Write $u^\dagger=K^\dagger f$. Every <least-squares solution> is $u^\dagger+w$ for some $w\in\mathcal N(K)$, while $u^\dagger\in\mathcal N(K)^\perp$. Decompose $u_0=P_{\mathcal N(K)}u_0+P_{\mathcal N(K)^\perp}u_0$ using <orthogonal projections>. Then the <Pythagorean identity> gives
$$
\|u^\dagger+w-u_0\|^2
=\|u^\dagger-P_{\mathcal N(K)^\perp}u_0\|^2
+\|w-P_{\mathcal N(K)}u_0\|^2.
$$
The first term is independent of $w$, and the second vanishes at exactly one point. Hence the unique <nearest least-squares solution> is
$$
\boxed{u_0^\dagger=K^\dagger f+P_{\mathcal N(K)}u_0.}
$$
Geometrically this is the <orthogonal projection> of $u_0$ onto the closed <affine subspace> of <least-squares solutions>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 326 / 1 / 1 / d / Solution

{title2=$u_0^\dagger=K^\dagger f+P_{\mathcal N(K)}u_0$}

The <orthogonal projection> of a prescribed point $u_0$ onto the <affine subspace> of <least-squares solutions> is unique. It retains $u_0$'s <null space> component and chooses the determined component from the <Moore–Penrose inverse of an operator>.

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 Incoming links: Nearest least-squares solution