Timeline for Similarity classes of invertible matrix
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Aug 8, 2016 at 7:25 | comment | added | Geoff Robinson | You have to read the whole sentence! The matrix $A$ is diagonalizable over some extension of $\mathbb{Z}/p\mathbb{Z}$ since $k$ is prime to $p$. A matrix is diagonalizable (over some extension field) if and only if its minimum polynomial has no repeated factors. But the condition about the multiplicative order of the eigenvalues of $A$ is also necessary. | |
| Aug 8, 2016 at 1:09 | comment | added | N. SNANOU | ok thank you so much. But can you explain why "the matrix $A \in {\rm GL}_{n}(\mathbb{Z}/p\mathbb{Z})$ will have order $k$ if and only if the minimum polynomial $f(x) \in \left(\mathbb{Z}/p\mathbb{Z}\right) [x]$ of $A$ is !!multiplicity free!!", | |
| Aug 7, 2016 at 21:56 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
clarification
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| Aug 7, 2016 at 20:38 | comment | added | Geoff Robinson | I have added a note in the answer. | |
| Aug 7, 2016 at 20:37 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Expanded answer
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| Aug 7, 2016 at 20:22 | comment | added | Geoff Robinson | A splitting field for $X^{k}-1$ is ${\rm GF}(p^{m})$, where $m$ is defined as in the question. But there is still work to do. | |
| Aug 7, 2016 at 20:08 | comment | added | N. SNANOU | Yes because a matrix is similar to exactly one rational canonical form, then it suffices to see the decomposition of the plynomial $X^k-1$ in $\mathbb{Z}/p\mathbb{Z}[X]$ that I am not sure what do about it. | |
| Aug 7, 2016 at 13:49 | comment | added | Geoff Robinson | This is now a question about rational canonical form. The answer is in some ways a routine computation though it gets a little messy. | |
| Aug 7, 2016 at 13:30 | comment | added | N. SNANOU | yes, thank you very much. Have you any idea about conjugacy classes of elements of order $k$ in ${\rm GL}_n(\mathbb{Z}/p\mathbb{Z})$?. | |
| Aug 6, 2016 at 9:53 | history | answered | Geoff Robinson | CC BY-SA 3.0 |