 Solution

Let $e_1=(1,0,\dots ,0{)}^T$. The displayed lower-bidiagonal matrix $Q$ satisfiesSince $P=Q^{-1}$, the Preconditioned conjugate gradient method usesSolving $Qz=e_1$ givesThe rank-one matrix $I+z{z}^T$ has eigenvalue $1+n$ in the direction $z$ and eigenvalue $1$ on the $(n-1)$-dimensional orthogonal complement of $z$. It therefore has only two distinct eigenvalues. Part (b) gives