Let $(\sigma_j,u_j,v_j)$ be a <singular system of a compact operator> $A:X\to Y$, so
$$
Av_j=\sigma_ju_j,
\qquad
A^*u_j=\sigma_jv_j,
\qquad
\sigma_j>0.
$$
The <Moore–Penrose inverse of an operator> has domain
$$
\mathcal D(A^\dagger)
=\operatorname{ran}A\mathbin\oplus
(\operatorname{ran}A)^\perp
$$
and acts by
$$
\boxed{
A^\dagger y
=\sum_j\frac{\langle y,u_j\rangle}{\sigma_j}v_j},
$$
with the orthogonal component of $y$ sent to zero. Equivalently, its domain consists of the data satisfying the <Picard criterion>. It obeys
$$
AA^\dagger y=P_{\overline{\operatorname{ran}A}}y,
\qquad
A^\dagger Ax=P_{(\ker A)^\perp}x.
$$

If $Ax=y$ is exactly solvable, every solution is $x^\dagger+z$ with $z\in\ker A$, and
$$
\boxed{x^\dagger=A^\dagger y}
$$
is the unique <minimum-norm least-squares solution>. It recovers the component of the original $x$ orthogonal to the null space; no data can determine the null-space component.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 335 / 3 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-335/3/i/solution