I got the following statement to prove: Let $A \in \mathbb R^{n \times n}$ be a (column-)stochastic matrix, i.e. $A \geq 0$ (which means $a_{ij} \geq 0$ $\forall i, j \in \{1, \dots ,n\}$) and $\sum_{i=1}^n a_{ij} = 1$ $\forall j \in \{1, \dots ,n\}$. Then there is a vector $x \in \mathbb R^n \setminus \{0\}$ such that $x \geq 0$ (which means $x_i \geq 0$ $\forall i, \in \{1, \dots ,n\}$) and $Ax = x$.
I'm this far already. For the adjoint $A^*$ of $A$ and $y = (1, \dots, 1) \in \mathbb R^n$ we have $A^*y = y$. So $1$ is an eigenvalue of $A^*$ and it follows that $1$ is also an eigenvalue of $A$, since $\lambda \in \sigma(A) \Leftrightarrow \overline{\lambda} \in \sigma(A^*)$. Hence there exists a $x \in \mathbb R^n \setminus \{0\}$ with $Ax = x$.
The only problem I got is to show that $x \geq 0$. So I would be thankful for helpful advices.
Consider the map $f$ defined by $x \mapsto \frac{Ax}{\sum_i (Ax)_i}$ defined on the (topological) disk $D$ that consists of vectors $x$ satifying $x_1 \ge 0, x_2 \ge 0, \ldots, x_n \ge 0, x_1 + x_2 + \ldots + x_n = 0$ (i.e., $D$ is the standard simplex in the positive octant).
Then $$ f : D \to D $$ is a continuous map of a closed disk to itself (this requires a sentence or two of proof...how do we know all entries of $Ax$ are positive? How do we know they're not all zero so that the division makes sense?), and hence has a fixed point, by the Brouwer theorem. This fixed point is a positive eigenvector for $A$.