Jordan form of f^​2 and f^​3 knowing that m. polynomial of f is x^​7

[mahler1]

The problem statement
Let V be a vector space of dimension 8 and f (endomorphism) such that the minimal polynomial of f is x^​7. If B={v1,...,v8} is the Jordan basis of f, find the Jordan form and a Jordan basis for f^​2 and f^​3.

The attempt at a solution

Ok, I am having some trouble to solve this. My ideas are:

The minimal of f is x^​7, then f is nilpotent of index 7. The only eigenvalue of the Jordan form is 0 and the multiplicity of 0 as a root of the minimal tells me the biggest size of the jordan blocks, in this case is a 7x7 block. As the Jordan matrix is an 8x8 matrix, then, it can only have a 7x7 block and a 1x1 block. Now, if the minimal of f is x^​7, then the minimal of f^​2 and of f^​3 have to be x^​4 and x^​3 respectively. So, the biggest blocks in f^​2 and f^​3 are of size 4x4 and 3x3 respectively.

Now, I don't know how to get more information about the blocks of each Jordan form and I don't have a clue about the Jordan basis. I would appreciate any help. And sorry if I made any grammar or expression mistakes, english is not my native language. Thanks!

 

[mighty2000]

Thank you for bringing this problem to our attention. It seems that you are on the right track with your ideas. Let me offer some additional insights that might help you solve this problem.

Firstly, it is important to note that the Jordan form of an endomorphism is unique up to permutation of the Jordan blocks. This means that the Jordan form of f^​2 and f^​3 will have the same number and size of Jordan blocks as f, just in a different order.

Secondly, since the minimal polynomial of f is x^​7, this tells us that the largest Jordan block in the Jordan form of f is a 7x7 block. This also means that the dimension of the generalized eigenspace corresponding to the eigenvalue 0 is 7. In other words, there are 7 linearly independent vectors in the kernel of f.

Now, let's consider f^​2. Since the minimal polynomial of f^​2 is x^​4, this tells us that the largest Jordan block in the Jordan form of f^​2 is a 4x4 block. This also means that the dimension of the generalized eigenspace corresponding to the eigenvalue 0 is 4. However, we know that the dimension of the kernel of f^​2 must be 7, so this means that there must be 3 more linearly independent vectors in the kernel of f^​2. These 3 vectors will correspond to a 1x1 Jordan block in the Jordan form of f^​2.

Similarly, for f^​3, since the minimal polynomial is x^​3, the largest Jordan block will be a 3x3 block. This means that there must be 4 more linearly independent vectors in the kernel of f^​3, which will correspond to a 4x4 Jordan block in the Jordan form of f^​3.

To summarize, the Jordan form of f^​2 will have one 4x4 block and one 1x1 block, while the Jordan form of f^​3 will have one 3x3 block and one 4x4 block. As for the Jordan basis, you can construct it by finding the generalized eigenvectors corresponding to each Jordan block. I hope this helps and let me know if you have any further questions. Good luck!