# 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 $A x = x$.

I am stuck for weeks with the following problem:

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

This was on the first problem set on a course of linear algebra based on the book written by Hoffman & Kunze. We haven't seen eigenvalues and eigenvectors yet. So, while any solution that uses anything more advanced than the first 3 chapters of that book is welcome (it may incentivize me to study something!), it does not solve the problem.

Can anyone help? Thanks!

EDIT: My question was marked as a exact duplicate of a question by Igor Caetano Diniz. While that is the exact same question, that post has one wrong answer and one answer that has a theorem that I haven't studied yet. So it doesn't solve my problem.

• Which topics are covered in the first 3 chapters of the book? Jul 2 '16 at 22:33
• Linear equations, vector spaces and linear transformations Jul 2 '16 at 22:43
• Edited stating why I think that my question is not a duplicate. Jul 2 '16 at 23:03
• The answers there don't seem suitable, I agree.
– Hoot
Jul 3 '16 at 0:51
• I have written an outline of an elementary proof that involves only the concept of linear independence. You may see if it's useful. Jul 3 '16 at 1:50

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 $Ax=x$.