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Suppose I want to construct an $n\times n$ matrix ${\bf A}$ such that ${\bf A}^n={\bf I}$. Matrices that have period $n$ and admit such property are permutation matrices. However, I was wondering if there is a general methodology to obtain such matrices.

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closed as not a real question by Scott Morrison Dec 22 '09 at 18:20

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

1. What is your field? 2. What is "period n"? – Ho Chung Siu Dec 22 '09 at 17:01
I've closed this, as the question doesn't seem to make that much sense. Certainly n x n matrices over C with "period n" aren't necessarily permutation matrices. – Scott Morrison Dec 22 '09 at 19:21
Moreover, not all permutations matrices have "period n". E.g. n=5, and permute the first three terms and the last two. – Theo Johnson-Freyd Dec 22 '09 at 19:40

I'm guessing you didn't mean for the size of the matrix and the period to be equal, so let's assume that the matrix is k-by-k. For any such matrix, the eigenvalues must be nth roots of unity. Then you can construct families of such matrices by picking k different nth roots of unity, and then conjugating this by any invertible matrix. To be more explicit, pick k different numbers of the form $\omega_j = \exp(2 \pi i a_j/n)$ where each aj is an integer between 0 and n-1 of your choice, for j=1,...,k. Then form the matrix $\Lambda$ whose diagonal elements are $\Lambda_{jj} = \omega_j$, and pick an arbitrary invertible matrix $S$ and form $S \Lambda S^{-1}$.

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Let ${\bf A}\in \mathbb{R}^n$ and period $n$ stand for the fact that $\underbrace{{\bf A}\cdot{\bf A}\cdot\ldots\cdot{\bf A}}_{n \text{ times}}={\bf I}$

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(This is supposed to be a reply to the comment above) – anadim Dec 22 '09 at 17:19
I think you mean the elements of A are in $\mathbb{R}$... And you should add this to the question, rather than as an answer, which you can do by editing your post. – Steve Flammia Dec 22 '09 at 17:34
I know, but for some reason I lost the open id associated with my first account I used to ask. So I created a second which I used to reply here. If I could access my first one I would mark your answer as an "answer" for my question :). Thanks a lot for answering. – anadim Dec 22 '09 at 17:38
I'm confused, so what does "Matrices that have period n and admit such property" mean? (From your reply here it seems that period n = the property in your first sentence?) – Ho Chung Siu Dec 22 '09 at 17:40
Sorry to confuse you. I am calling period the property that if you multiply the initial matrix with itself $n-1$ times you will get the identity matrix. – anadim Dec 22 '09 at 17:46

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