We often come across with a argument in linear algebra to prove that geometric multiplicity of the Perron eigenvalue is $1$ in proof of Perron's Theorem:
Suppose $ψ$ is the positive eigenvector of $ρ(A)$ and $ψ_0$ is a linearly independent eigenvector of the eigenvalue $ρ(A)$. We can assume that $ψ_0$ is real; otherwise we take real and imaginary parts, and the parts are still eigenvectors, because $A$ and $ρ(A)$ are real. One of them must be linearly independent of $ψ$. Let $c > 0$ be chosen so that $ψ − cψ_0$ is nonnegative and at least one entry is zero. It is not the zero vector, because $ψ, ψ_0$ are linearly independent.
But nowhere I found the method how to choose $c > 0$ o that $ψ − cψ_0$ is nonnegative and at least one entry is zero.
How can we choose such a $c$?
Without loss of generality, assume $\psi_0$ has at least one positive entry (else replace $\psi_0$ with $-\psi_0$). Let $I$ be the set of indices $i$ such that the $i$th entry of $\psi_0$ is positive. Let $c$ be the minimum of all the ratios $\psi_i / (\psi_0)_i$ for $i \in I$ (here $\psi_i$ denotes the $i$th entry of $\psi$, and similarly for $(\psi_0)_i$). It is easy to check that this $c$ satisfies the required conditions.