Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 335 / 3 / iv / Solution

On each left singular vector $u_j$, use $A^{\ast }{u}_j={\sigma }_j{v}_j$ and $(I-A^{\ast }A)v_j=(1-{\sigma }_j^2)v_j$. The finite geometric series givesThus with $\alpha =1/n$, the dimensionless Landweber spectral filter isThe full inverse-coefficient convention instead uses $[1-(1-{\sigma }^2{)}^{1/\alpha }]/\sigma$. With general step $\tau$, replace ${\sigma }^2$ inside the power by $\tau {\sigma }^2$.

In the normalized problem, $0\le f_n(\sigma )\le \min (n{\sigma }^2,1)$, so the inverse coefficient is at most $\min (n\sigma ,1/\sigma )\le \sqrt{n}$. This verifies directly why finite iteration is stable and why admitting progressively smaller singular values eventually amplifies noise.