Assume we are given a symmetric,non-negative, irreducible matrix $A \in \mathbb{R}^{n \times n}$.
Then by the Perron-Frobenius theorem, there exists an eigenvector $v \in \mathbb{R}^n$ for the eigenvalue $\sigma(A)>0$ and the eigenvector $v$ is positive.
Now we take some $x \in \mathbb{R}^n, c \in \mathbb{R}$ and construct the matrix $B \in \mathbb{R}^{n+1 \times n+1}$ as
\begin{equation} B = \begin{pmatrix} A & x \\ x^T & c \end{pmatrix}. \end{equation}
My question is whether we can decide based on $x,c$ if the matrix $B$ will have also have a Perron-Frobenius eigenpair, that is some $w>0$, such that $Bw = \sigma(B) w$ where $\sigma(B)>0$.
Some notes on this:
Obviously, if $x>0$ then we still get an irreducible non-negative matrix so this case is trivial. Another easy-to-see case is when $x^T v = 0, c = \sigma(A)$ then one simply append $v$ by $1$ to get a solution for $B$.
Equivalently, one can ask the question when $B$ is 'eventually positive' that is, if there exists some $k$ such that $B^l>0$ for all $l>k$, but this does not seem to be an easy to answer question
By the Cauchy-interlacing-theorem we have that $\lambda_1 \leq \mu_1 \leq \lambda_2 \leq \mu_2 ... \leq \lambda_{n-1} \leq \mu_{n-1} \leq \lambda_n = \sigma(B)$, where $\lambda_i, \mu_i$ are the eigenvalues of $A,B$ respectively. Together with the last note, this should at least tell us that we need $c \geq 0$. This is because from the above we have that $\operatorname{tr}(B) - \lambda_1 \geq \operatorname{tr}(A)$. So if now $c <0$, then $\operatorname{tr}(B) < \operatorname{tr}(A)$ which means $\lambda_1<0$ but then $B$ can not be eventually positive.
