Show that a matrix has strictly positive real eigenvalues topology exercise Let $M$ be a $3 \times 3$ matrix with strictly positive real entries.  Let 
$$S \ = \   \left\{ \left. \ \left(\begin{array}{c} x \\ y \\ z\end{array}\right) \in \mathbb{R}^3 \  \right| \ x \geq 0, \ y \geq 0, \ z\geq 0, \ x^2+y^2+z^2=1 \right\}.$$
By considering the map  $f:S \to S$ defined by  $\mathbf{x} \mapsto M \textbf{x} / |\!| M \textbf{x}|\!|$, show that $M$ has a strictly positive real eigenvalue.
I am looking for a careful explanation, would be super helpful to me. This is a study question on an old exam, since some people wanted to know the origin of the question. I am quite confused by the suggestion "considering the map.." so wanted to ask here.

Note that the set $S$ is homeomorphic to a closed unit disk and $f:S \to S$ is continuous. Thus the Brouwer fixed-point theorem says that there is $v\in S$ so that $f(v) = v$. Thus 
$$ Mv = \|Mv\| v.$$
As $v\neq 0$, $\|Mv\|$ is an eigenvalue of $M$. It's strictly positive (you should have checked that when defining $f$) and thus you are done. 
