Positive eigenvectors for nonnegative matrices Let $ A $ a nonnegative matrix (i.e. all the entries of $ A $ are real, nonnegative) of dimensions $ n \times n $. Is it true that the conditions:


*

*$ A x_1 = \lambda_1 x_1 $

*$ A x_2 = \lambda_2 x_2 $

*$ x_1 \gg 0, x_2 \gg 0 $ (i.e. all the components of $ x_1 $ and $ x_2 $ are strictly positive)


imply $ \lambda_1 = \lambda_2 $? (obviously $ \lambda_1, \lambda_2 $ are real scalars, while $ x_1, x_2 $ are real vectors).

Yes. This is because when $A\ge0$, its spectral radius $\rho(A)$ is the only eigenvalue (over $\mathbb C$) that can possibly possess a positive eigenvector. (That doesn't mean $\rho(A)$ always has a positive eigenvector --- consider e.g. $A=$ the $2\times2$ nontrivial nilpotent Jordan block. Nor does it mean that $\rho(A)$ is the only eigenvalue that has a nonnegative eigenvector --- consider e.g. $A=\operatorname{diag}(1,0)$.)
Suppose the contrary that $Av=\lambda v$ for some $v>0$ and $\lambda\in\mathbb C\setminus\{\rho(A)\}$. Then $\lambda$ must be a positive real number. Hence $0\le\lambda<\rho(A)$. Let $B=\frac1{\rho(A)}A$ and $b=\frac{\lambda}{\rho(A)}$. Then $Bv=bv$ and $0\le b<1$. It follows that
$$
\lim_{m\to\infty}v^TB^mv=\lim_{m\to\infty}b^m\|v\|^2=0.\tag{1}
$$
Since $v^TB^mv$ is a positively weighted sum of all entries of a nonnegative matrix, $(1)$ implies that $\lim_{m\to\infty}B^m=0$. Consequently,
$$
\lim_{m\to\infty}\rho(B)^m=\lim_{m\to\infty}\rho(B^m)=\rho\left(\lim_{m\to\infty}B^m\right)=0.
$$
But this is a contradiction because $\rho(B)=\rho\left(\frac1{\rho(A)}A\right)=1$.
Edit. Here is a simpler proof if one knows Perron-Frobenius theorem. Let $Av=\lambda v$ for some $\lambda>0$ and $v>0$. By Perron-Frobenius theorem $A$ has a nonnegative left eigenvector $u$ for the eigenvalue $\rho(A)$. Then $\rho(A)u^Tv=(u^TA)v=u^T(Av)=\lambda u^Tv$. Since $u\ne0,\,u\ge0$ and $v>0$, the product $u^Tv$ is positive. Therefore $\rho(A)=\lambda$, i.e., the only possible eigenvalue that has a positive eigenvector is $\rho(A)$.
