= Solution
A <singular system of a compact operator> consists of positive <singular values> $\sigma_i$ and <orthonormal> vectors $v_i\in X$, $u_i\in Y$, with
$$
\boxed{Av_i=\sigma_i u_i,\qquad A^*u_i=\sigma_i v_i.}
$$
The $v_i$ form an <orthonormal basis> of $(\ker A)^\perp$, and the $u_i$ form one of $\overline{\operatorname{ran}A}$. There are finitely many positive <singular values> for finite-rank $A$; otherwise $\sigma_i\to0$. Obtain the system by applying the <spectral theorem for compact self-adjoint operators> to $A^*A$, choosing $A^*Av_i=\sigma_i^2v_i$, and setting $u_i=Av_i/\sigma_i$. The <inner product> identities verify orthonormality and the displayed relations.
With <inner products> linear in the first argument,
$$
Ax=\sum_i\sigma_i(x,v_i)u_i,\qquad
A^*y=\sum_i\sigma_i(y,u_i)v_i.
$$
The <normal equation for a linear inverse problem> with the <Tikhonov regularization> penalty acts separately on each $v_i$. Hence the <singular-system Tikhonov filter> gives \b[the regularized solution]
$$
\boxed{x_\alpha=\sum_i\frac{\sigma_i}{\sigma_i^2+\alpha}(y,u_i)v_i.}
$$
This norm-convergent series follows from <Bessel's inequality> and the bound $\sigma_i/(\sigma_i^2+\alpha)\leq1/(2\sqrt\alpha)$. Data orthogonal to $\overline{\operatorname{ran}A}$ are annihilated by $A^*$, and the minimizing solution has no component in $\ker A$. The same formula with $y^\delta$ reconstructs noisy data. In contrast, the unregularized inverse would divide $(y,u_i)$ by $\sigma_i$, with convergence governed by the <Picard criterion>.
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