= Solution
For a <bounded linear operator> $K:U\to V$ between <Hilbert spaces>, the <Moore–Penrose inverse of an operator> assigns to each admissible datum its unique <minimum-norm least-squares solution>. Its domain is
$$
\boxed{\mathcal D(K^\dagger)=\operatorname{ran}K\oplus(\operatorname{ran}K)^\perp.}
$$
Write $f=Ku+h$, where $u\in(\ker K)^\perp$ and $h\in(\operatorname{ran}K)^\perp=\ker K^*$. Then $K^\dagger f=u$. Restricting to $(\ker K)^\perp$ removes the ambiguity among solutions differing by an element of the <kernel>. The <orthogonal decomposition> also shows that $KK^\dagger f$ is the <orthogonal projection> of $f$ onto $\overline{\operatorname{ran}K}$, and $K^\dagger Ku$ is the <orthogonal projection> of $u$ onto $(\ker K)^\perp$. The inverse is defined on a dense domain and need not be a <bounded linear operator> when the range is not closed.
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