Geometric Proof of Perron-Frobenius

Note that Jordan form implies that there is invariant 2-plane $P$ containing $v$. If there is $w\in P$ s.t. $Tw=0$, then it is a contradiction by considering a linear map $T$ on $C$.

Hence we have a diffeomorphism $f: S^1\rightarrow S^1,\ f(w )=\frac{T^2w}{|T^2w|}$ s.t. $f(\pm v)=\pm v$ (cf. By orientation, we consider $T^2$).

Define a vector field $X =d(x,f(x)) e$ where $d$ is a metric on $S^1$ and $e$ is unit vector pointing $f(x)$ at $x\in S^1$. By continuity, there is a fixed point of $f$, which is not in $\{ \pm v\}$.

[old] In further, considering $C$, $ f$ sends a compact set $ S^1-C-(-C)$ onto $S^1-C-(-C)$.

For $x_1\in S^1-C-(-C)$, let $x_n=f(x_{n-1})$. By compactness, $x_n\rightarrow x$ so that $f(x_n)\rightarrow f(x)$. That is, $f$ has a fixed point. It is a contradiction.

I don't understand the figure in the paper, but we may argue as follows.

Suppose the eigenvalue $\lambda>0$ with eigenvector $v>0$ is not simple. Then there exists some vector $u$, which is independent of $v$, such that $Tu=\lambda u+cv$ for some $c\in\{0,1\}$. Pick a real number $t$ such that $w=tv-u\in\partial\overline C$. This is always possible because $w>0$ when $t\to+\infty$ and $w<0$ when $t\to-\infty$. Note that $w\ne0$ because $u,v$ are linearly independent.

Then $Tw=\lambda w$ when $c=0$ and $Tw=\lambda w-v$ when $c=1$. The former case is impossible because every nonnegative eigenvector of the nonnegative and irreducible $T$ must be positive but $w$ is not. The latter case is impossible because $Tw$ is nonnegative but $\lambda w-v$ is not.