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复旦大学谢启鸿高等代数每周一题[2021A09]参考解答

2021-11-22 22:22 作者:CharlesMa0606  | 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2021A09]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2021A09]设A为列满秩的m%5Ctimes%20n实矩阵.

(1)求证:A%5E%5Cprime%20A为非异阵.

(2)设A的第一列元素全为1,令P%3DA%5Cleft(A%5E%5Cprime%20A%5Cright)%5E%7B-1%7DA%5E%5Cprime,求证:P的所有主对角元素都大于等于%5Cfrac%7B1%7D%7Bm%7D.

(1)我们使用Cauchy-Binet公式,可以得到:

%5Cleft%7CA%5E%5Cprime%20A%5Cright%7C%3D%5Csum_%7B1%5Cle%20i_1%3C%5Ccdots%3Ci_n%5Cle%20m%7D%7BA%5E%5Cprime%5Cleft(%5Cbegin%7Bmatrix%7D1%262%26%5Ccdots%26n%5C%5Ci_1%26i_2%26%5Ccdots%26i_n%5C%5C%5Cend%7Bmatrix%7D%5Cright)A%5Cleft(%5Cbegin%7Bmatrix%7Di_1%26i_2%26%5Ccdots%26i_n%5C%5C1%262%26%5Ccdots%26n%5C%5C%5Cend%7Bmatrix%7D%5Cright)%7D

%3D%5Csum_%7B1%5Cle%20i_1%3C%5Ccdots%3Ci_n%5Cle%20m%7D%7BA%5Cleft(%5Cbegin%7Bmatrix%7Di_1%26i_2%26%5Ccdots%26i_n%5C%5C1%262%26%5Ccdots%26n%5C%5C%5Cend%7Bmatrix%7D%5Cright)%5E2%7D%3E0

从而A%5E%5Cprime%20A为非异阵.

或者我们可以考虑线性方程组Ax%3D0A%5E%5Cprime%20Ax%3D0的解的关系,若A%5E%5Cprime%20Ax%3D0,则x%5E%5Cprime%20A%5E%5Cprime%20Ax%3D0,即%5Cleft(Ax%5Cright)%5E%5Cprime%5Cleft(Ax%5Cright)%3D0,从而Ax%3D0.反过来显然.

于是线性方程组Ax%3D0A%5E%5Cprime%20Ax%3D0的同解,即A%5E%5Cprime%20Ax%3D0只有零解,于是A%5E%5Cprime%20A是非异阵.

(2)注意到PA%3DA%2CA%5E%5Cprime%20P%3DA%5E%5Cprime%2CP%5E%5Cprime%3DP%2CP%5E2%3DP,于是根据A的第一列元素全为1可知P的每行每列之和都为1,并且

P_%7Bii%7D%3D%5Cleft(PP%5E%5Cprime%5Cright)_%7Bii%7D%3D%5Csum_%7Bj%3D1%7D%5E%7Bm%7DP_%7Bij%7D%5E2%5Cgeq%5Cfrac%7B1%7D%7Bm%7D%5Cleft(%5Csum_%7Bj%3D1%7D%5E%7Bm%7DP_%7Bij%7D%5Cright)%5E2%3D%5Cfrac%7B1%7D%7Bm%7D.

或者我们也注意到对任何%5Calpha%5Cneq0%5Cin%20R%5Em%EF%BC%8C%5Calpha%5E%5Cprime%20P%5Calpha%3D%5Cleft(A%5E%5Cprime%5Calpha%5Cright)%5E%5Cprime%5Cleft(A%5E%5Cprime%20A%5Cright)%5E%7B-1%7D%5Cleft(A%5E%5Cprime%5Calpha%5Cright)%5Cgeq0,从而P是半正定阵,于是P的所有主子式非负.考虑大于等于二阶主子阵中绝对值最大的元素,若其不在对角线,则可得出绝对值最大的元素所在的二阶主子式是负值,这是不可能的,于是每个主子阵的绝对值最大的元素都在对角线上,这也就说明了对角线元素不会比非对角线元素小,从而由P的每行每列之和都为1可以得到对角线元素大于等于%5Cfrac%7B1%7D%7Bm%7D.

%5BQ.E.D%5D

  文末附上图片格式的解法,有需要的读者可以自行取用,仅供学习交流

问题2021A09


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