Determining when an $n \times n$ matrix has a spectral radius (i.e., dominant eigenvalue) $\geq 1$? Equations
I have the following irreducible matrix:
$A=\begin{pmatrix}
  a_1 & a_2 & ... & a_{n-1} & a_n\\ 
  b_1 & 0 & ... & 0 & 0\\
0 & b_2 & 0 & ... & 0\\
\vdots & \ddots & \ddots & \ddots & \vdots\\
0 & 0 & ... & b_{n-1} & 0
\end{pmatrix}$,
where $a_i \geq 0$ $\forall \ i \in \{1,...,n-1\}$; $a_n>0$; and $0 < b_i \leq 1$ $\forall \ i \in \{1,...,n-1\}$.
Question
Is there a way to determine (in terms of $a_i$ and $b_i$) when this matrix has a spectral radius (i.e., dominant eigenvalue) greater than or equal to 1?
Comments
I suspect there is a theorem of some sort that I am not aware of that can help with this problem. Any suggestions would be greatly appreciated!
 A: Assume $\lambda$ is an eigenvalue of $A$ and $x=\{x_k\}_{k=1}^n$ a corresponding nonzero eigenvector. Then
$$b_{k-1}x_{k-1}=\lambda x_{k},\qquad k=2,\ldots,n$$ Hence $x_n\neq 0.$ We may assume that $x_n=1.$ Then
$$x_k={\lambda^{n-k}\over b_{k}b_{k+1}\ldots b_{n-1}},\quad 1\le k\le n-1$$
Moreover
$$\langle a,x\rangle =\lambda x_1$$ hence
$$\sum_{k=1}^{n-1} {a_k\over b_{k}b_{k+1}\ldots b_{n-1}}\lambda^{n-k}+a_n=\lambda {\lambda^{n-1}\over b_1b_2\ldots b_{n-1}} $$ Thus
$$\lambda^n= \sum_{k=2}^{n} a_k\,b_{1}b_2\ldots b_{k-1}\lambda^{n-k}+a_1\lambda^{n-1}\quad (*)$$
The equation admits positive solutions as LHS  vanishes at $0,$ while RHS is positive at $0,$ and LHS dominates RHS at infinity.
Let $\lambda_0$ denote the maximal positive solution.
Assume $\lambda$ is any solution of $(*).$ Then
$$|\lambda|^n\le  \sum_{k=2}^{n} a_k\, b_{1}b_2\ldots b_{k-1}|\lambda|^{n-k}+a_1|\lambda|^{n-1}$$ Hence $|\lambda|\le \lambda_0,$ which means $\lambda_0$ coincides with the spectral radius.
We have $\lambda_0\ge 1$ if and only if
$$1\le  \sum_{k=2}^{n} a_k\, b_{1}b_2\ldots b_{k-1}+a_1\quad (**)$$
Remark The assumptions suggested by @Ben Grossman $a_i+b_i\le 1,\ 1\le i\le n-1$ and $a_n\le 1$ imply that the spectral radius is less  than or equal to $1,$  in view of the Perron-Frobenius theorem.
This also follows directly from $(**)$ as for $a_k=1-b_k,$ $1\le k\le n-1,$ and $a_n=1$ we obtain  that RHS of $(**)$ is equal $1.$
