复旦大学谢启鸿高等代数每周一题[2021A05]参考解答

2021-10-31 17:58 作者:CharlesMa0606 0人读过 | 我要投稿

本文是本人给出的2021年复旦大学谢启鸿高等代数的每周一题[问题2021A05]的解答

题目来自于复旦大学谢启鸿教授在他的博客提供的每周一题练习

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

本文仅供学习交流，如有错误恳请指正！

[问题2021A05]设 $A%2CB$ 为 $n$ 阶方阵,满足:

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

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

解(方法一，变形配凑)

移项可得 $A%3DAB-%5Cleft(a_mB%5Em%2Ba_%7Bm-1%7DB%5E%7Bm-1%7D%2B%5Ccdots%2Ba_1B%5Cright)$ 两边左乘 $B$ ，有

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

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

注意到我们令 $B%3DB-I_n%2BI_n$ 代入题中式子可得

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

从而 $B-I_n$ 可逆，于是 $AB%3DBA$ .

$%5BQ.E.D%5D$

(方法二，类似方法一，但直接代入 $B%3DB-I_n%2BI_n$ )

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

二项式展开后移项，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 $B$ 和 $I_n$ 的矩阵多项式,从而与 $B-I_n$ 可交换，而 $B-I_n$ 和 $A-f%5Cleft(B%5Cright)$ 也可交换

从而

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

$%5BQ.E.D%5D$

(方法三，利用多项式理论)

设多项式 $f%5Cleft(x%5Cright)%3Da_mx%5Em%2Ba_%7Bm-1%7Dx%5E%7Bm-1%7D%2B%5Ccdots%2Ba_1x%2Cg%5Cleft(x%5Cright)%3Dx-1$ ，因为 $a_m%2Ba_%7Bm-1%7D%2B%5Ccdots%2Ba_1%5Cneq0$ ，从而 $f%5Cleft(1%5Cright)%5Cneq0$ ，于是 $%5Cleft(f%5Cleft(x%5Cright)%2Cg%5Cleft(x%5Cright)%5Cright)%3D1$ .

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

我们代入矩阵 $B$ ，可以得到： $f%5Cleft(B%5Cright)u%5Cleft(B%5Cright)%2Bg%5Cleft(B%5Cright)v%5Cleft(B%5Cright)%3DI_n$

由题， $Ag%5Cleft(B%5Cright)%3Df%5Cleft(B%5Cright)$ ，从而 $Ag%5Cleft(B%5Cright)u%5Cleft(B%5Cright)%2Bg%5Cleft(B%5Cright)v%5Cleft(B%5Cright)%3DI_n$ ，有：

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

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

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

从而

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