4. Solve the following linear equations, where are parameters and:
solutionAssuming that the coefficient matrix of the above system of equations is , then it can be obtained by the summation method or the reduced order formula of the determinant
Three situations are discussed below.
Case 1 If , then the original system of equations is equivalent to a linear equation
And , so we might as well discuss as (other situations are completely similar), at this time the basic solution system of the original system of equations is , , , .
Case 2If , then it is obviously a non-zero solution to the system of equations. Let's use the elementary transformation method to find . Add all the last columns of to the first column, and then multiply the first row by and add them to the next rows in turn:
Obviously, the rank of the matrix on the right is equal to , so . We can also use the reduced order formula of the matrix rank to find . Suppose , then , so from the Advanced White Paper Example 3.73, we know that,
So . According to the theory of solving linear equations, it can be known that the dimension of the solution space of the original system of equations is equal to , so the basic solution system of the original system of equations is .
Case 3If and , then , so according to Cramer's rule, the original system of equations has only zero solution.
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5. Assume that the square matrix of order satisfies and , prove: is a singular matrix. Prove that from can be obtained and , so . So therefore , that is, is a singular matrix.
Note: This question is similar to Example 2.18 of the Gaoda White Paper.############
6. Assume a square matrix of order , and the following determinants are called sequential principal subforms of :
Assume , prove: Several row and column swaps can be performed on , so that the first sequential principal subexpression of the resulting matrix is all non-zero, and the subsequent sequential principal subexpressions are all zero.
proveAssuming that is its row block, its maximal independent group can be swapped to the previous row through row swapping, so we might as well assume that is the maximal independent group of the row vectors of 3.13 It can be seen that, , , are also linearly independent. Note that
Therefore, one column can be selected from the rear column to form a maximum independent group of column vectors together with the front column, and this column can be exchanged with the th column. Therefore, it is possible to assume that the front column of the matrix is a maximum independent group of its column vectors, so
It can be obtained from the above induction method that after several row swaps and column swaps, the first sequential principal subformula of the matrix is non-zero. And , so it can be seen from the subformula discrimination method of matrix rank that the last sequential principal formula of is all zero.
NoteThis question is similar to the proof of theorem 3.6.2 (subformulation method of matrix rank) in advanced textbooks.
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7. A given non-negative real number satisfies the following two conditions: (i);
(ii) There exists a total permutation of , such that .
For any , let
Define order square matrix as:
Proof:
(1) The solution space of the linear equations is , where ;
(2) The solution space of the linear equations is , where ;
(3) Assume , then the necessary and sufficient conditions for having a solution are .
Proof method one(1) Obviously is included in the solution space of . On the contrary, for any solution of , there exists , so that . Consider the equation of , notice that , so we can get
Therefore, the above inequality takes the equal sign, that is, if , then there must be . For the convenience of writing, we might as well assume the above. From condition (ii), it can be seen that , so is also the maximum value. Repeat the above discussion for the third equation, and we can get that is also the maximum value. Continue this discussion, and finally we can get
Therefore, the conclusion is proved.
(2) Obviously is included in the solution space of . Then from (1) and the solution theory of linear equations, we can see that , so , and thus the dimension of the solution space is equal to , so the solution space of is .
(3) The conclusion can be drawn from Example 3.99 of the Advanced White Paper and (2).
Proof Method 2: Notice that , therefore assume
Then is a real symmetric matrix and the sum of the elements in each row and column is zero.
(1) Prove that the solution space of is , which is equivalent to proving that the solution space of is . Obviously is included in the solution space of . On the contrary, taking any solution of , then we have
Note that and , so from the above formula we can get
Therefore, all the inequality signs in can only take the equal sign, that is, if , then there must be . By assuming that there is a complete permutation of , such that , , , , we can get
Therefore, the conclusion is proved.
(2) Prove that the solution space of is , which is equivalent to proving that the solution space is , which is also equivalent to proving that the solution space of is , and this is the conclusion of (1).
(3) The conclusion can be drawn from Example 3.99 of the Advanced White Paper and (2).
Note: This question is similar to the eighth question in the final exam of Grade 19 Advanced Algebra I. ############ References
[1] Advanced textbooks: Advanced Algebra (Fourth Edition), edited by Xie Qihong, Yao Musheng, and Wu Quanshui, Fudan University Press, 2022.
[2] High generation white paper: Advanced Algebra (Fourth Edition), edited by Xie Qihong and Yao Musheng, Fudan University Press, 2022.
Machine-translated from the Chinese original 复旦大学数学学院23级高等代数I期中考试精选大题解答 (published 2023-12-14). WeChat original ↗