8. (10 points)Assume that the overall eigenvalues of the complex square matrix of order are , then the spectral radius of is defined as . Let be a real square matrix of order, so that is a positive semidefinite real symmetric matrix. It is proved that:
proveThe key point proved in this problem is the following lemma, which is to use the maximum and minimum eigenvalues of the real symmetric matrix (Hermite matrix) to estimate the upper and lower bounds of the values of the accompanying real quadratic form (Hermite type).
Lemma (a replica of Example 9.46 from the Advanced White Paper)Assume that is a Hermite matrix of order, in fact, the eigenvalue is
Then for any dimensional complex column vector , we have
Below we give three different proof methods. Among them, the function of the perturbation method is to reduce the positive semi-definite situation to the positive definite situation, thereby highlighting the core part of the proof. Of course, we can also use the properties of the positive semi-definite array to replace the perturbation method technique.
Proof 1 (block symmetric elementary transformation + perturbation method)It can be seen from the conditions that is a positive semi-definite real symmetric matrix. Let us first deal with the case where is a positive definite real symmetric matrix. Let the overall eigenvalues of
Then the overall eigenvalues of and are
Consider the following block symmetric elementary transformation (contract transformation):
It can be seen that is a positive semidefinite matrix, and it is also a positive semidefinite matrix. Taking any eigenvalue and its eigenvector of , we have and . Naturally regard as a positive semidefinite Hermite matrix, then we have
From the above lemma, we can get
Note that , thus , that is, , so . We use the perturbation method to deal with the situation of positive semi-definite. For any positive real number, it is a positive definite matrix, and at this time, is also a positive definite matrix. From the conclusion of the positive definite situation, we can know
Order and you will get .
Proof 2 (Orthogonally similar standard form + properties of positive semidefinite matrix)It can be known from the orthogonal similarity standard form theory of real symmetric matrices that there are orthogonal matrices such that
where is the positive eigenvalue of . Denote , then is a positive definite real diagonal matrix. Consider the following orthogonal similarity transformation:
It can be seen from that is a positive semidefinite matrix, the matrix on the right side of the above formula is also a positive semidefinite matrix. According to the property 1 of the positive semidefinite matrix (Example 8.43 of the Advanced White Paper), we can get , . In particular, if , that is, it can be obtained that , obviously the conclusion holds, so the following may be assumed to be . Notice that , so the overall eigenvalues of are composed of the overall eigenvalues of and , so that . Notice that , so . Notice is a positive semidefinite matrix and is a positive definite matrix, then from the discussion of the positive definite situation in Proof 1, we can get , so that .
Of course, we can also prove directly without using the proof of the positive definite case of Proof 1 (that is, without citing the above lemma). First, from the positive semi-definite of , we can get the positive semi-definite of . Secondly, taking any eigenvalue of and its eigenvector of , that is, and . Naturally regarded as a positive semi-definite Hermite matrix, we have
Finally, note that is not all zero, so there must be some such that , that is, , and thus .
Proof 3 (Estimation of Hermite type values)Naturally regarded as a positive semi-definite Hermite matrix, then for any , we have
Then . Taking any non-zero eigenvalue and its eigenvector of , there are and . In the above formula (1), let , , where , then we have
From the above lemma, we can see that, , so , and thus .
This can also be done. When is a positive definite matrix, let , , in the above formula (1), then we have
The conclusion can be obtained from the above lemma (similar to the discussion in Proof 1). When is a positive semi-definite, the perturbation method can be used for transition. In fact, this method is the Hermite version corresponding to Proof 1 (matrix method). Readers are asked to think about the correspondence by themselves.
NoteThe positive definite version of this question is: Let be a real square matrix of order, so that is a positive definite real symmetric matrix, then .
References
[1] Advanced Algebra White Paper: Edited by Yao Musheng and Xie Qihong, Learning Method Guide: Advanced Algebra (Third Edition), Fudan University Press, 2015.
Machine-translated from the Chinese original 复旦大学2019--2020学年第二学期19级高等代数II期末考试第八大题 (published 2020-10-29). WeChat original ↗