powers of a Markov matrix equals identity

Question

A Markov matrix is defined as a non-negative square matrix whose column sums are all $1$. How to find a Markov matrix $A$ such that $A^7$ is the identity matrix (and the lower powers are not the identity)?

Answer 1

$A$ and $A^{-1}=A^6$ have only $\geq 0$ entries. Then $A$ is a pseudo permutation.

Prove the above assertion and deduce an explicit solution of your exercise.

Answer 2

I'm writing in Matlab code so that you can verify this easily. The matrix A=[0 1; 1 0] satisfies $A^2=I$. This matrix represents two states that switching populations: S1->S2 and S2->S1.

Similarly, to get $A^3=I$, we can let S1->S2, S2->S3, S3->S1. This permutation has order $3$. A=[0 0 1; 1 0 0; 0 1 0] will do the job.

Therefore, one example of $A^7=I$ will be a matrix of $7\times 7$ which represents a permutation of order $7$.