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復(fù)旦大學(xué)謝啟鴻高等代數(shù)每周一題[2021A05]參考解答

2021-10-31 17:58 作者:CharlesMa0606  | 我要投稿

本文是本人給出的2021年復(fù)旦大學(xué)謝啟鴻高等代數(shù)的每周一題[問題2021A05]的解答

題目來自于復(fù)旦大學(xué)謝啟鴻教授在他的博客提供的每周一題練習(xí)

(鏈接:https://www.cnblogs.com/torsor/p/15329047.html)

本文僅供學(xué)習(xí)交流,如有錯(cuò)誤懇請(qǐng)指正!

[問題2021A05]設(shè)A%2CBn階方陣,滿足:

AB%3DA%2Ba_mB%5Em%2Ba_%7Bm-1%7DB%5E%7Bm-1%7D%2B%5Ccdots%2Ba_1B

其中a_m%2Ba_%7Bm-1%7D%2B%5Ccdots%2Ba_1%5Cneq0.求證: AB%3DBA.

(方法一,變形配湊)

移項(xiàng)可得A%3DAB-%5Cleft(a_mB%5Em%2Ba_%7Bm-1%7DB%5E%7Bm-1%7D%2B%5Ccdots%2Ba_1B%5Cright)兩邊左乘B,有

%5Cleft(AB-BA%5Cright)%5Cleft(B-I_n%5Cright)%3DO

%5Cleft(AB-BA%5Cright)%5Cleft(B-I_n%5Cright)%3DO

注意到我們令B%3DB-I_n%2BI_n代入題中式子可得

f%5Cleft(A%5Cright)%5Cleft(B-I_n%5Cright)%3D%5Cleft(a_m%2Ba_%7Bm-1%7D%2B%5Ccdots%2Ba_1%5Cright)I_n%5Cneq0

從而B-I_n可逆,于是AB%3DBA.

%5BQ.E.D%5D

(方法二,類似方法一,但直接代入B%3DB-I_n%2BI_n)

A%5Cleft(B-I_n%2BI_n%5Cright)%3DA%2Ba_m%5Cleft(B-I_n%2BI_n%5Cright)%5Em%2Ba_%7Bm-1%7D%5Cleft(B-I_n%2BI_n%5Cright)%5E%7Bm-1%7D%2B%5Ccdots%2Ba_1%5Cleft(B-I_n%2BI_n%5Cright)

二項(xiàng)式展開后移項(xiàng),有

%5Cleft(A-f%5Cleft(B%5Cright)%5Cright)%5Cleft(B-I_n%5Cright)%3D%5Cleft(a_m%2B%5Ccdots%2Ba_1%5Cright)I_n%2C(a_m%2Ba_%7Bm-1%7D%2B%5Ccdots%2Ba_1%5Cneq0)

其中f%5Cleft(B%5Cright)形式上是只有BI_n的矩陣多項(xiàng)式,從而與B-I_n可交換,而B-I_nA-f%5Cleft(B%5Cright)也可交換

從而

%5Cleft(A-f%5Cleft(B%5Cright)%5Cright)%5Cleft(B-I_n%5Cright)%3D%5Cleft(B-I_n%5Cright)%5Cleft(A-f%5Cleft(B%5Cright)%5Cright)%5CRightarrow%20AB%3DBA.

%5BQ.E.D%5D

(方法三,利用多項(xiàng)式理論)

設(shè)多項(xiàng)式f%5Cleft(x%5Cright)%3Da_mx%5Em%2Ba_%7Bm-1%7Dx%5E%7Bm-1%7D%2B%5Ccdots%2Ba_1x%2Cg%5Cleft(x%5Cright)%3Dx-1,因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=a_m%2Ba_%7Bm-1%7D%2B%5Ccdots%2Ba_1%5Cneq0" alt="a_m%2Ba_%7Bm-1%7D%2B%5Ccdots%2Ba_1%5Cneq0">,從而f%5Cleft(1%5Cright)%5Cneq0,于是%5Cleft(f%5Cleft(x%5Cright)%2Cg%5Cleft(x%5Cright)%5Cright)%3D1.

那么%5Cexists%20u%5Cleft(x%5Cright)%2Cv%5Cleft(x%5Cright)%2Cs.t.f%5Cleft(x%5Cright)u%5Cleft(x%5Cright)%2Bg%5Cleft(x%5Cright)v%5Cleft(x%5Cright)%3D1

我們代入矩陣B,可以得到:f%5Cleft(B%5Cright)u%5Cleft(B%5Cright)%2Bg%5Cleft(B%5Cright)v%5Cleft(B%5Cright)%3DI_n

由題,Ag%5Cleft(B%5Cright)%3Df%5Cleft(B%5Cright),從而Ag%5Cleft(B%5Cright)u%5Cleft(B%5Cright)%2Bg%5Cleft(B%5Cright)v%5Cleft(B%5Cright)%3DI_n,有:

g%5Cleft(B%5Cright)%5Cleft(Au%5Cleft(B%5Cright)%2Bv%5Cleft(B%5Cright)%5Cright)%3DI_n

從而g%5Cleft(B%5Cright)可逆,A%3Df%5Cleft(B%5Cright)g%5E%7B-1%7D%5Cleft(B%5Cright),由f%5Cleft(B%5Cright)%2Cg%5Cleft(B%5Cright)可交換,可得

f%5Cleft(B%5Cright)g%5Cleft(B%5Cright)%3Dg%5Cleft(B%5Cright)f%5Cleft(B%5Cright)

從而

A%5Cleft(B-I_n%5Cright)%3DAg%5Cleft(B%5Cright)%3Df%5Cleft(B%5Cright)%3Dg%5Cleft(B%5Cright)f%5Cleft(B%5Cright)g%5E%7B-1%7D%5Cleft(B%5Cright)%3Dg%5Cleft(B%5Cright)A%3D%5Cleft(B-I_n%5Cright)A

于是AB%3DBA.

%5BQ.E.D%5D

(1)方法一和方法二利用矩陣的操作,使用湊因子法“具體地”給出了B-I_%7Bn%7D的逆陣(實(shí)際上文中并沒有詳細(xì)寫出),從而證明AB%3DBA,而方法三利用多項(xiàng)式理論證明了它的逆陣的存在性,沒有繁雜的計(jì)算,顯示出了多項(xiàng)式理論在解決某些問題時(shí)的強(qiáng)大

(2)本文可能與up主@SCHEME_maths 給出的解答類似,事實(shí)上本文的解法的思考以及撰寫是獨(dú)立完成的,但復(fù)旦大學(xué)高等代數(shù)每周一題2021A05思路分析與三種證法 - 嗶哩嗶哩 (bilibili.com)一文給出了解決這個(gè)問題的清晰思路,值得讀者參考.

(3)文末附上圖片格式的解法,有需要的讀者可以自行取用,僅供學(xué)習(xí)交流

問題2021A05


復(fù)旦大學(xué)謝啟鴻高等代數(shù)每周一題[2021A05]參考解答的評(píng)論 (共 條)

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