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Is this proof of Perron's theorem correct, and if so is it original?
Then by positivity we have $A^2 \Psi > \rho A \Psi$, which means that by continuity there is some $\varepsilon > 0$ with $A^2 \Psi \ge (\rho + \varepsilon) A \Psi$. … Then $\rho (\Psi + c \psi) = A(\Psi + c \psi) > 0$ by positivity, but it has one zero entry, which is a contradiction. So there's no other linearly independent eigenvector. …