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Singular value decomposition for compact operators

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ and $K$ be real or complex Hilbert spaces (Hilbert space), let $T \in B (H, K)$ be a compact operator (Compact linear operator), let $∣ T ∣$ and the singular values $s_{n} (T)$ be as in the absolute-value definition (Absolute value and singular values of a compact operator). Let $J = {1, 2, 3, \dots}$ when $ran T$ is infinite-dimensional. When $ran T$ is finite-dimensional, put $r : = \dim ran T \in N$ (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis) and let $J = {1, \dots, r}$ , interpreted as $\emptyset$ when $r = 0$ . Thus $J$ indexes exactly the positive singular values counted with multiplicity, which we write as $(s_{j})_{j \in J}$ in nonincreasing order. Then:

there are orthonormal families $(e_{j})_{j \in J}$ in $(\ker T)^{⊥}$ and $(f_{j})_{j \in J}$ in $\overline{ran T}$ , indexed by exactly $J$ , with $∣ T ∣ e_{j} = s_{j} e_{j}$ and $f_{j} = s_{j}^{- 1} T e_{j}$ for every $j \in J$ ;
for every $x \in H$ the series converges in norm and $T x = \sum_{j \in J} s_{j} ⟨ x, e_{j} ⟩ f_{j},$ and its finite partial sums $T_{n} : = \sum_{j \leq n} s_{j} ⟨ \cdot, e_{j} ⟩ f_{j}$ satisfy $∥ T - T_{n} ∥ \leq s_{n + 1}$ for every $n$ with $n + 1 \in J$ (and $T_{n} = T$ for $n \geq r$ when $r < + \infty$ );
the linear map $U$ defined on the span of ${e_{j} : j \in J}$ by $U e_{j} : = f_{j}$ , extended by continuity to $(\ker T)^{⊥}$ and by zero on $\ker T$ , is a partial isometry with $T = U ∣ T ∣, U^{*} U = P_{(\ker T)^{⊥}}, U^{*} U is the orthogonal projection onto (\ker T)^{⊥},$ and $U U^{*}$ the orthogonal projection onto $\overline{ran T}$ ;
the zero-padded sequence $(s_{n} (T))_{n \geq 1}$ is not used to index the orthonormal systems: the systems carry exactly the index set $J$ of the positive singular values, and the terms $s_{n} (T) = 0$ beyond the rank in the finite-rank case are numerical padding only.

Facts & Assumptions

Given: Countable Choice, compact $T : H \to K$ , its absolute value $∣ T ∣$ , the index set $J$ of the positive singular values with multiplicity, the finite dimension $r = \dim ran T$ when the range is finite-dimensional, and the zero-padded sequence $(s_{n} (T))$ .

[A1]

Absolute value and finite rank. $∣ T ∣$ is compact, self-adjoint and positive with $∣ T ∣^{2} = T^{*} T$ , $∥ ∣ T ∣ x ∥ = ∥ T x ∥$ and $\ker ∣ T ∣ = \ker T$ ; the positive singular values with multiplicity are the positive eigenvalues of $∣ T ∣$ with multiplicity. They are finite in number exactly when $ran T$ is finite-dimensional, and otherwise form a countably infinite list. In the finite-dimensional case the isometric linear bijection $Φ : ran ∣ T ∣ \to ran T$ , $Φ (∣ T ∣ x) = T x$ , gives $\dim ran ∣ T ∣ = \dim ran T = r$ (Absolute value and singular values of a compact operator, Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis). No value $\dim V = \infty$ is used.

[A2]

Spectral theorem for $∣ T ∣$ . The nonzero eigenvalues of $∣ T ∣$ are positive, have finite-dimensional eigenspaces $E_{λ}$ , are mutually orthogonal across distinct $λ$ , and their closed span is $(\ker ∣ T ∣)^{⊥} = \overline{ran ∣ T ∣}$ ; moreover $\ker ∣ T ∣ = \ker T$ and $H = (\ker T)^{⊥} \oplus \ker T$ , so the closed span of the eigenspaces is $(\ker T)^{⊥}$ (Spectral theorem for compact self adjoint operators, Eigenspaces of a self adjoint operator are orthogonal, Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism, Orthogonality and the orthogonal complement, Orthogonal decomposition by a closed subspace).

[A3]

Bases and expansion. Every finite-dimensional eigenspace $E_{λ}$ has an orthonormal basis (Every finite-dimensional real or complex inner product space has an orthonormal basis); an orthonormal family is complete in a closed subspace in the case, and only in the case, that the finite-subset net of Fourier sums converges there, with Parseval and Bessel inequalities available (Fourier expansion in a Hilbert space, Parseval equivalences for an orthonormal family, The finite Bessel inequality and best approximation by a finite orthonormal family, Orthonormal families, complete orthonormal systems and Hilbert bases).

[A4]

Continuity. Bounded operators are continuous and satisfy $∥ S x ∥ \leq ∥ S ∥ ∥ x ∥$ ; limits are unique (A bounded linear operator between normed spaces, For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ).

[A5]

Countable Choice supplies, for the at most countable eigenvalue list, one orthonormal basis of each finite-dimensional eigenspace (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Finite, countably infinite, countable, uncountable).

Proof

technique · direct

Given: Countable Choice, the compact $T$ , its absolute value $∣ T ∣$ , the index set $J$ and the singular values $s_{j}$ , the finite integer $r$ when $ran T$ is finite-dimensional, and the eigenspaces $E_{λ}$ of $∣ T ∣$ for positive eigenvalues $λ$ .

1.1A1A2A3A5

Choosing the left system. By [A2] the positive eigenvalues of $∣ T ∣$ are precisely the positive singular values with multiplicity, and their eigenspaces are finite-dimensional with closed span $(\ker T)^{⊥}$ ; listing those eigenvalues with multiplicity as $(s_{j})_{j \in J}$ and choosing by [A5] an orthonormal basis of each $E_{λ}$ gives an orthonormal family $(e_{j})_{j \in J}$ with $∣ T ∣ e_{j} = s_{j} e_{j}$ for every $j$ , whose closed linear span is $(\ker T)^{⊥}$ , and the terms $s_{j} > 0$ are in nonincreasing order.

2.1step 1.1A1

The right system is orthonormal. For $j \in J$ put $f_{j} : = s_{j}^{- 1} T e_{j}$ , which lies in $ran T \subseteq \overline{ran T}$ and is well defined because $s_{j} > 0$ . For $i, j \in J$ , using $∣ T ∣^{2} = T^{*} T$ and the eigenvector property of [step 1.1], $⟨ f_{i}, f_{j} ⟩ = \frac{1}{s_{i} s_{j}} ⟨ T e_{i}, T e_{j} ⟩ = \frac{1}{s_{i} s_{j}} ⟨ T^{*} T e_{i}, e_{j} ⟩ = \frac{1}{s_{i} s_{j}} ⟨ ∣ T ∣^{2} e_{i}, e_{j} ⟩ = \frac{s_{i}^{2}}{s_{i} s_{j}} ⟨ e_{i}, e_{j} ⟩ = δ_{i j},$ so $(f_{j})_{j \in J}$ is orthonormal.

3.1step 1.1step 2.1A1A3A4

The expansion. Let $x \in H$ . By [A2] write $x = m + n$ with $m \in (\ker T)^{⊥}$ and $n \in \ker T$ . Since $(e_{j})$ is complete in $(\ker T)^{⊥}$ [step 1.1], the Fourier expansion [A3] gives $m = \sum_{j \in J} ⟨ m, e_{j} ⟩ e_{j} = \sum_{j \in J} ⟨ x, e_{j} ⟩ e_{j}$ as a norm limit of finite-subset partial sums, and then continuity of $T$ [A4] gives $T x = T m = \sum_{j \in J} ⟨ x, e_{j} ⟩ T e_{j} = \sum_{j \in J} s_{j} ⟨ x, e_{j} ⟩ f_{j}$ , because $T n = 0$ and the image net of the finite partial sums converges. Moreover for finite $F \subseteq J$ the remainder is $∥ (T - \sum_{j \in F} s_{j} ⟨ \cdot, e_{j} ⟩ f_{j}) x ∥^{2} = \sum_{j \notin F} s_{j}^{2} ∣ ⟨ x, e_{j} ⟩ ∣^{2} \leq s_{n + 1}^{2} ∥ x ∥^{2}$ whenever $F \supseteq {1, \dots, n}$ , by orthonormality [step 2.1] and Bessel [A3], so the partial sums $T_{n}$ of the statement satisfy $∥ T - T_{n} ∥ \leq s_{n + 1}$ for $n + 1 \in J$ , and $T_{n} = T$ for $n \geq r$ in the finite-rank case because then $s_{j} = 0$ for $j > r$ and every index in $J$ is $\leq r$ .

4.1step 2.1step 3.1

The right system spans the range closure. Each $f_{j} = s_{j}^{- 1} T e_{j}$ lies in $ran T$ by [step 2.1], so the closed linear span $N : = \overset{}{span} {f_{j} : j \in J}$ is contained in $\overline{ran T}$ ; conversely [step 3.1] exhibits every $T x$ as the norm limit of finite linear combinations of the $f_{j}$ , so $ran T \subseteq N$ and hence $\overline{ran T} = N$ .

5.1step 1.1step 2.1step 4.1A1A2A4

The partial isometry and $T = U ∣ T ∣$ . Define $U$ first on the linear span $V$ of ${e_{j} : j \in J}$ by $U (\sum_{j \in F} c_{j} e_{j}) : = \sum_{j \in F} c_{j} f_{j}$ for finite $F$ . This is well defined because $(e_{j})$ is linearly independent as an orthonormal family, and it is isometric, since by [step 2.1] $∥ \sum c_{j} f_{j} ∥^{2} = \sum ∣ c_{j} ∣^{2} = ∥ \sum c_{j} e_{j} ∥^{2}$ ; by [step 1.1] the closure of $V$ is $(\ker T)^{⊥}$ , so $U$ extends uniquely to a bounded linear operator, still denoted $U$ , on $(\ker T)^{⊥}$ with $∥ U m ∥ = ∥ m ∥$ for all $m \in (\ker T)^{⊥}$ and $U ((\ker T)^{⊥}) = \overset{}{span} {f_{j}} = \overline{ran T}$ by [step 4.1]. Extend $U$ to $H = (\ker T)^{⊥} \oplus \ker T$ by $U = 0$ on $\ker T$ ; then $U$ is bounded and, because $∣ T ∣$ is self-adjoint with $∣ T ∣ e_{j} = s_{j} e_{j}$ , $U ∣ T ∣ e_{j} = U s_{j} e_{j} = s_{j} f_{j} = T e_{j}$ for every $j$ and $U ∣ T ∣ = 0 = T$ on $\ker T = ∣ T ∣^{- 1} (0)$ [A1], so $U ∣ T ∣ = T$ by continuity on the closed span of $\ker T$ and the $e_{j}$ , which is $H$ by [A2]. Finally $U^{*} U$ and $U U^{*}$ : for $x, y \in H$ one has $⟨ U x, U y ⟩ = ⟨ P x, P y ⟩$ where $P$ is the orthogonal projection onto $(\ker T)^{⊥}$ , because $U$ is isometric on $(\ker T)^{⊥}$ and vanishes on $\ker T$ , so $⟨ U^{*} U x, y ⟩ = ⟨ P x, y ⟩$ and $U^{*} U = P$ ; dually, for $y \in K$ the vector $U^{*} y \in (\ker T)^{⊥}$ is characterised by $⟨ U^{*} y, z ⟩ = ⟨ y, U z ⟩$ for all $z \in (\ker T)^{⊥}$ , so $U U^{*} y = y$ for $y \in \overline{ran T}$ and $U U^{*} y = 0$ for $y ⊥ \overline{ran T}$ , that is $U U^{*}$ is the orthogonal projection onto $\overline{ran T}$ .

6.1step 1.1step 2.1step 3.1step 4.1step 5.1A1∎

Conclusion. Claim 1 is [step 1.1] and [step 2.1]; claim 2 is [step 3.1], whose index set is $J$ by construction; claim 3 is [step 5.1] together with [step 4.1]. Claim 4 is the indexing discipline used throughout: $J$ indexes the positive singular values with multiplicity and is $\emptyset$ only for $T = 0$ , when the finite dimension is $r = 0$ ; it is finite exactly when the range is finite-dimensional. In that case the vanishing terms $s_{n} (T) = 0$ with $n > r$ are numerical padding and index no vector.

Singular value decomposition for compact operators

Statement

Facts & Assumptions

Proof

Depends on

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