Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 72 / 3 / a / Solution

For a bounded linear operator $A:X\to Y$ between Hilbert spaces, let $N=\ker A$ and $R=\mathrm{ran}A$. The Moore–Penrose inverse of an operator is defined onFor $y=Ax+z$, with $z\in R^{\perp }$, defineThis is independent of the chosen preimage $x$, since two preimages differ by a vector in $N$. Equivalently, it is the inverse of the restriction of $A$ to $N^{\perp }$, applied to the component of the data in $R$, and is zero on $R^{\perp }$.

The generalized solution $x^{†}=A^{†}y$ is the unique minimum-norm least-squares solution. Its residual is orthogonal to the range, giving the operator normal equationThe operator normal equation alone leaves an arbitrary null-space component; the second condition fixes the minimum-norm representative. The identities $A^{†}A=P_{N^{\perp }}$ and $A{A}^{†}=P_{\overline{R}}$ hold on their appropriate domains.

For an infinite-rank compact operator, the range need not be closed, and $A^{†}$ is generally unbounded. Data outside $R\oplus R^{\perp }$ need not have any least-squares solution at all, even though they can be approximated by range elements. This domain qualification is crucial in part (d); the Moore–Penrose notation does not turn an ill-posed inverse into an everywhere-defined bounded operator.