TITLE: The intuition behind the Hilbert projective metric and the Perron Frobenius Theorem
QUESTION [9 upvotes]: Recently I have read a proof of the Perron Frobenius Theorem for positive aperiodic matrices. In this proof, the trick is to put a metric in the "positive quadrant" of $\mathbb{R}^n$, $\mathbb{R}^{n}_+$, in order to make the map $\mathbb{R}^n_+\ni p\mapsto pA $ a $\lambda$-contraction, $0<\lambda<1,$ and in this way applies the Contraction Fixed Point Theorem to obtain the only one fixed point for $A.$
To obtain this kind of result, $\mathbb{R}^n_+$ is "projectivezed" and the
 $\textbf{Projective Hilbert Metric}$ is considerate.  https://en.wikipedia.org/wiki/Hilbert_metric
This proof was given by G. Birkhoff, http://www.jstor.org/stable/1992971 
My Question: What is  the intuition behind the Hilbert Projective metric, there exists an easy way to see that the projective metric $d$ is the right metric  to make $(\mathbb{R}^n_+\ni p\mapsto pA, d) $ a  $\lambda$-contraction ?

REPLY [2 votes]: There is actually a paper written on this topic by E. Kohlberg and W. Pratt, named The  contraction  mapping  approach  tothe  Perron-Frobenius  theory  :  why  Hilbert’s  metric ? published in Math. Oper. Res.7.2 (1982). They explain there why the Hilbert metric is contracting.
They also prove something which is worth noting here. Let $K$ be a convex cone in $\mathbb{R}^d$. Say that $K$ is pointed if $K\cap -K=\{0\}$. A projective metric on a pointed cone $K$ is an application $D:K\times K \to \mathbb{R}_{\geq 0}$ which is symmetric, satisfies the triangle inequality and the following separation condition : if $D(x,y)=0$, then $x=\lambda y$ for some $\lambda\geq 0$. Typically, the Hilbert metric is a projective metric on the pointed cone of non-negative vectors.
Letting $K$ be a cone, say that a linear transformation $A$ is positive (with respect to $K$) if $Ax\in K$ whenever $x\in K$. When given a projective metric $D$ on $K$, define then
$$k_D(A)=\mathrm{sup}  \left ( \frac{D(Ax,Ay)}{D(x,y)},D(x,y)>0 \right ).$$
Then, $A$ induces a contraction on $K$ for the metric $D$ if and only if $k_D(A)<1$.
E. Kohlberg and W. Pratt define in general a Hilbert metric on any pointed cone $K$, which is the same as the metric described by @VaughnClimenhaga. They prove that any positive linear transformation $A$ induces a contraction for this Hilbert metric. They also prove the following.
Theorem (Kohlberg, Pratt) Let $K$ be a pointed cone in $\mathbb{R}^d$ and let $d$ be the Hilbert metric on $K$ and let $D$ be another projective metric on $K$.
Let $A$ be a positive linear transformation.
Then, there exists a function $f:\mathbb{R}\to\mathbb{R}$ such that for any $x,y$, $D(x,y)=f(d(x,y))$.
Moreover, $k_d(A)\leq k_D(A)$.
Consequently, not only the Hilbert metric yields a contraction, but this is the optimal metric doing so (since it has the best possible contraction coefficient).