Absolute value and singular values of a compact operator

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Let $H$ and $K$ be real or complex Hilbert spaces (Hilbert space) and let $T\in \mathcal{B}(H,K)$ be a compact operator (Compact linear operator, A bounded linear operator between normed spaces), with Hilbert adjoint $T^{\ast }\in \mathcal{B}(K,H)$ (The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities).

The absolute value. The operator $T^{\ast }T\in \mathcal{B}(H)$ is compact, since it is the composite of the compact $T$ with the bounded $T^{\ast }$ (Compositions with a compact operator are compact); it is self-adjoint, because $(T^{\ast }T{)}^{\ast }=T^{\ast }{T}^{\ast \ast }=T^{\ast }T$ (Hilbert-adjoint identities); and it is positive, because $⟨T^{\ast }Tx,x⟩=⟨Tx,Tx⟩=\mathrm{\parallel }Tx{\mathrm{\parallel }}^2\ge 0$ for every $x\in H$ (Self-adjoint, positive, unitary and normal operators). The absolute value of $T$ is the unique compact self-adjoint positive operator $\mathrm{\mid }T\mathrm{\mid }:=(T^{\ast }T{)}^{1\mathrm{/}2}$ with $\mathrm{\mid }T{\mathrm{\mid }}^2=T^{\ast }T$, whose existence and uniqueness are the preceding square-root lemma (Positive square root of a compact positive operator). It satisfies $\mathrm{\parallel }\mathrm{\mid }T\mathrm{\mid }x{\mathrm{\parallel }}^2=⟨\mathrm{\mid }T{\mathrm{\mid }}^{2}x,x⟩=⟨T^{\ast }Tx,x⟩=\mathrm{\parallel }Tx{\mathrm{\parallel }}^2(x\in H),$ so in particular $\parallel \mathrm{\mid }T\mathrm{\mid }\parallel =\mathrm{\parallel }T\mathrm{\parallel }$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum) and $\ker \mathrm{\mid }T\mathrm{\mid }=\ker T$, since $\mathrm{\parallel }\mathrm{\mid }T\mathrm{\mid }x\mathrm{\parallel }=\mathrm{\parallel }Tx\mathrm{\parallel }$ for every $x$.

The singular values. By the spectral theorem for $\mathrm{\mid }T\mathrm{\mid }$ (Spectral theorem for compact self adjoint operators) the nonzero eigenvalues of $\mathrm{\mid }T\mathrm{\mid }$ form a finite or countably infinite set of positive reals, each with finite multiplicity, and for every real $\epsilon >0$ only finitely many of them exceed $\epsilon$; positivity rules out negative eigenvalues and $0$ already corresponds to the kernel. The multiset of positive singular values of $T$ is the multiset of positive eigenvalues of $\mathrm{\mid }T\mathrm{\mid }$, counted with multiplicity (Eigenvalues, eigenvectors, eigenspaces $E_{\lambda }(T)=\ker (T-\lambda I)$, and the spectrum ${\sigma }_F(T)$ of an endomorphism).

The ordered singular-value sequence. The distinct positive singular values are listed with positive labels in decreasing order ${\mu }_1>{\mu }_2>\cdots$ as follows: if the multiset of positive eigenvalues is empty (equivalently $\mathrm{\mid }T\mathrm{\mid }=0$, equivalently $T=0$), the list is empty; otherwise ${\mu }_1:=\max \{\lambda >0:\lambda$ is an eigenvalue of $\mathrm{\mid }T\mathrm{\mid }\}$, a maximum and not merely a supremum, because a supremum value not attained would be an accumulation point different from $0$; having chosen ${\mu }_1,\dots ,{\mu }_k$, put ${\mu }_{k+1}:=\max \{\lambda >0:\lambda$ is an eigenvalue of $\mathrm{\mid }T\mathrm{\mid }$ and $\lambda <{\mu }_k\}$ whenever that set is nonempty, and stop otherwise. Each step is legitimate by the finiteness-above-thresholds property above, and an infinite list satisfies ${\mu }_k\to 0$ (otherwise its decreasing limit would be a nonzero accumulation point). Writing $d_k$ for the multiplicity of ${\mu }_k$ (a positive integer), the zero-padded singular-value sequence is $s_1\ge s_2\ge s_3\ge \cdots$ where $s_1=\cdots =s_{d_1}={\mu }_1$, $s_{d_1+1}=\cdots =s_{d_1+d_2}={\mu }_2$, and so on; if the multiset is finite with total multiplicity $r=d_1+\cdots +d_m$, one sets $s_n:=0$ for every $n>r$, and if $T=0$ one sets $s_n:=0$ for every $n\ge 1$. The number $s_n$ is written $s_n(T)$ and called the $n$-th singular value of $T$.

Zero-based domain and positive labels. Set $s_0(T):=\mathrm{\parallel }T\mathrm{\parallel }=s_1(T)$. Thus the numerical sequence is the function $n\mapsto s_n(T)$ on all of $\mathbb{N}$, including zero. The positive-labelled tail $(s_{m+1}(T){)}_{m\in \mathbb{N}}$ is the multiplicity-counting list constructed above. The auxiliary initial value $s_0$ is not an additional entry of the eigenvalue multiset, does not index a singular vector, and is excluded from multiplicity counts and singular-value sums, which use $n\ge 1$. The full sequence satisfies $s_0\ge s_1\ge s_2\ge \cdots$ and tends to zero. This preserves the page's positive rank labels while giving convergence statements a zero-based domain.

Rank and the finiteness of the list. The map $\mathrm{\Phi }:\mathrm{ran}\mathrm{\mid }T\mathrm{\mid }\to \mathrm{ran}T$ given by $\mathrm{\Phi }(\mathrm{\mid }T\mathrm{\mid }x):=Tx$ is well defined and linear, because $\mathrm{\mid }T\mathrm{\mid }x=\mathrm{\mid }T\mathrm{\mid }{x}^{\mathrm{\prime }}$ forces $x-x^{\mathrm{\prime }}\in \ker \mathrm{\mid }T\mathrm{\mid }=\ker T$ and hence $Tx=T{x}^{\mathrm{\prime }}$; it is injective, because $Tx=0$ gives $x\in \ker T=\ker \mathrm{\mid }T\mathrm{\mid }$ and $\mathrm{\mid }T\mathrm{\mid }x=0$; it is surjective onto $\mathrm{ran}T$ because $T=\mathrm{\Phi }\circ \mathrm{\mid }T\mathrm{\mid }$; and it is isometric, $\mathrm{\parallel }Tx\mathrm{\parallel }=\mathrm{\parallel }\mathrm{\mid }T\mathrm{\mid }x\mathrm{\parallel }$. Hence $\mathrm{ran}T$ is finite-dimensional if and only if $\mathrm{ran}\mathrm{\mid }T\mathrm{\mid }$ is finite-dimensional; in that case the linear bijection $\mathrm{\Phi }$ gives $\dim \mathrm{ran}T=\dim \mathrm{ran}\mathrm{\mid }T\mathrm{\mid }$ (Finite-dimensional vector space, and its dimension ${\dim }_{F}V$; infinite-dimensional means having no finite basis). Consequently, whenever $T$ has finite rank, the positive singular values with multiplicity number exactly $\dim \mathrm{ran}T$, the rank of $T$, and all later $s_n$ vanish, so the sequence is zero-padded. If $T$ does not have finite rank, the multiset of positive singular values is infinite and countable (Finite, countably infinite, countable, uncountable) and $s_n>0$ for every $n$, with $s_n\downarrow 0$; in particular finite rank of $T$ is characterised by the eventual vanishing $s_n=0$ for all sufficiently large $n$, and conversely such eventual vanishing forces finite rank. The sequence $(s_n)$ is numerical data only: no orthonormal system is selected here, and the zero padding is not an indexing of any family of vectors. The unordered multiset determines $(s_n)$ uniquely, so $(s_n)$ is well defined, and $s_1(T)=\mathrm{\parallel }\mathrm{\mid }T\mathrm{\mid }\mathrm{\parallel }=\mathrm{\parallel }T\mathrm{\parallel }$.