Krein-Rutman for cones with empty interior My question concerns the following theorem (a finite-dimensional version of Krein-Rutman):


  Let $V$ be a finite dimensional real normed space and $C \subseteq V$ a closed cone (i.e. a convex subset which is closed under multiplication with positive scalars). Suppose $C$ has nonempty interior. Then for every linear map $T : V \rightarrow V$ with $T(C) \subseteq C$ there exists an eigenvector $v \in C$ of $T$ whose corresponding eigenvalue is the spectral radius of $T$.


I was wondering if the cone really has to have nonempty interior for this to work. It is clear to me that I cannot hope to get an eigenvector corresponding to the spectral radius, but maybe still an eigenvector to some nonnegative eigenvalue?

My idea is as follows. If $C$ is any closed cone in $V$ and $T : V \rightarrow V$ leaves $C$ invariant, then it also leaves invariant the subspace $V_0 = C - C$  which is the subspace of $V$ generated by $C$. Considering $C$ as a cone of $V_0$, $C$ has nonempty interior, because $V_0$ has a basis $\{v_1,\dots,v_n\} \subseteq C$ and the convex hull of $\{0,v_1,\dots,v_n\}$ is an $n$-simplex contained in $C$ which definitely has nonempty interior in the $n$-dimensional space $V_0$. Applying the above theorem (with $V$ replaced by $V_0$) to the restriction $T|_{V_0} : V_0 \rightarrow V_0$ we get an eigenvector $v \in C$ with eigenvalue the spectral radius of $T|_{V_0}$. 

Is this argument correct? I have seen this kind of theorem in the literature only with the condition of a nonempty interior, so I was wondering if I might have made some mistake here.
E.g. in $\mathbb R^3$ let $C = \{(x,0,0): x \ge 0\}$.  Any matrix of the form
$$T =  \pmatrix{a &amp; 0 &amp; 0\cr
            0 &amp; b &amp; c\cr
            0 &amp; d &amp; e\cr}$$
with $a &gt; 0$ leaves $C$ invariant, but if the spectral radius is the spectral radius of $\pmatrix{b &amp; c\cr d &amp; e\cr}$ there is no eigenvector $v \in C$ for that spectral radius.
E.g. in $\mathbb R^3$ let $C = \{(x,0,0): x \ge 0\}$.  Any matrix of the form
$$T =  \pmatrix{a &amp; 0 &amp; 0\cr
            0 &amp; b &amp; c\cr
            0 &amp; d &amp; e\cr}$$
with $a &gt; 0$ leaves $C$ invariant, but if the spectral radius is the spectral radius of $\pmatrix{b &amp; c\cr d &amp; e\cr}$ there is no eigenvector $v \in C$ for that spectral radius.