Unit close disc to prove a matrix algebra identity? I need to prove that every $3 \times 3$ matrix with real positive entries has one eigenvector with a positive eigenvalue. Now, how do I prove this using the fact that the set $B=\{x=(x_1,x_2,x_3)\in \mathbb{R}^3 : x_i \geq 0 \}$ and also $|x|=1$ is homeomorphic with the closed unit disc?
 A: The strategy to use the Brouwer fixed-point theorem is obvious, right?
So, you're just looking for a continuous function on the closed unit disc in the plane. I'm going to use $\Bbb R^{3+}$ as an ad-hoc notation for $\{(x_1,x_2,x_3)\mid x_i\geq 0\}$. And also I'm using $B:=\{x\in \Bbb R^{3+}\mid |x|=1\}$ as I think you intended.
The hypothesis that the matrix you start with has positive entries says that it is a transformation $T$ that maps $\Bbb R^{3+}$ onto itself. It doesn't map $B$ onto itself: in general vectors get stretched. But we can correct this by composing with the continuous map $n:\Bbb R^{3+}\setminus{\{0\}}\to B$ which just maps $x\mapsto \frac{x}{|x|}$. In summary, $nT:B\to B$ is a continuous map.
Let $h$ be a homeomorphism of $\Bbb R^{3+}$ onto the closed unit disc $D$ in the plane. Then $hnTh^{-1}$ is a continuous function on $D$, so it has a fixed point, say $x$.
Then $hnTh^{-1}(x)=x$ implies $nT(h^{-1}(x))=h^{-1}(x)$, so $nT$ has a fixed point too. Can you see why $h^{-1}(x)$ is an eigenvector? What can you say about its eigenvalue?
