Singular values of a compact operator

Remarks

The list is well defined by the spectral theorem for compact self-adjoint operators, and $s_n(T) \to 0$. One has $s_1(T) = \norm{T}$, and more generally $$s_n(T) = \min \set{\norm{T - F} : F \text{ of rank less than } n},$$ the distance from $T$ to the finite-rank operators of rank less than $n$. There are orthonormal families $(u_n)$ in $H$ and $(v_n)$ in $K$ such that $T x = \sum_n s_n(T) \ip{x}{u_n} v_n$ for all $x \in H$, the Schmidt decomposition of $T$. Example: if $T$ is diagonal in an orthonormal basis $(e_n)$ of $\ell^2$, with $T e_n = d_n e_n$ and $d_n \to 0$, then $\abs{T} e_n = \abs{d_n} e_n$, so the singular values are the numbers $\abs{d_n}$ arranged in decreasing order. Non-example: the identity of an infinite-dimensional Hilbert space is not compact, so it has no singular values in this sense; the eigenvalue $1$ of $\abs{I} = I$ has infinite multiplicity. For matrices, this is the usual notion of singular value.