Let $\lambda_i$ denote the eigenvalues of the matrix $B$ and let $\rho(A)$ and $\rho(B)$ denote the spectral radii of the matrices $A$ and $B$, respectively. Suppose that $A,B$ are nonnegative matrices (a matrix with every entry $\geq 0$) with $$\rho(A) < \lambda_2 < \lambda_1 = \rho(B)$$ where $B$ is a rank-$1$ correction of $A$, in other words, $\operatorname{rank} (B-A) \leq 1$. In the problem it is known that $\lambda_1$ is simple. What I'd like to affirm is that exists a value of $s^*$ close to $1$ such that $A_s = A + s(B-A)$ has a simple eigenvalue (presumably the one strictly closer to $\rho(B)$, i.e $\lambda_2$, since then I'd like to replace $B$ with $s^*$).
I don't think the problem is true in general since with $B$ a multiple of $A$ doesn't seem true. In this setting I also have that the characteristic polynomial of $A_s$ is the convex combination of that of $A$ and $B$. Any help would be appreciated.
It is straight-forward to prove the weaker claim that after perturbing $s$, if needed, we may take $\lambda_2$ to have odd degree. It may be possible to tweak the argument in the two paragraphs below to include derivative information of $p_s$ and get to a case of $\lambda_2$ being simple. However as the final paragraph makes clear, this is unnecessary when one is familiar with winding number basics from complex analysis like the Argument Principle.
Where $\lambda_i$ are roots of characteristic polynomial $p_1$ (which the authors define so as to be monic) and $\lambda_1$ is the biggest positive root and is already known to be simple and $\lambda_2\gt \rho(A)$ is the next largest positive root
$p_s = p_0 + s\cdot (p_1-p_0) = (1-s)\cdot p_0 + s\cdot p_1$ for $s\in [0,1]$
where $p_0(x)\gt 0$ for $x \gt \rho(A)$ and $p_1(x)\gt 0$ for $x\gt \lambda_1$, $\lambda_1$ is a simple root, hence $p_1(x) \lt 0$ for $x\in (\lambda_2,\lambda_1)$
Suppose $\lambda_2$ has even multiplicity $m$, with $m=2$ in particular as our Base Case. This means $p_1\big([\lambda_2-\delta, \lambda_2+\delta]\big)\leq 0$, with equality only at $\lambda_2$, for some $\delta \gt 0$ -- otherwise $p_1\big( (\lambda_2, \lambda_1)\big)\gt 0$ and combining this with $p_1(x)\gt 0$ for $x\gt \lambda_1$ would mean $\lambda_1$ has even degree. Then for $s \in (1-\epsilon,1 )$ for any $\epsilon \gt 0$ small enough
$p_s\big(\lambda_2 - \delta\big)\lt 0$ and $p_s\big(\lambda_2 + \delta\big)\lt 0$ but $p_s\big(\lambda_2\big)\gt 0$ so Intermediate Value Theorem tells us there now at least two distinct roots $\in \big(\lambda-\delta, \lambda+\delta\big)$. In particular with $m=2$ there are exactly two simple real roots in a neighborhood of $\lambda_2$ and we now call the largest of these $\lambda_2$ (I prefer $\lambda_2^{(s)}$ but that isn't the authors' convention). For the inductive case i.e. even $m\geq 4$ we see $p_s$ has some finite number $\geq 2$ of distinct real roots in neighborhood of $\lambda_2$ and again re-label the largest one of these $\lambda_2$. This (re-labeled) $\lambda_2$ is either of odd degree or of even degree $\leq m-2$, so by strong induction hypothesis we may perturb $s$ finitely many times to recover a new $\lambda_2$ of odd degree.
Thus page 753 runs the same except line 6 says "Thus, $\lambda_1$ is simple and $\lambda_2 \lt \lambda_1$. The root $\lambda_2$ can be considered of odd degree, or we otherwise replace the matrix $B=A_1$ with the matrix $A_s$ for some $s$ close enough to $1$... The updated matrices possess the same properties: the rank of their difference is still not greater than $1$, and $B$ has two real eigenvalues greater than $\rho(A)$. It is so because $\rho(A)$ changes continuously with the matrix and so does $\lambda_1$ as a simple root of $p_1$. As for $\lambda_2$ the Argument Principle (/Rouche) tells us that the interior (/bounded component) of $\gamma(t) = \lambda_2 + r\cdot\exp\big(2\pi i \cdot t\big)$ for $t\in[0,1]$ for $r\gt 0$ small enough has the same number of roots with multiplicity for $p_1$ and the characteristic polynomial $q$ of $B+\Delta$ whenever $\Delta$'s components are small enough, and since the multiplicity of $\lambda_2$ is odd, and non-real roots must come in conjugate pairs, this means there is at least one real root in a r-neighborhood of $\lambda_2$ and we re-label the largest of these real roots in such neighborhood as $\lambda_2$." The rest of the proof of Theorem 1 finishes identically.