Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 40C / d / Solution

The displayed matrix is an upper bidiagonal matrix. For ${\lambda }_1=1$, the right eigenvector equation $(A-I)v=0$ successively givesWe may therefore take the unit right eigenvector to be $v=e_1$.

For a left eigenvector, $(A^T-I)u=0$ givesSince ${\lambda }_i-1=-1/i$,Choosing the initial scale $u_1=1$ producesAfter normalization,The eigenvalue sensitivity is thereforewhere $n!$ is the factorial. Hence