I have been stuck on this problem for quite some time now and, unfortunately, appear to have given up. Perhaps the minds on this page will help me out.
Given an $n\times n$ matrix D, where $$D=\begin{bmatrix} a_{1,\ 1} & a_{1,\ 2} & \cdots & a_{1,\ n-1} & 0\\ b_{2,\ 1} & b_{2,\ 2} & \cdots & b_{2,\ n-1} & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ m-1_{n-1,\ 1} & m-1_{n-1,\ 2} & \cdots & m-1_{n-1,\ n-1} & 0\\ m_{n,\ 1} & m_{n,\ 2} & \cdots & m_{n,\ n-1} & \kappa\\ \end{bmatrix},$$
such that $\kappa \in \mathbb{R}$, prove there exists at least one $\lambda=\kappa$.
For extra credit, prove that the eigenvector corresponding to $\kappa$, $E_{\lambda=\kappa}$, is equal to the n$^\text{th}$ unit vector in $\mathbb{R}^n$.
Using the definition of matrix multiplication, you can just check directly that $De_n$ ($e_n$ is the unit vector with $1$ at the $n$-th position) is $\kappa e_n$.