7. (10 points) Assume that the square matrix of order on the number field satisfies and , prove: .
Proof We give four different proofs.
Proof 1 (Theory of solving linear equations)If is not different, then , and thus . If , then , and thus . Assume , then according to the solution theory of linear equations, it can be known that the dimension of the solution space of is equal to , so we might as well assume , where is the basic solution system of . It can be seen that each column vector of belongs to the solution space of , and thus is proportional to , so there exists , so that . By the same reason, we can know that there exists , so that . From the commutativity of the matrix trace, we can get
So
Proof 2 (eigenvalue theory)If is not different, then , and thus . If , then , and thus . Assume that , then the dimension of the eigensubspace with respect to eigenvalue 0 is equal to , so we might as well assume , where is the eigenvector of the eigenvalue of . It can be seen that each column vector of belongs to the eigensubspace, and thus is proportional to , so there exists such that. Since , therefore any eigensubspace of is The invariant subspace of . In particular, is the invariant subspace of , so there exists , such that , that is, is the eigenvector with respect to the eigenvalue of . Note that and , so from Example 7.28 of the Advanced White Paper (using the reduced-order formula of eigenvalues), we know that the eigenvalues of are all zero (this conclusion can also be derived from the eigenvalues of and ), so , and thus , therefore .
Proof 3 (Application of Cayley-Hamilton Theorem)It can be seen from the proof of Example 6.62 in the Advanced White Paper: If the characteristic polynomial of
then there is
From Example 6.15 of the Gaoda White Paper, we can see:
So
Proof 4 (Application of Jordan Standard Type)If is not different, then , and thus . If , then , and thus . Assume that . Note that the conditions and conclusions of the problem do not change under simultaneous similar transformations , (where ), so we might as well assume from the beginning that is the Jordan standard form, where is a block diagonal matrix composed of Jordan blocks with non-zero eigenvalues, and is a block diagonal matrix composed of Jordan blocks with zero eigenvalues. Let
is the corresponding block, then it can be known from
By non-exclusive, we get , and . Note that the geometric multiplicity of the eigenvalue is , so it is composed of a Jordan block, we assert that . Using proof by contradiction, if , then
Therefore, it is a contradiction! It can also be proved in this way, noting that is the algebraic multiplicity of the eigenvalue of , if , then the eigenvalue of is , , so the eigenvalue of is only , and the other eigenvalues are all zero, thus, it is a contradiction! Therefore,
Among them
It can be seen that the following rows of are equal to zero, so , and thus .
NoteIt can be seen from Proof 1 and Proof 3 that the conclusion of this question can be improved to: Assume that the square matrix of order on the number field satisfies , then the necessary and sufficient condition of is or .
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期末考试第七大题 (published 2020-10-12). WeChat original ↗