= Solution
For admissible data, the <Moore–Penrose inverse of an operator> selects the <minimum-norm least-squares solution>:
$$
\boxed{x^\dagger\in(\ker A)^\perp,\qquad \|Ax^\dagger-y\|=\inf_{x\in X}\|Ax-y\|,\qquad A^\dagger y=x^\dagger.}
$$
Its domain is $\operatorname{ran}A\oplus(\operatorname{ran}A)^\perp$. When the equation is consistent it is the solution of $Ax=y$ orthogonal to the kernel; for inconsistent admissible data it solves the <normal equation for a linear inverse problem> and projects away the component orthogonal to the range.
Take a <singular system of a compact operator> satisfying $Av_j=\sigma_ju_j$, $A^*u_j=\sigma_jv_j$. The generalized inverse is
$$
\boxed{x^\dagger=\sum_j\frac{\langle y,u_j\rangle}{\sigma_j}v_j.}
$$
The <Picard criterion> requires the squared coefficients to be summable. For an infinite-rank compact operator, $\sigma_j\to0$, so $\|A^\dagger u_j\|=1/\sigma_j\to\infty$ even though $\|u_j\|=1$. Hence $A^\dagger$ is unbounded. Finite-rank compact operators do not have this obstruction. Data failing the Picard condition need not have a least-squares minimizer at all, even when the infimum residual is zero.
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