= Solution
Since one eigenvalue is one, the other is the determinant. Direct calculation gives
$$
\boxed{\vartheta=\frac{\mu(\lambda+\alpha)}{\lambda(\mu+\beta)},\qquad v_\vartheta=\begin{pmatrix}b\\1\end{pmatrix},\qquad b=\frac{\alpha(\mu+\beta)}{\beta(\lambda+\alpha)}>0.}
$$
Indeed the first row requires $(1+\alpha/\lambda-\vartheta)b=\alpha/\lambda$, and the second row then follows from the characteristic equation. At $\vartheta=1$, $b=1$ and the eigenvectors coincide; the nonidentity matrix has a repeated eigenvalue and a Jordan block.
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