Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 2 / 26I / d

Since one eigenvalue is one, the other is the determinant. Direct calculation givesIndeed 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.