This article will give a detailed analysis and promotion of two advanced algebra test questions in the mathematics paper A of the 12th National College Students Mathematics Competition Preliminary Competition.
The third big questionAssume that are both orthogonal matrices of order 2020, and the solution space dimension of the homogeneous linear system of equations is 3. Question: Is it possible for the matrices to be similar? Prove your conclusion.
Idea analysisIf the matrices are similar, that is, there is a non-heterogeneous matrix , such that , then and . More generally, there are the same characteristic polynomials, and thus the same eigenvalues, and the trace and determinant of the matrix are just two coefficients in the characteristic polynomials (possibly different signs). Furthermore, note that are all orthogonal matrices (also real normal matrices), so from the orthogonal similarity standard form theory of real normal matrices (Theorem 9.7.3 of Fudan Advanced Textbook), it can be seen that, Orthogonal similarity has the same eigenvalues and is similar. With these analyses, it is not difficult to find that the conditions of the problem should not be sufficient to prove that there are the same eigenvalues. Therefore, we can try to prove that is not similar, and the common technique for dealing with this type of problem is to verify or . Which one we choose to start with for proof may determine how far we can go.
It is always a difficult problem to deal with two matrices at the same time. If multiplication is commutative, then has many simultaneous properties, such as simultaneous upper triangulation, simultaneous orthogonal normalization and simultaneous unitary diagonalization, etc. (Refer to the relevant chapters of the Fudan Advanced White Paper or Lecture 6 of the exercise class teaching video). Unfortunately, the conditions of the problem are not multiplication commutative. If only one of them becomes an orthogonal similarity standard form or a unitary similarity standard form, Then it is found that the simplification effect is not very good. These thoughts remind us: we should make full use of the properties of orthogonal matrices for simplification. All order orthogonal matrices form a group (this is one of the most essential properties of orthogonal matrices), that is, the product of orthogonal matrices, the inverse matrix of orthogonal matrices and the unit matrix are all orthogonal matrices. Using this property, two orthogonal matrices can be turned into one orthogonal matrix, which has a sufficient simplification effect (this technique has appeared many times in the high-generation white paper), Then, we can obtain by using the analysis of the eigenvalues of the orthogonal matrix, so that .
Proof of the third major questionIt can be seen that is an orthogonal matrix and is also an orthogonal matrix. From the non-isotropy of , it can be seen that the linear equations and have the same solution, so the solution space dimension of the linear equation is 3. In other words, the geometric multiplicity of the orthogonal matrix with respect to the eigenvalue 1 is equal to 3. Since the orthogonal matrix is also a unitary matrix, it can be diagonalized, so the geometric multiplicity of the eigenvalues is equal to the algebraic multiplicity, so the algebraic multiplicity with respect to the eigenvalue 1 is also equal to 3. It can be seen from the basic properties of an orthogonal matrix that its eigenvalues are composed of real eigenvalues and conjugate imaginary eigenvalues with modulus length 1. Since the order of is even, conjugate imaginary eigenvalues appear in pairs, and the eigenvalue 1 is odd-numbered, so the eigenvalues are also odd-numbered, so , that is, there is , so they must be dissimilar.
The above discussion on the eigenvalues of orthogonal matrices is very interesting. Similar discussions can prove the following two generalizations. The first one is the generalization of question 18 of the weekly question of Grade 19 Advanced Generation II, given by Kuang Qi, Grade 19 of Fudan School of Mathematics.
Promotion 1Assume that are all orthogonal matrices of order, satisfying and not the eigenvalues of . Prove: is the eigenvalue of .
proveAssume that is an orthogonal matrix, then the basic properties of the orthogonal matrix can be used to assume that the overall eigenvalues of are , , , where . If is not an eigenvalue of , then the eigenvalues have , so . Returning to this question, we can know from the conditions that , , so . Note that is still an orthogonal matrix, so must be an eigenvalue of .
The second one is a generalization of the third major question above. The following is a more general form.
Promotion 2 (Example 9.103 of the Advanced White Paper)Assume that are all orthogonal matrices of order, prove: if and only if is an odd number.
proveThe proof in the white paper uses the orthogonal similarity standard form theory of orthogonal matrices. Here we only use the basic properties of orthogonal matrices and the complex diagonalization to prove it. Turn the two orthogonal matrices into one orthogonal matrix , so we only need to prove: If is an orthogonal matrix of order, then if and only if is an odd number. From the basic properties of orthogonal matrices, it can be seen that the eigenvalues of are composed of real eigenvalues and conjugate imaginary eigenvalues with modulus 1, so equal and only if the algebraic multiplicity of the eigenvalue is odd. Since the orthogonal matrix is also a unitary matrix, it can be diagonalized, so the geometric multiplicity of the eigenvalue is equal to the algebraic multiplicity, so the algebraic multiplicity of the eigenvalue is odd if and only if the geometric multiplicity of the eigenvalue is odd, so the conclusion is proved.
Note 1It can be seen from Generalization 2 that if the order is an even number, then the solution space dimension of the linear equation system is an odd number, and therefore must be dissimilar. This is the generalization of the third major question.
The fourth major questionThe degree term (possibly 0) of a nonvalued unary degree polynomial (after merging similar terms) is called the second term. Find the first polynomials of all 2020 degree complex coefficients, such that for each complex root of , there is a nonvalued first polynomial of complex coefficients and , such that the second term coefficients of , and and are equal.
Idea analysisJudging from the test questions of the previous 11 national competitions, the content of polynomials is by no means the focus of the examination (in fact, polynomials are not the focus of advanced algebra teaching). The strange condition given in the question "the first polynomial and the second coefficient of are equal" reminds us: This question only uses the cloak of polynomials. In fact, it hopes that candidates can find all the complex roots of polynomials through Vieta's theorem and important methods in advanced algebra, and this important method is the solution theory of linear equations. If candidates can figure this out, the problem-solving process will become suddenly clear.
Answer to the fourth questionAssume all the complex roots of are , then it can be seen that, and all the complex roots combined are exactly . According to Vieta's theorem, the second coefficient of is equal to the sum of all complex roots multiplied by ; The remaining elements are or . Let , then we can get a system of linear equations
By simply finding the solution to the linear system of equations , we can obtain all the complex roots of the polynomial, thus completely determining . According to the theory of solving linear equations, we first need to calculate the rank of the coefficient matrix , for which we prove the following proposition.
Proposition 1Assume that is a square matrix of order, and its main diagonal elements are all 0, and the remaining elements are or , then it is a non-heterogeneous matrix.
proveThis is Proposition 1 in the teaching paper [4], where three proof methods are given: The first proof method is to use the combinatorial definition of determinants to discuss; the proof method two is to use the addition 2 transformation to discuss; the proof method three is to use the modulo 2 congruence to discuss. Due to the limited length of the article, I will not go into details of the proof here, and readers are asked to refer to the teaching paper [4]. It is worth mentioning that the official answer to the fourth major question uses two methods to deal with this proposition. Transformation method and modulo 2 congruence method.
It can be seen from Proposition 1 that the 2020-order square matrix in the fourth major question is a non-heterogeneous matrix, so the system of linear equations has only zero solution, that is, , so .
Promotion 3Find the first-order polynomials of all degree complex coefficients , such that for each complex root of , there exists a non-valued first-order polynomial of complex coefficients and , such that
(1) ;
(2) The second coefficient of and is equal;
(3) When is an odd number, the degrees of and are equal.
Brief explanationWhen is an even number, it can be seen from Proposition 1 that the system of linear equations has only zero solution, so . When is an odd number, the equal degree of and ensures that is the basic solution system of the system of linear equations (this conclusion happens to be the topic discussed in the teaching paper [4]), so , and thus , where is an arbitrary complex number.
In general, the two advanced test questions in Paper A of the Mathematics Category of the 12th National College Student Mathematics Contest Preliminary Competition are not difficult, but they better examine students' mastery and application of the theory of solving linear equations (including determinant theory, etc.) and the related properties of orthogonal matrices (including eigenvalues and complex diagonalizability, etc.). They should be considered more suitable competition questions.
This article provides such a detailed analysis of problem-solving ideas, and also wants to remind readers: the proficient application of important ideas, methods and techniques in advanced algebra not only needs to be achieved by doing a certain number of exercises (that is, brushing them), but also needs to carefully think about problem-solving ideas and repeatedly measure the feasibility of various methods before doing each exercise. Only in this way can we truly improve our problem-solving ability and innovation ability.
References
[1] Advanced algebra textbooks: edited by Yao Musheng, Wu Quanshui, and Xie Qihong, Advanced Algebra (Third Edition), Fudan University Press, 2014.
[2] Advanced Algebra White Paper: Edited by Yao Musheng and Xie Qihong, Learning Method Guide: Advanced Algebra (Third Edition), Fudan University Press, 2015.
[3] Xie Qihong, Fudan University Advanced Algebra Exercise Course Teaching Video, https://www.bilibili.com/video/av90771191/
[4] Xie Qihong, Shao Meiyue, Looking at several techniques in advanced algebra from an elementary problem, Advanced Mathematics Research, 2013, 16(4), 53--55.
Machine-translated from the Chinese original 第十二届全国大学生数学竞赛初赛数学类A卷两道高代试题的思路分析及其推广 (published 2021-01-12). WeChat original ↗