 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 isWrite $f=Ku+h$, where $u\in (\ker K{)}^{\perp }$ and $h\in (\mathrm{ran}K{)}^{\perp }=\ker K^{\ast }$. Then $K^{†}f=u$. Restricting to $(\ker K{)}^{\perp }$ removes the ambiguity among solutions differing by an element of the kernel. The orthogonal decomposition also shows that $K{K}^{†}f$ is the orthogonal projection of $f$ onto $\overline{\mathrm{ran}K}$, and $K^{†}Ku$ 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.