7. (10 points) Suppose is a linear transformation on a dimensional linear space, , where are all -invariant subspaces. Prove:
(1) For any positive integer,
Both are -invariant subspaces;
(2) There exists a positive integer , such that for any , , , and
prove(1) It is easy to verify that and are both subspaces of . Take any , that is, since is an invariant subspace, so , thus
That is, , so is an -invariant subspace. Note that the above formula also tells us: , that is, , so there is an inclusion relationship of the subspace:
Take any , where , since is an invariant subspace, so , thus
That is, is an invariant subspace. Note that the above formula also tells us: , so there is an inclusion relationship of the subspace:
(2.1) From the above discussion, it can be seen that there is the following inclusion relationship of -invariant subspace:
There are thus inequalities in dimensions:
Note that the increasing and decreasing integer (subspace dimension) sequences are sandwiched between two finite intervals, so they cannot have infinite strict increases and infinite strict decreases. Therefore, there are positive integers, so that for any,
That is, starting from the index , the dimensions of the two types of subspaces remain unchanged, and then based on the inclusion relationship of the subspaces: for any ,
Proof method 1: We follow the definition of direct sum and prove it in two steps
(2.2) Prioritize . Take any , that is, exist , such that and , so , so there is , thus . There is another proof method. Limit the linear transformation to , and get the linear map , which is a surjection. From (2.1) we can also know that , so is a linear isomorphism. Note that in the above discussion, , thus , so .
(2.3) Proof again. Take any , suppose , where , , then , that is, exists, such that , that is . Let , then , thus
Proof method twoWe can use the dimension formula to replace (2.2) or (2.3) above to prove the direct sum, but this dimension formula is not obvious and requires strict proof.
(2.4) Proof
Limiting the linear transformation to , we obtain the linear mapping . Note that , , thus we can obtain from the dimension formula of the linear mapping
Since , , therefore , that is, , thus
due to
Therefore, we can get from the formula
From the two equations of and , and then using the dimension formula of , we can get the proof.
Proof method threeWe can also directly prove that . Limiting the linear transformation to the invariant subspace, we obtain the linear transformation . From the conclusion of Example 4.33 of the Advanced White Paper, we can see that: there is a positive integer such that for any , , , and
Note that it is easy to verify from the definition, so
Therefore, the above inclusion relationship can only take the equal sign, that is, there are and .
NoteWhen , this question is Example 4.33 of Gao Dai White Paper. Therefore, this question and its proofs 1 and 2 are the generalization of Example 4.33 of Gao Dai White Paper and its proof. You can carefully compare the similarities between the two. The second question can take a more refined value (discussed using the drawer principle like Example 4.32 of Gao Dai White Paper), but as far as the conclusion of this question is concerned, There is no need for such a detailed discussion.
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期末考试第七大题 (published 2020-11-03). WeChat original ↗