8. (10 points)Let be a real symmetric matrix of order, satisfying: the sum of the elements in each row is equal to zero, and the non-main diagonal elements are less than or equal to zero. Assume the indicator set , the two indicators are called connected, if there is a column of indicators, such that , , , . Let the indicator set be divided into connected branches , that is, different indicators in the same connected branch are connected, and the indicators between different connected branches are not connected. Proof:
proveThrough row swapping and symmetric column swapping, the connected indicators can be put together, so as to offset (conform to) the real symmetric matrix , where is a real symmetric matrix of order, satisfying: the sum of the elements in each row is equal to zero, and the non-principal diagonal elements are less than or equal to zero. If is proved, then
Therefore, we only need to prove: For a real symmetric matrix that satisfies the conditions of the question, when all indicators are connected, it is enough. According to the solution theory of linear equations, this is equivalent to proving that: the dimension of the solution space of homogeneous linear equations is equal to .
On the one hand, assuming , we can know from the problem conditions. On the other hand, if we take any solution of , that is, , then we have
Using the inequalities and we can get:
Therefore, the equality sign of the above inequality holds true if and only if: if , then there must be . Since all indicators are connected, there must be , that is . Therefore, , thus .
NoteThis question is adapted from a lemma in algebraic surface theory, that is, if a curve (effective divisor) on a smooth projective algebraic surface shrinks to a point or is a fiber of the surface fibrosis, then the real symmetric matrix composed of the intersection numbers between its irreducible branches is a negative semi-definite matrix. In this question, from the second half of the above proof, it can be directly obtained that is a positive semi-definite matrix. Of course, it can also be proved like this: for any , are all strictly diagonally dominant matrices and all main diagonal elements are greater than zero, then it can be seen from Example 8.28 of the Advanced White Paper that for any , they are all positive definite matrices, and furthermore, it can be seen from Example 8.40 of the Advanced White Paper that is a positive semi-definite matrix.
Machine-translated from the Chinese original 复旦大学2019--2020学年第一学期19级高等代数I期末考试第八大题 (published 2020-11-03). WeChat original ↗