Let $A$ and $x$ be $n \times n$ and $n \times 1$ matrices, all entries real and strictly positive. Assume that $A^2 x = x$. Show that $A x = x$.

Solution 1:

The Perron-Frobenius Theorem says, among others, the following: A square matrix $A$ with strictly positive entries has a positive eigenvalue $\rho$ and corresponding eigenvector $y$ with strictly positive entries. All other eigenvalues of $A$ are strictly smaller in absolute value, and the corresponding eigenvectors do not have all positive entries.

In the situation at hand we are given a matrix $A$ with positive entries. Let $\rho$ and $y$ be as in the theorem. The matrix $A^2$ has positive entries as well, and $y$ is an eigenvector of $A^2$ with corresponding eigenvalue $\rho^2$. Since we are told that the all-positive vector $x$ is an eigenvector of $A^2$ with corresponding eigenvalue $1$, it follows from the uniqueness part of the theorem that in fact $\rho^2=1$ and $x=\lambda y$ for a $\lambda>0$. This allows to conclude that $Ax=x$.