# How to construct matrices with periodicity [closed]

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

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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 $S \Lambda S^{-1}$.
Let ${\bf A}\in \mathbb{R}^n$ and period $n$ stand for the fact that $\underbrace{{\bf A}\cdot{\bf A}\cdot\ldots\cdot{\bf A}}_{n \text{ times}}={\bf I}$
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
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