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.

closed as not a real question by Scott Morrison♦ Dec 22 '09 at 18:20
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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 kbyk. 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 a_{j} is an integer between 0 and n1 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 $\underbrace{{\bf A}\cdot{\bf A}\cdot\ldots\cdot{\bf A}}_{n \text{ times}}={\bf I}$ 

