Stabilizer group of generalized determinant

TL;DR

This paper investigates the stabilizer group of a generalized determinant, a linear combination of the determinant and permanent, providing insights into its algebraic symmetry properties.

Contribution

It introduces the concept of a generalized determinant and determines its stabilizer group, a novel contribution to algebraic symmetry analysis.

Findings

Identifies the stabilizer group of the generalized determinant.

Provides a characterization of the symmetry group.

Extends understanding of determinant and permanent combinations.

Abstract

In this paper, we consider linear combination of determinant and permanent, which we call generalized determinant, and determine the stabilizer group of it.

Equations

$${\det }^{(\alpha ,\beta )}(A):=\alpha {\det }_{A_n}(A)+\beta {\det }_{{\bar{A}}_n}(A).$$

$${\det }^{(\alpha ,\beta )}(Y_{l_1{l}_2}^{k_1{k}_2})$$

$$\frac{\partial }{\partial y_{pq}}{\det }^{(\alpha ,\beta )}(X_{l_1\cdots l_r}^{k_1\cdots k_r})=\sum _{s,t=1}^n\frac{\partial x_{st}}{\partial y_{pq}}\frac{\partial }{\partial x_{st}}{\det }^{(\alpha ,\beta )}(X_{l_1\cdots l_r}^{k_1\cdots k_r})$$

$$\frac{\partial }{\partial y_{pq}}{\det }^{(\beta ,\alpha )}(X_{l_1\cdots l_r}^{k_1\cdots k_r})=\sum _{s,t=1}^n\frac{\partial x_{st}}{\partial y_{pq}}\frac{\partial }{\partial x_{st}}{\det }^{(\beta ,\alpha )}(X_{l_1\cdots l_r}^{k_1\cdots k_r}).$$

$$T(X)=C*P\begin{pmatrix}x_{11}&&&\mbox{\lower 4.73611pt\hbox to0.0pt{\hss\smash{\huge$*$}}}\\ &\ddots&&\\ &&x_{n-1,n-1}&\\ \mbox{\hbox to0.0pt{\smash{\huge$*$}\hss}\quad}&&&x_{nn}\end{pmatrix}Q$$

$$T(X)=C\ast PXQ(\mathrm{sgn}(\sigma )\mathrm{sgn}(\tau )=1).$$

$$T(X)=C\ast P{}^{t}XQ(\mathrm{sgn}(\sigma )\mathrm{sgn}(\tau )=1),$$

$$c_{1w(1)}\dots c_{nw(n)}=1(\forall w\in A_n)$$

$$c_{1w(1)}\dots c_{nw(n)}=1(\forall w\in {\bar{A}}_n)$$

$$c_{1w(1)}\dots c_{nw(n)}=1(\forall w\in S_n)$$

$$\left(\begin{array}{ccc}{c}_{11}& c_{12}& c_{13}\\ c_{21}& c_{22}& c_{23}\\ c_{31}{c}_{44}& c_{32}{c}_{44}& c_{33}{c}_{44}\end{array}\right)=\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& u& v\\ \frac{1}{a_{12}v}& \frac{1}{a_{13}{a}_{21}}& \frac{1}{a_{11}u}\end{array}\right).$$

$$c_{44}=\frac{1}{a_{12}{a}_{31}v},c_{32}=\frac{a_{12}{a}_{31}v}{a_{13}{a}_{21}},c_{33}=\frac{a_{12}{a}_{31}v}{a_{11}u}.$$

$$a_{14}\cdot v\cdot \frac{a_{12}{a}_{31}v}{a_{13}{a}_{21}}\cdot c_{41}=1\therefore c_{41}=\frac{a_{13}{a}_{21}}{a_{12}{a}_{14}{a}_{31}{v}^2}.$$

$$a_{12}\cdot c_{24}\cdot \frac{a_{12}{a}_{31}v}{a_{11}u}\cdot \frac{a_{13}{a}_{21}}{a_{12}{a}_{14}{a}_{31}{v}^2}=1\therefore c_{24}=\frac{a_{11}{a}_{14}uv}{a_{12}{a}_{13}{a}_{21}}$$

$$a_{13}\cdot u\cdot c_{34}\cdot \frac{a_{13}{a}_{21}}{a_{12}{a}_{14}{a}_{31}{v}^2}=1\therefore c_{34}=\frac{a_{12}{a}_{14}{a}_{31}{v}^2}{a_{13}^2{a}_{21}u}.$$

$$a_{13}\cdot \frac{a_{11}{a}_{14}uv}{a_{12}{a}_{13}{a}_{21}}\cdot a_{31}\cdot c_{42}=1\therefore c_{42}=\frac{a_{12}{a}_{21}}{a_{11}{a}_{14}{a}_{31}uv}$$

$$a_{14}\cdot a_{21}\cdot \frac{a_{12}{a}_{31}v}{a_{11}u}\cdot c_{42}=1\therefore c_{42}=\frac{a_{11}u}{a_{12}{a}_{14}{a}_{21}{a}_{31}v}.$$

$$u=\pm \frac{a_{12}{a}_{21}}{a_{11}}.$$

$$a_{11}\cdot \frac{a_{11}{a}_{14}uv}{a_{12}{a}_{13}{a}_{21}}\cdot \frac{a_{12}{a}_{31}v}{a_{13}{a}_{21}}\cdot c_{43}=1\therefore c_{43}=\frac{a_{13}^2{a}_{21}^2}{a_{11}^2{a}_{14}{a}_{31}u{v}^2}$$

$$a_{14}\cdot u\cdot a_{31}\cdot c_{43}=1\therefore c_{43}=\frac{1}{a_{14}{a}_{31}u}.$$

$$v=\pm \frac{a_{13}{a}_{21}}{a_{11}}.$$

$$C=\left(\begin{array}{cccc}1& 1& 1& 1\\ 1& {\epsilon }_u& {\epsilon }_v& {\epsilon }_u{\epsilon }_v\\ 1& {\epsilon }_v& {\epsilon }_u{\epsilon }_v& {\epsilon }_u\\ 1& {\epsilon }_u{\epsilon }_v& {\epsilon }_u& {\epsilon }_v\end{array}\right)\ast {\left(\frac{a_{i1}{a}_{1j}}{a_{11}}\right)}_{1\le i,j\le 4}$$

$$C=\left(\begin{array}{cccc}1& 1& 1& 1\\ 1& {\epsilon }_u& {\epsilon }_v& {\epsilon }_u{\epsilon }_v\\ 1& {\epsilon }_u{\epsilon }_v& {\epsilon }_u& {\epsilon }_v\\ 1& {\epsilon }_v& {\epsilon }_u{\epsilon }_v& {\epsilon }_u\end{array}\right)\ast {\left(\frac{a_{i1}{a}_{1j}}{a_{11}}\right)}_{1\le i,j\le 4}$$

$$C={\left(\frac{a_{i1}{a}_{1j}}{a_{11}}\right)}_{1\le i,j\le 4}$$

$$C={\left(\frac{a_{i1}{a}_{1j}}{a_{11}}\right)}_{1\le i,j\le n}$$

$$\left(\begin{array}{cccc}{c}_{11}& c_{12}& c_{13}& c_{14}\\ c_{21}& c_{22}& c_{23}& c_{24}\\ c_{31}& c_{32}& c_{33}& c_{34}\\ c_{41}{c}_{55}& c_{42}{c}_{55}& c_{43}{c}_{55}& c_{44}{c}_{55}\end{array}\right)=\left(\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ a_{21}& {\epsilon }_u\frac{a_{21}{a}_{12}}{a_{11}}& {\epsilon }_v\frac{a_{21}{a}_{13}}{a_{11}}& {\epsilon }_u{\epsilon }_v\frac{a_{21}{a}_{14}}{a_{11}}\\ a_{31}& {\epsilon }_v\frac{a_{31}{a}_{12}}{a_{11}}& {\epsilon }_u{\epsilon }_v\frac{a_{31}{a}_{13}}{a_{11}}& {\epsilon }_u\frac{a_{31}{a}_{14}}{a_{11}}\\ z& {\epsilon }_u{\epsilon }_v\frac{z{a}_{12}}{a_{11}}& {\epsilon }_u\frac{z{a}_{13}}{a_{11}}& {\epsilon }_v\frac{z{a}_{14}}{a_{11}}\end{array}\right)$$

$$\left(\begin{array}{ccc}{c}_{11}& \cdots & c_{1,n-1}\\ ⋮& \ddots & ⋮\\ c_{n-2,1}& \cdots & c_{n-2,n-1}\\ c_{n-1,1}{c}_{nn}& \cdots & c_{n-1,n-1}{c}_{nn}\end{array}\right)=\left(\begin{array}{cccc}{a}_{11}& \cdots & \cdots & a_{1,n-1}\\ ⋮& \frac{a_{21}{a}_{12}}{a_{11}}& \dots & \frac{a_{21}{a}_{1,n-1}}{a_{11}}\\ ⋮& ⋮& \ddots & ⋮\\ a_{n-2,1}& \frac{a_{n-2,1}{a}_{12}}{a_{11}}& \dots & \frac{a_{n-2,1}{a}_{1,n-1}}{a_{11}}\\ z& \frac{z{a}_{12}}{a_{11}}& \dots & \frac{z{a}_{1,n-1}}{a_{11}}\end{array}\right)$$

$$c_{nn}=\frac{z}{a_{n-1,1}},c_{n-1,j}=\frac{a_{n-1,1}{a}_{1j}}{a_{11}}(1\le j\le n-1).$$

$$a_{1n}\cdot \frac{a_{12}\cdots a_{1,n-1}{a}_{21}\cdots a_{n-1,1}}{a_{11}^{n-2}}\cdot c_{n1}=1$$

$$\therefore c_{n1}=\frac{a_{11}^{n-2}}{a_{12}\cdots a_{1n}{a}_{21}\cdots a_{n-1,1}}=\frac{a_{11}z}{a_{1n}{a}_{n-1,1}}.$$

$$z=\frac{a_{1n}{a}_{n-1,1}{a}_{n1}}{a_{11}}.$$

Full text

1. Introduction.

For an $n\times n$ matrix $A:=(a_{ij})_{1\leq i,j\leq n}$, we will define the even determinant $\mathrm{det}_{A_{n}}\!(A):=\sum_{\sigma\in A_{n}}\prod_{i=1}^{n}a_{i\sigma(i)}$ and the odd determinant $\mathrm{det}_{\bar{A}_{n}}\!(A):=\sum_{\sigma\in\bar{A}_{n}}\prod_{i=1}^{n}a_{i\sigma(i)}$ where $\bar{A}_{n}$ is the set $S_{n}\backslash A_{n}$. Let $K$ be a field of characteristic 0. Let $\alpha,\beta\in K$. We introduce the generalized determinant $\mathrm{det}^{(\alpha,\beta)}(A)$ of an $n\times n$ matrix $A$ by

[TABLE]

Let $X:=(x_{ij})_{1\leq i,j\leq n}$ where $x_{ij}$ are the standard basis of $\mathrm{Mat}_{n}^{*}$. Write $\mathrm{det}_{n}:=\mathrm{det}(X)$, $\mathrm{perm}_{n}:=\mathrm{perm}(X)$ and $\mathrm{det}^{(\alpha,\beta)}_{n}:=\mathrm{det}^{(\alpha,\beta)}(X)$. The stabilizer group $\mathrm{Stab}(f)$ of $f\in\mathrm{Sym}^{n}(\mathrm{Mat}_{n}^{*})$ is defined as $\mathrm{Stab}(f):=\{\,T\in GL(\mathrm{Mat}_{n}(K))\mid T\cdot f=f\,\}$. If $\alpha=-\beta\neq 0$, then $\mathrm{Stab}(\mathrm{det}^{(\alpha,-\alpha)})=\mathrm{Stab}(\mathrm{det}_{n})$, which had been determined by Frobenius [1] as follows.

Theorem 1** (Frobenius).**

It holds that $\mathrm{Stab}(\mathrm{det}_{n})=\{\,X\mapsto PXQ$ or $P{}^{t}\!XQ\mid\mathrm{det}P\,\mathrm{det}Q=1\,\}$. Here $P,Q\in GL_{n}(K)$.

On the other hand, if $\alpha=\beta\neq 0$, then $\mathrm{Stab}(\mathrm{det}^{(\alpha,\alpha)})=\mathrm{Stab}(\mathrm{perm}_{n})$, which was determined by Marcus and May [2] as follows.

Theorem 2** (Marcus and May).**

Let $n\geq 3$. It holds that $\mathrm{Stab}(\mathrm{perm}_{n})=\{\,X\mapsto LPXQR$ or $LP{}^{t}\!XQR\mid\mathrm{det}L\,\mathrm{det}R=1\,\}$ where $P$ and $Q$ are permutation matrices, $L$ and $R$ are diagonal matrices.

Our purpose is to relate the two results by $\mathbb{P}^{1}(K)$-family $\{\mathrm{det}^{(\alpha,\beta)}_{n}\mid[\alpha:\beta]\in\mathbb{P}^{1}(K)\}$. Our main result is the following.

Theorem 3**.**

Let $n\geq 5$ . If $\alpha\neq\pm\beta$, then $\mathrm{Stab}(\mathrm{det}^{(\alpha,\beta)}_{n})=\mathrm{Stab}(\mathrm{det}_{n})\cap\mathrm{Stab}(\mathrm{perm}_{n})$.

2. Notations.

We define a submatrix $X_{l_{1}\cdots l_{r}}^{k_{1}\cdots k_{r}}$ of X by $X_{l_{1}\cdots l_{r}}^{k_{1}\cdots k_{r}}:=(x_{k_{i}l_{j}})_{1\leq i.j\leq r}$. Let $P_{r}^{(\alpha,\beta)}(X):=(\mathrm{det}^{(\alpha,\beta)}(X_{l_{1}\cdots l_{r}}^{k_{1}\cdots k_{r}}))_{\begin{subarray}{c}1\leq k_{1}<\dots<k_{r}\leq n\\ 1\leq l_{1}<\dots<l_{r}\leq n\end{subarray}}$ be a $\binom{n}{r}\times\binom{n}{r}$ matrix. Specially, we write $P_{r}(X):=P^{(1,0)}_{r}(X)$ and $\overline{P}_{r}(X):=P^{(0,1)}_{r}(X).$

Lemma 4**.**

Let $n\geq 4$. If $\mathrm{det}^{(\alpha,\beta)}(T(X))=\mathrm{det}^{(\alpha,\beta)}(X)$, then $\mathrm{det}^{(\alpha,\beta)}(T(X)_{l_{1}l_{2}}^{k_{1}k_{2}})$ and $\mathrm{det}^{(\beta,\alpha)}(T(X)_{l_{1}l_{2}}^{k_{1}k_{2}})$ are expressible as a linear combination of $\mathrm{det}^{(\alpha,\beta)}(X_{l^{\prime}_{1}l^{\prime}_{2}}^{k^{\prime}_{1}k^{\prime}_{2}})$ and $\mathrm{det}^{(\beta,\alpha)}(X_{l^{\prime}_{1}l^{\prime}_{2}}^{k^{\prime}_{1}k^{\prime}_{2}})(1\leq k_{1}^{\prime},k_{2}^{\prime},l_{1}^{\prime},l_{2}^{\prime}\leq n)$ respecticely.

Proof.

Let $Y:=T(X)$. We can write each entry of $X=T^{-1}(Y)$ as $x_{st}=\sum_{p,q=1}^{n}g_{st}^{pq}y_{pq}$. We first prove the case $\mathrm{det}^{(\alpha,\beta)}(T(X)_{l_{1}l_{2}}^{k_{1}k_{2}})$. Since $n\geq 4$, there exists $\sigma\in A_{n}$ such that $\sigma(k_{1})=l_{1}$, $\sigma(k_{2})=l_{2}$. The permutation $(l_{1}~{}l_{2})\sigma\in\bar{A}_{n}$ satisfies $(l_{1}~{}l_{2})\sigma(k_{1})=l_{2}$ and $(l_{1}~{}l_{2})\sigma(k_{2})=l_{1}$. We compute

[TABLE]

To compute it, we use

[TABLE]

and

[TABLE]

We have $\frac{\partial x_{st}}{\partial y_{pq}}=g_{st}^{pq}\in K$ and $\frac{\partial}{\partial x_{st}}\mathrm{det}^{(\alpha,\beta)}(X_{l_{1}\cdots l_{r}}^{k_{1}\cdots k_{r}})$ is equal to $\mathrm{det}^{(\alpha,\beta)}(X_{l_{1}\cdots\hat{t}\cdots l_{r}}^{k_{1}\cdots\hat{s}\cdots k_{r}})$, $\mathrm{det}^{(\beta,\alpha)}(X_{l_{1}\cdots\hat{t}\cdots l_{r}}^{k_{1}\cdots\hat{s}\cdots k_{r}})$ or 0. Hence, differentiating $n-2$ times, the lemma follows.

We now turn to the case $\mathrm{det}^{(\beta,\alpha)}(T(X)_{l_{1}l_{2}}^{k_{1}k_{2}})$. Since $n\geq 4$, there exists $\sigma\in A_{n}$ such that $\sigma(k_{1})=l_{2},\,\sigma(k_{2})=l_{1}$. The permutation $(l_{1}~{}l_{2})\sigma\in\bar{A}_{n}$ satisfies $(l_{1}~{}l_{2})\sigma(k_{1})=l_{1},(l_{1}~{}l_{2})\sigma(k_{2})=l_{2}$. Hence the same proof works for $\mathrm{det}^{(\beta,\alpha)}(T(X)_{l_{1}l_{2}}^{k_{1}k_{2}})$. ∎

Lemma 5**.**

Let $A\in\mathrm{Mat}_{n}(K)$. If $P_{2}(A)=\overline{P}_{2}(A)=0$, then $A$ is 0, a row matrix or a column matrix.

Proof.

Consider $A\neq 0$. Without loss of generality we can assume $a_{11}\neq 0$. Then $P_{2}(A)=0$ implies $a_{ij}=0\ (2\leq\forall i,\,j\leq n)$. If there is $j\geq 2$ such that $a_{1j}\neq 0$, then $a_{i1}=0\,(2\leq\forall i\leq n)$ from $\overline{P}_{2}(A)=0$. Thus $A$ is a row matrix. If there is $i\geq 2$ such that $a_{i1}\neq 0$, then $a_{1j}=0\,(2\leq\forall j\leq n)$ from $\overline{P}_{2}(A)=0$. Thus $A$ is a column matrix. ∎

Lemma 6**.**

Define $F_{ij}:=T(E_{ij})$. If $\alpha\neq\pm\beta$, then the number of non-zero entries in $F_{ij}$ is one.

Proof.

By Lemma 4, both $\mathrm{det}^{(\alpha,\beta)}({F_{ij}}_{l_{1}l_{2}}^{k_{1}k_{2}})$ and $\mathrm{det}^{(\beta,\alpha)}({F_{ij}}_{l_{1}l_{2}}^{k_{1}k_{2}})$ are linear combination of $\mathrm{det}^{(\alpha,\beta)}({E_{ij}}_{l^{\prime}_{1}l^{\prime}_{2}}^{k^{\prime}_{1}k^{\prime}_{2}})$ and $\mathrm{det}^{(\beta,\alpha)}({E_{ij}}_{l^{\prime}_{1}l^{\prime}_{2}}^{k^{\prime}_{1}k^{\prime}_{2}})\ (1\leq k_{1}^{\prime},\,k_{2}^{\prime},\,l_{1}^{\prime},\,l_{2}^{\prime}\leq n)$ respecticely. We thus get $P_{2}^{(\alpha,\beta)}(F_{ij})=\alpha\,P_{2}(F_{ij})+\beta\,\overline{P}_{2}(F_{ij})=0$ and $P_{2}^{(\beta,\alpha)}(F_{ij})=\beta\,P_{2}(F_{ij})+\alpha\,\overline{P}_{2}(F_{ij})=0$. Since $\alpha^{2}-\beta^{2}\neq 0$, $P_{2}(F_{ij})=\overline{P}_{2}(F_{ij})=0$. Applying Lemma 5, we see that $F_{ij}$ is a row matrix or a column matrix.

Suppose that the number of non-zero entries in $F_{ij}$ is two or more. Let us assume that $F_{ij}$ is a row matrix with non-zero entries in the $i^{\prime}$th row. Since $F_{ij}+F_{it}=T(E_{ij}+E_{it}),\,F_{ij}+F_{tj}=T(E_{ij}+E_{tj})$, we have $P_{2}(F_{ij}+F_{it})=\overline{P}_{2}(F_{ij}+F_{it})=P_{2}(F_{ij}+F_{tj})=\overline{P}_{2}(F_{ij}+F_{tj})=0\,(1<\forall t\leq n)$. By Lemma 5, $F_{ij}+F_{it}$ and $F_{ij}+F_{tj}$ are row matrices or column matrices, so that $F_{it}$ and $F_{tj}$ are row matrices lying in the $i^{\prime}$th row. However $\dim\mathrm{span}\{E_{i1},E_{i2},\dots,E_{in},E_{1j},\dots,E_{nj}\}>\dim\mathrm{span}\{E_{i^{\prime}1},E_{i^{\prime}2},\dots,E_{i^{\prime}n}\}$, which contradicts the fact that $T$ is non-singular. The same proof works in the case that $F_{ij}$ is a column matrix. ∎

By Lemma 6, we have $T(E_{ij})=c_{ij}E_{i^{\prime}j^{\prime}}$. Since $T$ is non-singular, $c_{ij}\neq 0$ and $(i,j)\neq(s,t)$ implies $(i^{\prime},j^{\prime})\neq(s^{\prime},t^{\prime})$. Hereafter, we always assume $\alpha\neq\pm\beta$ and define maps $\mu,\lambda$ by $T(E_{ij})=c_{ij}E_{\mu(i,j)\lambda{(i,j)}}$.

Lemma 7**.**

There exist permutation matrices $P:=(\delta_{i\sigma(j)})_{1\leq i,j\leq n}$, $Q:=(\delta_{i\tau(j)})_{1\leq i,j\leq n}$ where $\mathrm{sgn}(\sigma)\,\mathrm{sgn}(\tau)=1$, and a matrix $C:=(c_{ij})_{1\leq i,j\leq n}\in M_{n,n}(K)$ with $\forall c_{ij}\neq 0$ such that $T(X)=C*PXQ$ or $T(X)=C*P{}^{t}\!XQ$ (the operation $*$ is the Hadamard product).

Proof.

We may assume that $\mu(1,1)=1$ and $\lambda(1,1)=1$ by swapping rows or columns even number of times, that is, the number of row and column transpositions are both even or both odd. Since $\mathrm{rank}(E_{11}+E_{22})=2$, we have $P_{2}(F_{11}+F_{22})\neq 0$ or $\overline{P}_{2}(F_{11}+F_{22})\neq 0$. It follows that $\mu(2,2)\geq 2$ and $\lambda(2,2)\geq 2$. Therefore, swapping rows or columns even number of times properly, we may assume that $\mu(2,2)=2$ and $\lambda(2,2)=2$. By continuing the same argument, we can assume $\mu(i,i)=i\,(1\leq\forall i\leq n)$ and $\lambda(i,i)=i\,(1\leq\forall i\leq n-2)$. There are two possibilities: (i) $\lambda(n-1,n-1)=n-1$ and $\lambda(n,n)=n$, (ii) $\lambda(n-1,n-1)=n$ and $\lambda(n,n)=n-1$. However, the case (ii) never happens because the coefficients of $x_{11}\dots x_{n-1n-1}x_{nn}$ in $\mathrm{det}^{(\alpha,\beta)}(T(X))$ and in $\mathrm{det}^{(\alpha,\beta)}(X)$ are different. Therefore, we conclude that

[TABLE]

where $\mathrm{sgn}(\sigma)\,\mathrm{sgn}(\tau)=1$. To continue the argument, we may assume $P=Q=I_{n}$ that is $\mu(i,i)=i$ and $\lambda(i,i)=i\,(1\leq\forall i\leq n)$ without loss of generality.

By $P_{2}(E_{11}+E_{12})=\overline{P}_{2}(E_{11}+E_{12})=0$, we get $\mu(1,2)=1$ or $\lambda(1,2)=1$. By $P_{2}(E_{22}+E_{12})=\overline{P}_{2}(E_{22}+E_{12})=0$, we also get $\mu(1,2)=2$ or $\lambda(1,2)=2$. Combining these, we obtain two possibilities: (I) $\mu(1,2)=1$ and $\lambda(1,2)=2$, (II) $\mu(1,2)=2$ and $\lambda(1,2)=1$.

Suppose first that (I) holds. Let $3\leq\gamma\leq n$. By $P_{2}(E_{11}+E_{1\gamma})=\overline{P}_{2}(E_{11}+E_{1\gamma})=0$, we get $\mu(1,\gamma)=1$ or $\lambda(1,\gamma)=1$. By $P_{2}(E_{12}+E_{1\gamma})=\overline{P}_{2}(E_{12}+E_{1\gamma})=0$, we also get $\mu(1,\gamma)=1$ or $\lambda(1,\gamma)=2$. Combining these gives $\mu(1,\gamma)=1$. By $P_{2}(E_{\gamma\gamma}+E_{1\gamma})=\overline{P}_{2}(E_{\gamma\gamma}+E_{1\gamma})=0$, we obtain $\lambda(1,\gamma)=\gamma$.

Let $\delta\neq 1$. By $P_{2}(E_{11}+E_{\delta 1})=\overline{P}_{2}(E_{11}+E_{\delta 1})=0$, we have $\mu(\delta,1)=1$ or $\lambda(\delta,1)=1$.　However $\mu(1,\gamma)=1\,(1\leq\gamma\leq n)$ gives $\mu(\delta,1)\neq 1$ as $T$ is non-singular. Hence $\lambda(\delta,1)=1$. By $P_{2}(E_{\delta\delta}+E_{\delta 1})=\overline{P}_{2}(E_{\delta\delta}+E_{\delta 1})=0$, we obtain $\mu(\delta,1)=\delta$.

Let $1<\gamma\neq\delta\leq n$. By $P_{2}(E_{\delta 1}+E_{\delta\gamma})=\overline{P}_{2}(E_{\delta 1}+E_{\delta\gamma})=0$, we get $\mu(\delta,\gamma)=\delta$ or $\lambda(\delta,\gamma)=1$ but the latter is impossible. By $P_{2}(E_{1\gamma}+E_{\delta\gamma})=\overline{P}_{2}(E_{1\gamma}+E_{\delta\gamma})=0$, we also get $\lambda(\delta,\gamma)=\gamma$.

By the above argument, we obtain $\mu(i,j)=i$ and $\lambda(i,j)=j\,(1\leq\forall i,\,j\leq n)$ that is

[TABLE]

The same proof works for the case (II) and we also obtain

[TABLE]

if (II) holds. ∎

Lemma 8**.**

If $n\geq 5$, then $\mathrm{rank}(C)=1$.

Proof.

Comparing the coefficients of $\mathrm{det}^{(\alpha,\beta)}(X)$ and $\mathrm{det}^{(\alpha,\beta)}(C*PXQ)$, we obtain

[TABLE]

Let us solve these simultaneous equations.

For preperation, we first consider the case $n=4$. Let $\beta=0$. By $c_{11}c_{22}(c_{33}c_{44})=1,\,c_{12}c_{23}(c_{31}c_{44})=1$ and $c_{13}c_{21}(c_{32}c_{44})=1$, we can write

[TABLE]

Set $a_{31}:=c_{31}$. Then we have

[TABLE]

Set $a_{14}:=c_{14}$. By $c_{14}c_{23}c_{32}c_{41}=1$, we have

[TABLE]

By $c_{12}c_{24}c_{33}c_{41}=1$ and $c_{13}c_{22}c_{34}c_{41}=1$, we have

[TABLE]

[TABLE]

By $c_{13}c_{24}c_{31}c_{42}=1$ and $c_{14}c_{21}c_{33}c_{42}=1$, we have

[TABLE]

[TABLE]

Combining these yields

[TABLE]

By $c_{11}c_{24}c_{32}c_{43}=1$ and $c_{14}c_{22}c_{31}c_{43}=1$, we have

[TABLE]

[TABLE]

Combining these yields

[TABLE]

Now, set $a_{41}:=c_{41}=\frac{a_{11}^{2}}{a_{12}a_{13}a_{14}a_{21}a_{31}}.$ Summarizing the above, we obtain

[TABLE]

where $\varepsilon_{u},\varepsilon_{v}\in\{+1,-1\}$ and $a_{11}\cdots a_{14}a_{11}\cdots a_{41}=a_{11}^{4}$. Conversely, the matrix C is a solution of the simultaneous equation (1).

Let $\alpha=0$. Interchanging the 3rd and the 4th row of (4), we obtain

[TABLE]

where $\varepsilon_{u},\varepsilon_{v}\in\{+1,-1\}$ and $a_{11}\cdots a_{14}a_{11}\cdots a_{41}=a_{11}^{4}$ as a solution of the simultaneous equation (2).

Let $\alpha\neq 0$ and $\beta\neq 0$. Combining (4) and (5), we obtain

[TABLE]

where $a_{11}\cdots a_{14}a_{11}\cdots a_{41}=a_{11}^{4}$ as a solution of the simultaneous equation (3).

Let us consider the simultaneous equations for $n\geq 5$. We prove that the solution of the each simultaneous equations (1),(2),(3) is expressible as

[TABLE]

where $a_{11}\cdots a_{1n}a_{11}\cdots a_{n1}=a_{11}^{n}$ respectively by induction of the matrix size $n$.

In the case $n=5$ and $\beta=0$, using the fact that any solution of (1) may be written as (4), we can write

[TABLE]

where $\varepsilon_{u},\varepsilon_{v}\in\{+1,-1\}$ and $a_{11}\cdots a_{14}a_{11}\cdots a_{31}z=a_{11}^{4}$. Set $a_{41}:=c_{41}$. Then $c_{55}=\frac{z}{a_{41}},\,c_{42}=\varepsilon_{u}\varepsilon_{v}\frac{a_{41}a_{12}}{a_{11}},\,c_{43}=\varepsilon_{u}\frac{a_{41}a_{13}}{a_{11}},\,c_{44}=\varepsilon_{v}\frac{a_{41}a_{14}}{a_{11}}$. Set $a_{15}:=c_{15}$. If $\varepsilon_{u}=-1$ or $\varepsilon_{v}=-1$, there exist $w\in A_{n}$ such that $w(1)=5,\,w(5)=1,\,c_{iw(i)}=\frac{a_{i1}a_{1w(i)}}{a_{11}}\,(2\leq i\leq 4)$ and $w^{\prime}\in A_{n}$ such that $w^{\prime}(1)=5,\,w^{\prime}(5)=1,\,c_{iw^{\prime}(i)}=-\frac{a_{i1}a_{1w^{\prime}(i)}}{a_{11}}\,(2\leq i\leq 4)$. Then $c_{15}c_{2w(2)}c_{3w(3)}c_{4w(4)}c_{51}\neq c_{15}c_{2w^{\prime}(2)}c_{3w^{\prime}(3)}c_{4w^{\prime}(4)}c_{51}$ and one of the two cannot be equal to 1, a contradiction. Thus $\varepsilon_{u}=\varepsilon_{v}=+1$. The similar consideration applies to the case $n=5$ and $\alpha=0$.

Therefore, assuming (7) to hold for $n-1$, we can write

[TABLE]

where $a_{11}\cdots a_{1,n-1}a_{11}\cdots a_{n-2,1}z=a_{11}^{n-1}$. Set $a_{n-1,1}:=c_{n-1,1}$. Then we have

[TABLE]

Set $a_{1n}:=c_{1n}$. By $c_{1n}(c_{2w(2)}\cdots c_{n-1w(n-1)})c_{n1}=1$ for $w\in A_{n}$ such that $w(1)=n,\,w(n)=1$,

[TABLE]

[TABLE]

Set $a_{n1}:=c_{n1}$. Then

[TABLE]

It follows that $c_{ij}=\frac{a_{i1}a_{1j}}{a_{11}}$ for $1\leq i,\,j\leq n-1$ and $(i,j)=(1,n)$, $(n,1)$, $(n,n)$ where $a_{11}\cdots a_{1n}a_{11}\cdots a_{n1}=a_{11}^{n}$. By $c_{in}(c_{1w(1)}\cdots\widehat{c_{in}}\cdots c_{n-1w(n-1)})c_{n1}=1$, we have

[TABLE]

[TABLE]

By $(c_{1w(1)}\cdots c_{n-1w(n-1)})c_{nj}=1$, we have

[TABLE]

[TABLE]

From the above, we obtain (7).

Each of $2\times 2$ minor determinants of C is 0, so that $\mathrm{rank}(C)=1$ follows. ∎

Proof of Theorem 3.

Let $T^{-1}\in\mathrm{Stab}(\mathrm{det}^{(\alpha,\beta)}_{n})$ and $\alpha\neq\pm\beta$. By Lemma 6 and 7, We can write $T(X)=C*PXQ$ or $T(X)=C*P{}^{t}\!XQ$ where $\text{sgn}(\sigma)\,\text{sgn}(\tau)=1$. By Lemma 8, we can write $c_{ij}=l_{i}r_{j}\,(1\leq i,\,j\leq n)$ where $l_{1}\cdots l_{n}r_{1}\cdots r_{n}=1$. Set $L:=\text{diag}(l_{1},\dots,l_{n})$ and $R:=\text{diag}(r_{1},\dots,r_{n})$. Then we obtain $T(X)=LPXQR$ or $T(X)=LP{}^{t}\!XQR$ where $\text{sgn}(\sigma)\,\text{sgn}(\tau)=1$, which proves the theorem. ∎