= Solution
The two-dimensional <Brouwer fixed-point theorem> states that every continuous map from a closed filled triangle to itself has a fixed point. Use the triangle $\Delta=\{x\in\mathbb R^3:x_i\geq0,\ \sum_i x_i=1\}$, which lies in a two-dimensional affine plane. Strict positivity of the matrix entries ensures $(Ax)_i>0$ for every $x\in\Delta$. Therefore
$$
T(x)=\frac{Ax}{\sum_i(Ax)_i}
$$
is a continuous self-map of $\Delta$, in fact into its interior. Its fixed point $x$ satisfies $Ax=\lambda x$, where $\lambda=\sum_i(Ax)_i>0$. Since $x=T(x)$ lies in the interior, \b[$A$ has an <eigenvector> all of whose entries are strictly positive]. This is <positive matrix eigenvector from simplex normalization>; it uses strict positivity, not merely nonnegative entries.
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