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For a Fredholm operator $L$ on a Hilbert space, invert its restriction from the kernel complement to its closed range, extending by zero on the range complement. The resulting bounded $B$ has $BL=I-P$ and $LB=I-Q$ with finite-rank projections. Adding a compact operator $K$ gives two-sided inverses modulo compact errors. Such a parametrix forces finite kernel, a lower bound on the kernel complement, closed range and finite adjoint kernel; the latter is the cokernel. Repeating with $-K$ proves the equivalence without assuming self-adjointness.

A bounded linear operator on a Hilbert space is Hilbert-Schmidt when ${\sum }_n\Vert T{e}_n{\Vert }^2<\infty$ for one, equivalently every, orthonormal basis. Every Hilbert-Schmidt operator is a compact operator.

A unit left singular vector of a matrix or compact operator $A$ is an eigenvector of $A{A}^{\ast }$ with eigenvalue ${\sigma }_j^2$. For ${\sigma }_j>0$, it pairs with a right singular vector $v_j$ through $A{v}_j={\sigma }_j{u}_j$ and $A^{\ast }{u}_j={\sigma }_j{v}_j$.

For a bounded operator $A:X\to Y$ between Hilbert spaces, the Moore–Penrose inverse maps admissible data to the unique minimum-norm least-squares solution. If $({\sigma }_j,u_j,v_j)$ is a singular system of a compact operator, thenon the data for which this series converges.

More precisely, its domain is $\mathrm{ran}A\oplus (\mathrm{ran}A{)}^{\perp }$. It vanishes on $(\mathrm{ran}A{)}^{\perp }$ and takes values in $(\ker A{)}^{\perp }$. An infinite-rank compact operator on infinite-dimensional spaces has singular values approaching zero, so its inverse on its range is generally unbounded; Picard criterion specifies admissible data.

The closed unit ball of a Hilbert space is compact in norm exactly when the space is finite-dimensional. Finite-dimensional compactness follows from the Heine-Borel theorem. In infinite dimension, an orthonormal sequence has pairwise distance $\sqrt{2}$ and no convergent subsequence. Restricting a compact operator that equals a nonzero multiple of the identity on a subspace therefore forces that subspace to be finite-dimensional.