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If $A$ has finite rank, the image of its unit ball is bounded in the finite-dimensional space $\mathrm{ran}A$. The closure of a bounded set in a finite-dimensional normed space is compact. Hence every bounded finite-rank operator is a compact operator.

Let $Q_n=P_n^{\ast }{P}_n$ be the orthogonal projection ontoThen $Q_n\to I$ strongly. Strong convergence is uniform on every compact subset: if $K$ is compact, cover it by finitely many small balls and use $\Vert I-Q_n\Vert \le 1$ at their centers. Since the closure of $A$ applied to the unit ball is compact,The adjoint of a compact operator is compact, so the same argument for $A^{\ast }$ givesSince $A_n=Q_{n}A{Q}_n$,ThereforeThis is the finite-section approximation of a compact operator.

Let $R_z=(A-zI{)}^{-1}$ be compact and let $w$ be another resolvent point. The resolvent identity givesorThe bracket is bounded and the product of a bounded operator with a compact operator is compact. Thus compactness at one resolvent point implies compactness at every resolvent point.

Fix such a $z$. Spectral mapping for the bounded compact operator $R_z$ givesEvery nonzero spectral point of a compact operator is an isolated eigenvalue of finite multiplicity, and zero is its only possible accumulation point. Hence a compact resolvent operator has only isolated eigenvalues of finite multiplicity, with no finite accumulation point. The spectrum is allowed to be empty.

A unit right singular vector of a matrix or compact operator $A$ is an eigenvector of $A^{\ast }A$ with eigenvalue ${\sigma }_j^2$. For ${\sigma }_j>0$, $u_j=A{v}_j/{\sigma }_j$ is the corresponding left singular vector. The largest singular value is the operator norm of $A$, attained on its corresponding right singular vectors.

A compact Hermitian operator has real nonzero eigenvalues of finite multiplicity, with zero as their only possible accumulation point, and an orthonormal eigenbasis after a basis of its kernel is included.

For nonzero compact operator $A$, step $0<\tau <2/\Vert A{\Vert }^2$, and data admitting a minimum-norm least-squares solution $x^{†}$, zero-initialized Landweber iteration satisfiesThe Picard criterion makes the majorant summable, so the dominated convergence theorem proves this limit. No mesh-independent or singular-value-independent strict contraction factor is required. At $\tau =2/\Vert A{\Vert }^2$, the top singular mode has multiplier $-1$ and may not converge. Without solvable projected data, the iterates need not have a limit in the solution Hilbert space, even when their residuals approach the least-squares infimum.