Understanding Karlin's Proof of Perron's Theorem.

Question

Reading a proof of Perron's Theorem I got stuck at some place:

We have proved that spectral radius $\rho$ is a simple eigenvalue. Now we have to check that there are no eigenvalues other than spectral radius of modulus $\rho$.

Consider $A − \epsilon I > 0$ for small $\epsilon > 0$. Its largest positive eigenvalue is $\rho − \epsilon$, which we have proved to be its spectral radius.

Translating this smaller circle back to the right by $\epsilon$ we see that all remaining eigenvalues of $A$ lie within the open disk $|λ| < ρ$.

I can't understand the above line(bold), Help Needed!

Ref: The Many Proofs and Applications of Perron's Theorem pg496.

Answer 1

Perhaps this will help: any eigenvalues of $A$ must satisfy $|\lambda| \leq \rho$ and $|\lambda - \epsilon| \leq \rho - \epsilon$. The only complex number which satisfies both $|\lambda| = \rho$ and $|\lambda - \epsilon| \leq \rho - \epsilon$ is $\lambda = \rho$.

Answer 2

Translating the smaller circle corresponds to undoing the operation $A - \varepsilon I$, in terms of the eigenvalues. That is, if in the complex plane you draw a point for each eigenvalue of $A - \varepsilon I$, then moving them all to the right by $\varepsilon$ gives you the eigenvalues of $A$. Since all of them are inside the circle of radius $\rho - \varepsilon$, when you move them, the only way there can be an eigenvalue in the boundary of the circle of radius $\rho$ is to have it at the positive real axis, i.e. at $\rho$.