﻿ 量子行列式及相关问题研究

# 量子行列式及相关问题研究Study on Quantum Determinant and Related Problems

Abstract: IBy generalizing the determinant of the product of ordinary matrix, the quantum determinant formula for the product of two quantum matrices is obtained. The generalized quantum determi-nant is defined, and the corresponding row and column expansion theorem and the generalized quantum determinant for matrix product are constructed.

1. 引言

2. 有关定义

$A=\left(\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ {a}_{21}& {a}_{22}& \cdots & {a}_{2n}\\ ⋮& ⋮& & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}\right)$

${|A|}_{q}=\underset{\sigma \in {S}_{n}}{\sum }{\left(-q\right)}^{-l\left(\sigma \right)}{a}_{1\sigma \left(1\right)}{a}_{2\sigma \left(2\right)}\cdots {a}_{n\sigma \left(n\right)}$ ,

$A=\left(\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ {a}_{21}& {a}_{22}& \cdots & {a}_{2n}\\ ⋮& ⋮& & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}\right)$

$\left\{\begin{array}{l}{a}_{ik}{a}_{im}={q}^{-1}{a}_{im}{a}_{ik},\\ {a}_{ik}{a}_{jk}={q}^{-1}{a}_{jk}{a}_{ik},\\ {a}_{im}{a}_{jk}={a}_{jk}{a}_{im},\\ {a}_{ik}{a}_{jm}-{a}_{jm}{a}_{ik}=\left({q}^{-1}-q\right){a}_{im}{a}_{jk}.\end{array}$

3. 主要结果

$b\underset{\sigma \in {S}_{n}}{\sum }{\left(-q\right)}^{-l\left(\sigma \right)}{a}_{1\sigma \left(1\right)}\cdots {a}_{i\sigma \left(i\right)}\cdots {a}_{i\sigma \left(j\right)}\cdots {a}_{n\sigma \left(n\right)}+\underset{\sigma \in {S}_{n}}{\sum }{\left(-q\right)}^{-l\left(\sigma \right)}{a}_{1\sigma \left(1\right)}\cdots {a}_{j\sigma \left(j\right)}\cdots {a}_{n\sigma \left(n\right)}$ ,

${\left(-q\right)}^{-l\left(\sigma \right)}{a}_{1\sigma \left(1\right)}\cdots {a}_{i\sigma \left(i\right)}\cdots {a}_{i\sigma \left(j\right)}\cdots {a}_{n\sigma \left(n\right)}$ .

$n=2$ 时，由定义1及定义2得，

$\underset{1\le {i}_{1}<{i}_{2}\le n}{\sum }{A}_{{i}_{1}{i}_{2}}{B}_{{i}_{1}{i}_{2}}={a}_{11}{a}_{22}{b}_{11}{b}_{22}+{a}_{12}{a}_{21}{b}_{12}{b}_{21}-{q}^{-1}{a}_{11}{a}_{22}{b}_{12}{b}_{21}-{q}^{-1}{a}_{12}{a}_{21}{b}_{22}{b}_{11}=D$ .

$D={|\begin{array}{cc}{a}_{11}{b}_{11}+\cdots +{a}_{1k}{b}_{1k}& {a}_{11}{b}_{21}+\cdots +{a}_{1k}{b}_{2k}\\ {a}_{21}{b}_{11}+\cdots +{a}_{2k}{b}_{1k}& {a}_{21}{b}_{21}+\cdots +{a}_{2k}{b}_{2k}\end{array}|}_{q}=\underset{1\le {i}_{1}<{i}_{2}\le k}{\sum }{A}_{{i}_{1}{i}_{2}}{B}_{{i}_{1}{i}_{2}}$ .

$\begin{array}{c}D={|\begin{array}{cc}\left({a}_{11}{b}_{11}+\cdots +{a}_{1k}{b}_{1k}\right)+{a}_{1k+1}{b}_{1k+1}& \left({a}_{11}{b}_{21}+\cdots +{a}_{1k}{b}_{2k}\right)+{a}_{1k+1}{b}_{2k+1}\\ \left({a}_{21}{b}_{11}+\cdots +{a}_{2k}{b}_{1k}\right)+{a}_{2k+1}{b}_{1k+1}& \left({a}_{21}{b}_{21}+\cdots +{a}_{2k}{b}_{2k}\right)+{a}_{2k+1}{b}_{2k+1}\end{array}|}_{q}\\ ={|\begin{array}{cc}{a}_{11}{b}_{11}+\cdots +{a}_{1k}{b}_{1k}& {a}_{11}{b}_{21}+\cdots +{a}_{1k}{b}_{2k}\\ {a}_{21}{b}_{11}+\cdots +{a}_{2k}{b}_{1k}& {a}_{21}{b}_{21}+\cdots +{a}_{2k}{b}_{2k}\end{array}|}_{q}+{|\begin{array}{cc}{a}_{11}{b}_{11}+\cdots +{a}_{1k}{b}_{1k}& {a}_{1k+1}{b}_{2k+1}\\ {a}_{21}{b}_{11}+\cdots +{a}_{2k}{b}_{1k}& {a}_{2k+1}{b}_{2k+1}\end{array}|}_{q}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{|\begin{array}{cc}{a}_{1k+1}{b}_{1k+1}& {a}_{11}{b}_{21}+\cdots +{a}_{1k}{b}_{2k}\\ {a}_{2k+1}{b}_{1k+1}& {a}_{21}{b}_{21}+\cdots +{a}_{2k}{b}_{2k}\end{array}|}_{q}+{|\begin{array}{cc}{a}_{1k+1}{b}_{1k+1}& {a}_{1k+1}{b}_{2k+1}\\ {a}_{2k+1}{b}_{1k+1}& {a}_{2k+1}{b}_{2k+1}\end{array}|}_{q}\\ =\underset{1\le {i}_{1}<{i}_{2}\le k}{\sum }{A}_{{i}_{1}{i}_{2}}{B}_{{i}_{1}{i}_{2}}+{A}_{1k+1}{B}_{1k+1}+\cdots +{A}_{kk+1}{B}_{kk+1}=\underset{1\le {i}_{1}<{i}_{2}\le k+1}{\sum }{A}_{{i}_{1}{i}_{2}}{B}_{{i}_{1}{i}_{2}}\end{array}$

${|A|}_{q}=\underset{1\le {i}_{1}<\cdots <{i}_{m}\le n}{\sum }{\left(-q\right)}^{-l\left(\sigma \right)-\left[{i}_{1}+{i}_{2}+\cdots +{i}_{m}-\frac{m\left(m+1\right)}{2}\right]}{a}_{1{i}_{\sigma \left(1\right)}}{a}_{2{i}_{\sigma \left(2\right)}}\cdots {a}_{m{i}_{\sigma \left(m\right)}}$ ,

$m=1$ 时，由定义3， ${|A|}_{q}=\underset{i=1}{\overset{n}{\sum }}{\left(-q\right)}^{1-i}{a}_{1i}$ ，结论成立。

$\begin{array}{c}{a}_{1{i}_{1}}A\stackrel{^}{1}{\stackrel{^}{i}}_{1}={\left(-q\right)}^{1-{i}_{1}}{a}_{1{i}_{1}}\underset{1\le {i}_{2}<\cdots <{i}_{m}\le n}{\sum }{\left(-q\right)}^{-l\left(\sigma \left(2\right),\cdots ,\sigma \left(m\right)\right)-\left[{i}_{2}+\cdots +{i}_{m}-\frac{m\left(m-1\right)}{2}-\left(m-1-{m}_{1}\right)\right]}{a}_{2{i}_{\sigma \left(2\right)}}\cdots {a}_{m{i}_{\sigma \left(m\right)}}\\ =\underset{1\le {i}_{2}<\cdots <{i}_{m}\le n}{\sum }{\left(-q\right)}^{-l\left(\sigma \right)-\left[{i}_{1}+{i}_{2}+\cdots +{i}_{m}-\frac{m\left(m+1\right)}{2}+{m}_{1}\right]}{a}_{1{i}_{1}}{a}_{2{i}_{\sigma \left(2\right)}}\cdots {a}_{m{i}_{\sigma \left(m\right)}}\end{array}$ ,

${|A|}_{q}=\underset{i=1}{\overset{n}{\sum }}{a}_{1i}{A}_{\stackrel{^}{1}\stackrel{^}{i}}=\underset{1\le {i}_{1}<\cdots <{i}_{m}\le n}{\sum }{\left(-q\right)}^{-l\left(\sigma \right)-\left[{i}_{1}+{i}_{2}+\cdots +{i}_{m}-\frac{m\left(m+1\right)}{2}\right]}{a}_{1{i}_{\sigma \left(1\right)}}{a}_{2{i}_{\sigma \left(2\right)}}\cdots {a}_{m{i}_{\sigma \left(m\right)}}$ .

${|A|}_{q}=\underset{1\le {i}_{1}<\cdots <{i}_{m}\le n}{\sum }{\left(-q\right)}^{-\left[{i}_{1}+{i}_{2}+\cdots +{i}_{m}-\frac{m\left(m+1\right)}{2}\right]}{M}_{{i}_{1}{i}_{2}\cdots {i}_{m}}$ ,

${|AB|}_{q}=\underset{1\le {i}_{1}<\cdots <{i}_{m}\le s}{\sum }{\left(-q\right)}^{-\left[{i}_{1}+{i}_{2}+\cdots +{i}_{m}-\frac{m\left(m+1\right)}{2}\right]}{|A{B}_{{i}_{1}{i}_{2}\cdots {i}_{m}}|}_{q}$ ,

${|AB|}_{q}=\underset{1\le {i}_{1}<\cdots <{i}_{m}\le n}{\sum }{\left(-q\right)}^{-\left[{i}_{1}+{i}_{2}+\cdots +{i}_{m}-\frac{m\left(m+1\right)}{2}\right]}{M}_{{i}_{1}{i}_{2}\cdots {i}_{m}}$ ,

NOTES

*通讯作者。

[1] 潘庆年. q-行列式及其性质[J]. 数学的实践与认识, 2000, 30(3): 358-362.

[2] 孙杰, 孙多. 行列式乘法的推广及应用[J]. 扬州教育学院学报, 2000(3): 76-77.

[3] 王立志. 一般矩阵的广义行列式[J]. 山西大学学报(自然科学版), 1995(3): 254-258.

[4] 郭忠海, 王立志. 关于矩阵乘积的广义行列式[J]. 忻州师范学院学报, 2003, 19(2): 46-48.

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