^2}^{*}\), and so, by multiplying \(\left(1+\gamma^{1-q}\right)\left(1+\gamma^{q-1}\right)\) on both sides of the Equation (46 ), we have \[\label{E24638} \begin{align} &b_{k-1,1}\left(1+\gamma^{i(1-q)}\right)\left(1+\gamma^{q-1}\right)+b_{k-1,1}^q\left(1+\gamma^{i(q-1)}\right)\left(1+\gamma^{1-q}\right)\\ =&-b_{k-1,0}^{q+1}\left[\left(1+\gamma^{1-q}\right)\left(1+\gamma^{q-1}\right)-\left(1+\gamma^{i(1-q)}\right)\left(1+\gamma^{i(q-1)}\right)\right]. \end{align}\tag{47}\] Next, by expanding and simplifying the Equation (47 ), we have \[\label{E24639} b_{k-1,1}\left(1+\gamma^{i(1-q)}\right)\left(1+\gamma^{q-1}\right)+b_{k-1,1}^q\left(1+\gamma^{i(q-1)}\right)\left(1+\gamma^{1-q}\right) =-b_{k-1,0}^{q+1}\gamma^{q-1}\left(\gamma^{(i-1)(q-1)}-1\right)\left(\gamma^{(i+1)(1-q)}-1\right).\tag{48}\] Now by multiplying \(\gamma^{i(q-1)}\) on both sides of the Equation (48 ), we have \[\begin{align} &b_{k-1,1}\left(\gamma^{i(q-1)}+1\right)\left(1+\gamma^{q-1}\right)+b_{k-1,1}^{q}\gamma^{(i-1)(q-1)}\left(1+\gamma^{i(q-1)}\right)\left(\gamma^{q-1}+1\right)\\ =&-b_{k-1,0}^{q+1}\left(\gamma^{(i-1)(q-1)}-1\right)\left(1-\gamma^{(i+1)(q-1)}\right), \end{align}\] which is equivalent to \[b_{k-1,1}\left(\gamma^{i(q-1)}+1\right)\left(1+\gamma^{q-1}\right)\left(1+b_{k-1,1}^{q-1}\gamma^{(i-1)(q-1)}\right)=b_{k-1,0}^{q+1}\left(\gamma^{(i-1)(q-1)}-1\right)\left(\gamma^{(i+1)(q-1)}-1\right),\] i.e., \(\varGamma_{1}=0\).

7 The proof of Proposition 10↩︎

We present the detailed proofs of \(6\) cases in Proposition 10.

case 1. For \(u,v\in \left\{1, 2, \ldots , k-2 \right\}\).

(1) For \(u=1\), it is easy to know that \(u+v-2\in \left\{ 0,1,\ldots ,k-3 \right\}\). Now by \(q-1\le k-3=q+j-3<2q-2\), we know that \(a_{1v} \neq 0\) if and only if \(u+v-2=0\) or \(q-1\), i.e., \(v=1\) or \(v=q\). Hence, \[a_{11}=q-1, a_{1q}=\left( q-1 \right) \beta ^{q\left( q-1 \right)}.\]

(2) For \(2\le u\le q-j+1=j+4\). it is easy to know that \(u+v-2\in \left\{ u-1, u, \ldots , u+k-4 \right\}\). Now by \(0<1\le u-1\le q-j<q-1\) and \(q-1< q+j-2 \le u+k-4=u+q+j-4\le 2q-3<2q-2\), we know that \(a_{uv} \neq 0\) if and only if \(u+v-2=q-1\), i.e., \(v=q-u+1\). Hence, \[a_{u(q-u+1)}=\left( q-1 \right) \beta ^{(u-1)+q(q-u)}=(q-1)\beta^{(q-1)(q-u+1)}.\]

(3) For \(j+5=q-j+2\le u \le q-1\), it is easy to know that \(u+v-2\in \left\{ u-1, u, \ldots , u+k-4 \right\}\). Now by \(0<q-j+1\le u-1\le q-2<q-1\) and \(2q-2\le u+k-4 \le q+k-5=2q+j-5=3q-j-8<3q-3\), we know that \(a_{uv} \neq 0\) if and only if \(u+v-2=q-1\) or \(2q-2\), i.e., \(v=q-u+1\) or \(2q-u\). Hence, \[a_{u(q-u+1)}=\left( q-1 \right) \beta ^{(u-1)+q(q-u)}=(q-1)\beta^{(q-1)(q-u+1)},\] \[a_{u(2q-u)}=\left( q-1 \right) \beta ^{(u-1)+q(2q-u-1)}=(q-1)\beta^{(q-1)(q-u)}.\]

(4) For \(u=q\), it is easy to know that \(u+v-2\in \left\{ q-1, q, \ldots , q+k-4 \right\}\). Now by \(2q-2\le q+k-4=2q+j-4 =3q-j-7<3q-3\), we know that \(a_{qv} \neq 0\) if and only if \(q+v-2=q-1\) or \(2q-2\), i.e., \(v=1\) or \(q\). Hence, \[a_{q1}=\left( q-1 \right) \beta ^{q-1} \text{ and } a_{qq}=\left( q-1 \right) \beta ^{(q-1)+q(q-1)}=(q-1)\beta^{q^2-1}=q-1.\]

(5) For \(q+1\le u\le k-2=q+j-2\), it is easy to know that \(u+v-2\in \left\{ u-1, u, \ldots , u+k-4 \right\}\). Now by \(q-1<q\le u-1\le q+j-3=2q-j-6<2q-2\) and \(2q-2< 2q+j-3 \le u+k-4=u+q+j-4\le 2q+2j-6=3q-9<3q-3\), we know that \(a_{uv} \neq 0\) if and only if \(u+v-2=2q-2\), i.e., \(v=2q-u\). Hence, \[a_{u(2q-u)}=\left( q-1 \right) \beta ^{(u-1)+q(2q-u-1)}=(q-1)\beta^{(q-1)(2q-u+1)}.\]

case 2. For \(u=k-1, v\in \left\{1, 2, \ldots , k-2 \right\}\), it’s easy to get \[\label{E3465} \begin{align} a_{\left( k-1 \right) v}&=\sum_{s=1}^{q-1}\left[ \left( \alpha _s\beta \right) ^{k-2+\left( v-1 \right) q}+b_{k-2,0}\left( \alpha _s\beta \right) ^{k+\left( v-1 \right) q}+b_{k-2,1}\left( \alpha _s\beta \right) ^{k+1+\left( v-1 \right) q} \right]\\ &=\sum_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k-2+\left( v-1 \right) q}+b_{k-2,0}\sum_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k+\left( v-1 \right) q}+b_{k-2,1}\sum_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k+1+\left( v-1 \right) q}. \end{align}\tag{49}\]

Now by Remark 2, we know that \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k-2+\left( v-1 \right) q}\neq 0\) if and only if \[\label{k-343v} k-3+v\equiv 0(\bmod (q-1)),\tag{50}\] \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k+\left( v-1 \right) q}\neq 0\) if and only if \[\label{k-143v} k-1+v \equiv 0(\bmod (q-1))\tag{51}\] and \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k+1+\left( v-1 \right) q}\neq 0\) if and only if \[\label{k43v} k+v\equiv 0(\bmod (q-1)).\tag{52}\]

Note that \(q\ge 7\), thus, any two congruences among 50 , 51 and 52 cannot hold simultaneously. Hence, any two of the sums \(\sum\limits_{s=1}^{q-1}\left(\alpha_s\beta\right)^{k-2+(v-1)q}\), \(\sum\limits_{s=1}^{q-1}\left(\alpha_s\beta\right)^{k+(v-1)q}\) and \(\sum\limits_{s=1}^{q-1}\left(\alpha_s\beta\right)^{k+1+(v-1)q}\) are not non-zeros simultaneously.

(1) For \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k-2+\left( v-1 \right) q}\), we have \(k+v-3\in \left\{ k-2,k-1,\ldots ,2k-5 \right\}\). Now by \(q-1<k-2=q+j-2=2q-j-5<2q-2\) and \(2q-2<2k-5=2q+2j-5=3q-8<3q-3\), we know that \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k-2+\left( v-1 \right) q}\neq 0\) if and only if \(k+v-3=2q-2\), i.e., \(v=q-j+1=j+4\). Hence, \[\label{E3466} \begin{align} a_{\left( k-1 \right) (j+4)}&=\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k-2+\left( j+3 \right) q}=(q-1)\beta^{k-2+(j+3)q}=(q-1)\beta^{q+j-2+(q-j)q}\\ &=(q-1)\beta^{(1-q)(j-1)}=(q-1)\beta^{(q-1)(1-j+q+1)}=(q-1)\beta^{(q-1)(j+5)}. \end{align}\tag{53}\]

(2) For \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k+\left( v-1 \right) q}\), we have \(k+v-1\in \left\{ k,k+1,\ldots ,2k-3 \right\}\). Now by \(q-1<k=q+j=2q-j-3<2q-2\) and \(2q-2<2k-3=2q+2j-3=3q-6<3q-3\), we know that \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k+\left( v-1 \right) q}\neq 0\) if and only if \(k+v-1=2q-2\), i.e., \(v=q-j-1=j+2\). Hence, \[\label{E3467} \begin{align} a_{\left( k-1 \right) (j+2)}&=b_{k-2,0}\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k+(j+1)q} =b_{k-2,0}(q-1)\beta^{k+(j+1)q}=b_{k-2,0}(q-1)\beta^{q+j+(q-j-2)q}\\ &=b_{k-2,0}(q-1)\beta^{(1-q)(j+1)}=b_{k-2,0}(q-1)\beta^{(q-1)(q-j)}=b_{k-2,0}(q-1)\beta^{(q-1)(j+3)}. \end{align}\tag{54}\]

(3) For \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k+1+\left( v-1 \right) q}\), we have \(k+v\in \left\{ k+1,k+2,\ldots ,2k-2 \right\}\). Now by \(q-1<k+1=q+j+1=2q-j-2<2q-2\) and \(2q-2<2k-2=2q+2j-2=3q-5<3q-3\), we know that \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k+1+\left( v-1 \right) q}\neq 0\) if and only if \(k+v=2q-2\), i.e., \(v=q-j-2=j+1\). Hence, \[\label{E3468} \begin{align} a_{\left( k-1 \right) (j+1)}&=b_{k-2,1}\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k+1+qj} =b_{k-2,1}(q-1)\beta^{k+1+qj}=b_{k-2,1}(q-1)\beta^{q+j+1+(q-j-3)q}\\ &=b_{k-2,1}(q-1)\beta^{(1-q)(j+2)}=b_{k-2,1}(q-1)\beta^{(q-1)(q-j-1)}=b_{k-2,1}(q-1)\beta^{(q-1)(j+2)}. \end{align}\tag{55}\]

case 3. For \(u=k, v\in \left\{1, 2, \ldots , k-2 \right\}\), it’s easy to get \[\label{E3469} \begin{align} a_{kv}&=\sum_{s=1}^{q-1}\left[\left( \alpha _s\beta \right) ^{k-1+\left( v-1 \right) q}+b_{k-1,0}\left( \alpha _s\beta \right) ^{k+\left( v-1 \right) q}+b_{k-1,1}\left( \alpha _s\beta \right) ^{k+1+\left( v-1 \right) q}\right]\\ &=\sum_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k-1+\left( v-1 \right) q}+ b_{k-1,0}\sum_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k+\left( v-1 \right) q} +b_{k-1,1}\sum_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k+1+\left( v-1 \right) q}. \end{align}\tag{56}\]

In the similar proof as that of Case 2, we know that \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k-1+\left( v-1 \right) q}\neq 0\) if and only if \(k-2+v\equiv 0(\bmod (q-1))\), and any two of the sums \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k-1+\left( v-1 \right) q}\), \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k+\left( v-1 \right) q}\) and \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k+1+\left( v-1 \right) q}\) are not non-zeros simultaneously.

(1) For \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k-1+\left( v-1 \right) q}\), we have \(k+v-2\in \left\{ k-1,k,\ldots ,2k-4 \right\}\). Now by \(q-1<k-1=q+j-1=2q-j-4<2q-2\) and \(2q-2<2k-4=2q+2j-4=3q-7<3q-3\), we know that \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k-1+\left( v-1 \right) q}\neq 0\) if and only if \(k+v-2=2q-2\), i.e., \(v=q-j=j+3\). Hence, \[\begin{align} a_{k (j+3)}&=\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k-1+\left( j+2 \right) q}=(q-1)\beta^{k-1+(j+2)q}=(q-1)\beta^{q+j-1+(q-j-1)q}\\ &=(q-1)\beta^{(1-q)j}=(q-1)\beta^{(q-1)(q+1-j)}=(q-1)\beta^{(q-1)(j+4)}. \end{align}\]

(2) For \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k+\left( v-1 \right) q}\), by the Equation (54 ), we can get \(a_{k (j+2)}=b_{k-1,0}(q-1)\beta^{(q-1)(j+3)}.\)

(3) For \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{k+1+\left( v-1 \right) q}\), by the Equation (55 ) , we can get \(a_{k (j+1)}=b_{k-1,1}(q-1)\beta^{(q-1)(j+2)}\).
case 4. For \(u\in {1, 2, \ldots , k-2}, v=k-1\), it’s easy to get \[\label{E34610} \begin{align} a_{u\left( k-1 \right)}&=\sum_{s=1}^{q-1}\left[\left( \alpha _s\beta \right) ^{q\left( k-2 \right) +u-1}+b_{k-2,0}^{q}\left( \alpha _s\beta \right) ^{qk+u-1}+b_{k-2,1}^{q}\left( \alpha _s\beta \right) ^{q\left( k+1 \right) +u-1}\right]\\ &=\sum_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{q\left( k-2 \right) +u-1}+b_{k-2,0}^{q}\sum_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{qk+u-1}+b_{k-2,1}^{q}\sum_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{q\left( k+1 \right) +u-1}. \end{align}\tag{57}\]

Now by Remark 2, we know that \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{(k-2)q+ u-1}\neq 0\) if and only if \[\label{k-343u} k-3+u\equiv 0(\bmod (q-1)),\tag{58}\] \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{qk+u-1}\neq 0\) if and only if \[\label{k-143u} k-1+u \equiv 0(\bmod (q-1))\tag{59}\] and \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{(k+1)q+u-1}\neq 0\) if and only if \[\label{k43u} k+u\equiv 0(\bmod (q-1)).\tag{60}\] Note that \(q\ge 7\), thus, any two congruences among 58 , 59 and 60 cannot hold simultaneously. Hence, any two of the sums \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{(k-2)q+ u-1}\), \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{qk+u-1}\) and \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{(k+1)q+u-1}\) are not non-zeros simultaneously.

(1) For \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{q\left( k-2 \right) +u-1}\), we have \(k+u-3\in \left\{ k-2,k-1,\ldots ,2k-5 \right\}\). Now by \(q-1<k-2=q+j-2=2q-j-5<2q-2\) and \(2q-2<2k-5=2q+2j-5=3q-8<3q-3\), we know that \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{q\left( k-2 \right) +u-1}\neq 0\) if and only if \(k+u-3=2q-2\), i.e., \(u=q-j+1=j+4\). Hence, \[\label{E34611} a_{(j+4)\left( k-1 \right) }=\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{(k-2)q+j+3 }=(q-1)\beta^{(k-2)q+j+3}=(q-1)\beta^{(q+j-2)q+q-j}=(q-1)\beta^{(q-1)(j-1)}.\tag{61}\]

(2) For \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{qk+ u-1 }\), we have \(k+u-1\in \left\{ k,k+1,\ldots ,2k-3 \right\}\). Now by \(q-1<k=q+j=2q-j-3<2q-2\) and \(2q-2<2k-3=2q+2j-3=3q-6<3q-3\), we know that \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{qk+u-1}\neq 0\) if and only if \(k+u-1=2q-2\), i.e., \(u=q-j-1=j+2\). Hence, \[\label{E34612} a_{(j+2)\left( k-1 \right) }=b_{k-2,0}^q\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{qk+j+1} =b_{k-2,0}^q(q-1)\beta^{qk+j+1}=b_{k-2,0}^q(q-1)\beta^{q(q+j)+q-j-2}=b_{k-2,0}^q(q-1)\beta^{(q-1)(j+1)}.\tag{62}\]

(3) For \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{q(k+1)+u-1 }\), we have \(k+u\in \left\{ k+1,k+2,\ldots ,2k-2 \right\}\). Now by \(q-1<k+1=q+j+1=2q-j-2<2q-2\) and \(2q-2<2k-2=2q+2j-2=3q-5<3q-3\), we know that \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{q(k+1)+u-1 }\neq 0\) if and only if \(k+u=2q-2\), i.e., \(u=q-j-2=j+1\). Hence, \[\label{E34613} a_{(j+1)\left( k-1 \right) }=b_{k-2,1}^q\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{(k+1)q+j} =b_{k-2,1}^q(q-1)\beta^{(k+1)q+j}=b_{k-2,1}^q(q-1)\beta^{(q+j+1)q+q-j-3}=b_{k-2,1}^q(q-1)\beta^{(q-1)(j+2)}.\tag{63}\]

case 5. When \(u\in \left\{1, 2, \ldots , k-2 \right\}, v=k\), it is easy to know that \[\label{E34614} \begin{align} a_{uk}&=\sum_{s=1}^{q-1}\left[\left( \alpha _s\beta \right) ^{q\left( k-1 \right) +u-1}+b_{k-1,0}^{q}\left( \alpha _s\beta \right) ^{qk+u-1}+b_{k-1,1}^{q}\left( \alpha _s\beta \right) ^{q\left( k+1 \right) +u-1}\right]\\ &=\sum_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{q\left( k-1 \right) +u-1}+b_{k-1,0}^{q}\sum_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{qk+u-1}+ b_{k-1,1}^{q}\sum_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{q\left( k+1 \right) +u-1} . \end{align}\tag{64}\]

In the similar proof as that of Case 4, we know that \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{q\left( k-1 \right) +u-1}\neq 0\) if and only if \(k-2+u\equiv 0(\bmod (q-1))\), and any two of the sums \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{q\left( k-1 \right) +u-1}\), \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{qk+u-1}\) and \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{q\left( k+1 \right) +u-1}\) are not non-zeros simultaneously.

(1) For \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{q\left( k-1 \right) +u-1}\), we have \(k+u-2\in \left\{ k-1,k,\ldots ,2k-4 \right\}\). Now by \(q-1<k-1=q+j-1=2q-j-4<2q-2\) and \(2q-2<2k-4=2q+2j-4=3q-7<3q-3\), we know that \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{q\left( k-1 \right) +u-1}\neq 0\) if and only if \(k+u-2=2q-2\), i.e., \(u=q-j=j+3\). Hence, \[a_{ (j+3)k}=\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{(k-1)q+ j+2}=(q-1)\beta^{(k-1)q+ j+2}=(q-1)\beta^{(q+j-1)q+q-j-1}=(q-1)\beta^{(q-1)j}.\]

(2) For \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{qk+u-1}\), by the Equation (62 ), we can get \(a_{(j+2) k}=b_{k-1,0}^q(q-1)\beta^{(q-1)(j+1)}\).

(3) For \(\sum\limits_{s=1}^{q-1}\left( \alpha _s\beta \right) ^{q\left( k+1 \right) +u-1}\), by the Equation (63 ), we can get \(a_{(j+1) k}=b_{k-1,1}^q(q-1)\beta^{(q-1)(j+2)}\).

case 6. For \(u,v \in \{k-1,k\}\), by directly calculating the Equation (4 ), we have \[\begin{align} \label{E34615} A_1&=\left( q-1 \right) b_{k-2,0}^{q+1}\beta ^{\left( q+1 \right) k} =\left( q-1 \right) b_{k-2,0}^{q+1}\beta ^{qk+q} =\left( q-1 \right) b_{k-2,0}^{q+1}\beta ^{q(q+j)+2q-j-3} =\left( q-1 \right) b_{k-2,0}^{q+1}\beta ^{(q-1)(j+2)};\\ A_2&=\left( q-1 \right) \left[ b_{k-2,0}b_{k-1,0}^{q}\beta ^{\left( q+1 \right) k}+b_{k-2,1}\beta ^{q\left( k-1 \right) +k+1} \right] =\left( q-1 \right) \left[ b_{k-2,0}b_{k-1,0}^{q}\beta ^{(q-1)(j+2)}+b_{k-2,1}\beta ^{q\left( q+j-1 \right) +2q-j-2} \right]\\ &=\left( q-1 \right) \left[ b_{k-2,0}b_{k-1,0}^{q}\beta ^{(q-1)(j+2)}+b_{k-2,1}\beta ^{(q-1)(j+1)} \right];\\ A_3&=\left( q-1 \right) \left[ b_{k-2,1}^{q}\beta ^{q\left( k+1 \right) +k-1}+b_{k-2,0}^{q}b_{k-1,0}\beta ^{\left( q+1 \right) k} \right] =\left( q-1 \right) \left[ b_{k-2,1}^{q}\beta ^{q\left( q+j+1 \right) +2q-j-4}+b_{k-2,0}^{q}b_{k-1,0}\beta ^{(q-1)(j+2)} \right] \\ &=\left( q-1 \right) \left[ b_{k-2,1}^{q}\beta ^{(q-1)(j+3)}+b_{k-2,0}^{q}b_{k-1,0}\beta ^{(q-1)(j+2)} \right] ; \\ A_4&=\left( q-1 \right) \left[ b_{k-1,1}^{q}\beta ^{q\left( k+1 \right) +k-1}+b_{k-1,0}^{q+1}\beta ^{\left( q+1 \right) k}+b_{k-1,1}\beta ^{q\left( k-1 \right) +k+1} \right]\\ &=\left( q-1 \right) \left[ b_{k-1,1}^{q}\beta ^{(q-1)(j+3)}+b_{k-1,0}^{q+1}\beta ^{(q-1)(j+2)}+b_{k-1,1}\beta ^{(q-1)(j+1)} \right]. \end{align}\tag{65}\]

References↩︎

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1. Corresponding author: Qunying Liao. Emails:3120193984@qq.com; liangzhongh0807@163.com; 2907245673@qq.com; qunyingliao@sicnu.edu.cn.↩︎

2. Chenlu Jia, Zhonghao Liang, Yue Huang and Qunying Liao are with College of Mathematical Sciences, Sichuan Normal University, Chengdu 610066, China.↩︎

3. This paper is supported by National Natural Science Foundation of China (12471494) and Natural Science Foundation of Sichuan Province (2024NSFSC2051).↩︎