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Trace decomposition through generalized eigenspaces and the invariant quotient

Statement

Assume the Axiom of Choice. Let $H$ be a separable complex Hilbert space and let $T : H \to H$ be trace class. Write $Λ (T) = {λ \in σ (T) : λ \neq 0}$ , let $G_{λ} (T)$ be the generalized eigenspace and let $m_{a l g} (λ; T) = \dim G_{λ} (T)$ . Define $M : = \overline{span {G_{λ} (T) : λ \in Λ (T)}}, P : = P_{M}, Q : = I - P,$ where $P_{M}$ is the Hilbert orthogonal projection onto $M$ . Put $T_{M} : = T ∣_{M}$ and $D : = (Q T Q) ∣_{M^{⊥}}$ . Then $T_{M}$ and $D$ are trace class, and $D$ has no nonzero spectral values (the assertion is vacuous if $M^{⊥} = {0}$ ). With traces taken on their displayed Hilbert spaces, ${tr}_{H} (T) = {tr}_{M} (T_{M}) + {tr}_{M^{⊥}} (D), {tr}_{M} (T_{M}) = \sum_{λ \in Λ (T)} m_{a l g} (λ; T) λ,$ and the eigenvalue sum is absolutely convergent. The extension $Q T Q : H \to H$ is quasinilpotent and has trace zero.

Facts & Assumptions

Given: AC; a separable complex Hilbert space $H$ ; and a trace-class operator $T : H \to H$ .

[A1]

AC selects from every family of nonempty sets (The Axiom of Choice).

[A2]

In ZF, $A C \Rightarrow D C \Rightarrow {A C}_{ω}$ (AC implies DC implies countable choice).

[A3]

A complex Hilbert space is a Banach space in its induced norm (Hilbert space).

[A4]

A trace-class operator is compact and bounded (Trace class operator).

[A5]

A compact operator on a complex Banach space has finite-dimensional generalized eigenspaces at its nonzero spectral values; every such value is an eigenvalue, and only finitely many spectral values have modulus at least any fixed $ε > 0$ (Riesz schauder spectrum of a compact operator).

[A6]

For each nonzero eigenvalue $λ$ , $G_{λ} (T)$ is the stabilized kernel of $(T - λ I)^{m}$ , is finite dimensional, and $m_{a l g} (λ; T) = \dim G_{λ} (T) = rank P_{λ}$ (Algebraic multiplicity of a nonzero compact-operator eigenvalue).

[A7]

For an isolated spectral value, the Riesz projection is a bounded idempotent commuting with $T$ , its range and kernel are closed invariant subspaces giving a direct sum, and the restriction spectra are the two spectral parts (Riesz spectral projection, Riesz spectral projection properties).

[A8]

A subspace $W$ is $T$ -invariant when $T (W) \subseteq W$ , and this invariance makes $\overset{ˉ}{T} (v + W) : = T (v) + W$ a well-defined linear operator on $H / W$ with canonical projection $π : H \to H / W$ satisfying $π T = \overset{ˉ}{T} π$ (Invariant subspaces, restrictions, and induced quotient operators, The quotient vector space $V / W$ and its canonical projection, Invariance makes the induced quotient operator well defined and linear, with $π T = \overset{ˉ}{T} π$ ).

[A9]

The quotient seminorm is $∥ x + W ∥ = \inf_{w \in W} ∥ x + w ∥$ and is a norm when $W$ is closed (The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M)), The quotient seminorm is a norm exactly when the subspace is closed).

[A10]

Under AC, if $X$ is Banach and $K : X \to X$ is compact, then $I - K$ is injective if and only if it is surjective, and either condition gives a bounded inverse (Fredholm alternative for identity minus compact).

[A11]

A trace-class operator remains trace class after composition on either side with bounded maps between Hilbert spaces (Trace class is a two sided Banach operator ideal).

[A12]

For every supplied Hilbert basis $E$ of a trace-class operator $S$ , its relative trace is ${tr}_{E} (S) = \sum_{e \in E} ⟨ S e, e ⟩$ ; the series is absolutely convergent, and the trace theorem identifies it with the basis-independent trace $tr (S)$ (Trace of a trace class operator, Trace is absolutely convergent and basis independent).

[A13]

A closed subspace of a Hilbert space is Hilbert; an orthogonal projection onto a closed subspace gives $H = M \oplus M^{⊥}$ , is self-adjoint, and is contractive (Orthogonal decomposition by a closed subspace, The Hilbert orthogonal projection onto a closed subspace, Hilbert projections are linear, self-adjoint and contractive).

[A14]

From a supplied dense sequence in a Hilbert space, Gram--Schmidt gives a finite or countable Hilbert basis (A Hilbert space with a dense sequence has a finite or countable orthonormal basis).

[A15]

For a trace-class compact operator, the eigenvalue sequence repeated by algebraic multiplicity can be listed so that $\sum_{j} ∣ λ_{j} (T) ∣ \leq ∥ T ∥_{1}$ (Weyl product and sum inequalities for compact operators).

[A16]

Every finite-dimensional nilpotent endomorphism has an ordered basis concatenating Jordan strings (Every finite-dimensional nilpotent endomorphism has a basis of Jordan strings); on a string $(v_{1}, \dots, v_{m})$ at $λ$ , $(T - λ I) v_{1} = 0$ and $(T - λ I) v_{j} = v_{j - 1}$ for $j \geq 2$ (Jordan blocks, Jordan strings, and their endpoints).

[A17]

If $S$ is trace class on a separable complex Hilbert space and $σ (S) \subseteq {0}$ , then $tr (S) = 0$ (A quasinilpotent trace-class operator has zero trace).

[A18]

The complex inner product is linear in its first argument and conjugate-linear in its second (Real and complex inner-product spaces and their induced length).

[A19]

A closed linear subspace of a Banach space is Banach in its induced norm (A closed subspace of a Banach space is Banach).

[A20]

For a bounded operator on a complex Banach space, $λ \notin σ (T)$ exactly when $λ I - T$ is bijective with bounded inverse (Spectrum and resolvent of a bounded operator).

[A21]

Boundedness of $T$ supplies a constant $C_{T} \geq 0$ with $∥ T x ∥ \leq C_{T} ∥ x ∥$ for every $x \in H$ (A bounded linear operator between normed spaces).

[A22]

Separability of $H$ supplies a dense sequence in $H$ (Separability: the existence of an at most countable dense subset).

Choice accounting: The exact assumption is AC. It supplies DC for Riesz--Schauder/Fredholm suppliers and AC $_{ω}$ for trace, projection and separable-basis suppliers. The Weyl list supplies an enumeration of the nonzero eigenvalue multiset; AC $_{ω}$ permits choosing Jordan-string bases for its countably many finite-dimensional generalized eigenspaces. The basis of $M^{⊥}$ is obtained by projecting a supplied dense sequence of $H$ and applying the choice-free Gram--Schmidt construction. No ambient basis is used without being supplied or constructed.

Source audit: Kostenko's §3.4.4 proof of Theorem 3.4.7 derives the spectral product and trace identity by invoking Theorem 3.4.5, the Hadamard minimal-type product formula. Van Neerven's Theorem 14.33 proof obtains the determinant spectral product from Theorem 14.43, whose proof invokes Lemma 14.42; Proposition 14.22 separately gives the eigenvalue absolute-sum bound and uses finite-dimensional invariant generalized-eigenspace sums. These routes are comparison only. This item proves the trace on $M$ using an adapted orthonormal basis and proves the compressed quotient has no nonzero spectrum using Riesz splitting and the compact Fredholm alternative. Kostenko's Theorem 3.4.7 proof and van Neerven's Proposition 14.22 and Theorem 14.33 arguments were read in full; no source premise is left unverified.

Proof

technique · direct

1.1A3A4algebra

If $H = {0}$ , then $M = M^{⊥} = {0}$ , all operators and traces in the claim are zero, and the eigenvalue sum is empty; hence assume $H \neq {0}$ .

2.1A3A4A6A13step 1.1algebra

Each $G_{λ} (T)$ is $T$ -invariant because $(T - λ I)^{m} T x = T (T - λ I)^{m} x = 0$ for $x \in G_{λ} (T)$ ; boundedness of $T$ then makes its closed span $M$ invariant. By [A13], $P = P_{M}$ and $Q = I - P$ are bounded orthogonal projections, $H = M \oplus N$ with $N = M^{⊥}$ , and both $M, N$ are closed Hilbert subspaces.

3.1A11step 2.1algebra

Let $i_{M} : M ↪ H$ and $i_{N} : N ↪ H$ be inclusions. Invariance gives $T_{M} = P T i_{M}$ , while $D = Q T i_{N}$ and $B : = Q T Q$ on $H$ ; by [A11] all three are trace class, and $B ∣_{M} = 0$ , $B ∣_{N} = D$ .

4.1A1A2A6A12A15A16A18step 3.1algebra

Let $(λ_{j})$ be the finite or countable nonzero eigenvalue list from [A15], repeated by algebraic multiplicity. For each distinct $λ$ , choose a Jordan-string basis of $G_{λ} (T)$ for $(T - λ I) ∣_{G_{λ} (T)}$ using [A6, A16]. These generalized eigenspaces are linearly independent: in a finite relation $\sum_{μ} x_{μ} = 0$ , $x_{μ} \in G_{μ}$ , applying $R_{λ} = \prod_{μ \neq λ} (T - μ I)^{m_{μ}}$ kills every other term, while each factor on $G_{λ}$ is $(λ - μ) I + N_{λ}$ with $N_{λ}$ nilpotent and hence invertible by a finite geometric sum; thus $x_{λ} = 0$ . Order the distinct eigenvalues by their first occurrence in $(λ_{j})$ and concatenate their string bases. Every finite initial span is $T$ -invariant, and its successive one-dimensional quotient acts by the corresponding eigenvalue. Applying Gram--Schmidt preserves these initial spans, so it gives a Hilbert basis $(e_{j})$ of $M$ with $T e_{j} - λ_{j}^{'} e_{j} \in span (e_{1}, \dots, e_{j - 1})$ , where $(λ_{j}^{'})$ is a reordering of $(λ_{j})$ . By [A12], [A18], and [A15], ${tr}_{M} (T_{M}) = \sum_{j} ⟨ T e_{j}, e_{j} ⟩ = \sum_{j} λ_{j}^{'} = \sum_{j} λ_{j}$ , and the sum is absolutely convergent. The empty and finite lists give the empty and finite bases.

4.2A8A9A13A21step 2.1step 3.1algebra

Let $X : = H / M$ with its quotient norm and let $π : H \to X$ be the canonical projection. By [A8, A9], $\overset{ˉ}{T} (x + M) = T x + M$ is well defined. For $m \in M$ , $T x + T m \in T x + M$ , so [A21] gives $∥ \overset{ˉ}{T} (x + M) ∥ \leq C_{T} ∥ x + m ∥$ ; taking the infimum over $m$ shows that $\overset{ˉ}{T}$ is bounded. The map $J : N \to X$ , $J (n) = n + M$ , is an isometric isomorphism: every coset has the representative $Q x \in N$ , and $∥ n + M ∥ = \inf_{m \in M} ∥ n - m ∥ = ∥ n ∥$ by orthogonality. For $D = Q T ∣_{N}$ , $J D = \overset{ˉ}{T} J$ .

5.1A12A13A14A22step 2.1step 3.1

By [A22] take a dense sequence $(h_{j})$ in $H$ . Contractivity of $P, Q$ makes $(P h_{j})$ dense in $M$ and $(Q h_{j})$ dense in $N$ , so [A14] supplies Hilbert bases $E_{M}$ of $M$ and $E_{N}$ of $N$ ; the basis of $M$ may be the adapted one from step 4.1. Their union is a Hilbert basis of $H$ . For $e \in E_{M}$ , $T e = T_{M} e$ ; for $f \in E_{N}$ , $T f - B f = P T f \in M ⊥ f$ , and $B e = 0$ . Summing the absolutely convergent diagonal series from [A12] over these two disjoint basis parts yields ${tr}_{H} (T) = {tr}_{M} (T_{M}) + {tr}_{H} (B)$ and ${tr}_{H} (B) = {tr}_{N} (D)$ .

5.2A3A4A8A10A13A19A20step 2.1step 3.1step 4.2algebra

Fix $λ \neq 0$ with $λ \notin σ (T)$ and set $A = λ I - T$ . By [A20], $A$ has a bounded inverse on $H$ . Its restriction to $M$ is injective; $A ∣_{M} = λ (I_{M} - T_{M} / λ)$ , where $T_{M}$ is compact by [A4, step 3.1], and $M$ is Banach by [A3, A13, A19]. The Fredholm alternative [A10] makes $A ∣_{M}$ surjective with bounded inverse. Consequently $A^{- 1} (M) = M$ and the induced quotient operator $\overset{ˉ}{A} = λ I_{X} - \overset{ˉ}{T}$ has the bounded inverse induced by $A^{- 1}$ .

5.3A3A4A5A6A7A8A9A10A19A20step 2.1step 3.1step 4.2algebra

Fix $λ \neq 0$ with $λ \in σ (T)$ . By [A5], $λ$ is an eigenvalue; the finiteness of every nonzero spectral annulus isolates it, so [A7] gives the Riesz projection $P_{λ}$ , with $G : = G_{λ} (T) = ran P_{λ}$ and $Y : = \ker P_{λ}$ . Put $M_{0} : = M \cap Y$ . For $x \in G_{μ} (T)$ with $μ \neq λ$ , $P_{λ} x \in G$ and $(T - μ I)^{m} P_{λ} x = P_{λ} (T - μ I)^{m} x = 0$ ; on $G$ , $T - μ I = (λ - μ) I + N_{λ}$ is invertible, so $P_{λ} x = 0$ , while $P_{λ}$ is the identity on $G$ . Continuity and the definition of $M$ give $P_{λ} (M) \subseteq G \subseteq M$ , hence $M = G \oplus M_{0}$ . If $Y \neq {0}$ , [A7] gives that $A_{Y} : = λ I_{Y} - T ∣_{Y}$ is boundedly invertible; its restriction to $M_{0}$ is injective. The subspace $M_{0}$ is closed in $M$ , so [A19] makes it Banach. The restriction $T ∣_{M_{0}}$ is compact because a bounded sequence in $M_{0}$ has a subsequence whose $T$ -images converge in $H$ , and the limit lies in $M_{0}$ by closedness. Thus [A10] makes $A ∣_{M_{0}} = λ (I_{M_{0}} - T ∣_{M_{0}} / λ)$ boundedly invertible and $A_{Y}^{- 1} (M_{0}) = M_{0}$ . Thus $A_{Y}$ and its inverse both preserve $M_{0}$ , so they induce mutually inverse bounded operators on $Y / M_{0}$ . The map $Φ : Y / M_{0} \to H / M$ , $y + M_{0} \mapsto y + M$ , is well defined and bounded: changing $y$ by an element of $M_{0}$ does not change its image, and $∥ y + M ∥ \leq ∥ y + M_{0} ∥$ . It is onto because $x + M = (I - P_{λ}) x + M$ and $(I - P_{λ}) x \in Y$ ; it is one-to-one because $Y \cap M = M_{0}$ . Its inverse is therefore $x + M \mapsto (I - P_{λ}) x + M_{0}$ . This inverse is well defined since $(I - P_{λ}) M \subseteq M_{0}$ , and bounded with norm at most $∥ I - P_{λ} ∥$ : for every $m \in M$ , $(I - P_{λ}) (x + m)$ represents the same image modulo $M_{0}$ , so taking the infimum over $m$ gives the bound. The map $Φ$ intertwines the induced operators because $A_{Y} y + M = A y + M$ for $y \in Y$ . Hence $\overset{ˉ}{A} = λ I_{X} - \overset{ˉ}{T}$ is boundedly invertible. If $Y = {0}$ then $P_{λ} = I_{H}$ , so $G = H \subseteq M$ , hence $M = H$ and $X = {0}$ , whose unique endomorphism is bijective, giving the same quotient conclusion.

6.1A20step 4.2step 5.2step 5.3algebra

Steps 5.2--5.3 show that $λ I_{X} - \overset{ˉ}{T}$ is boundedly invertible for every $λ \neq 0$ . By step 4.2, $λ I_{N} - D$ is also boundedly invertible. Since $B = 0_{M} \oplus D$ on the orthogonal sum $H = M \oplus N$ , the operator $λ I_{H} - B = λ I_{M} \oplus (λ I_{N} - D)$ has a bounded inverse for every $λ \neq 0$ . Thus [A20] gives $σ_{H} (B) \subseteq {0}$ ; the compression $D$ has no nonzero spectral values whenever $N \neq {0}$ .

7.1

Apply [A17] to the trace-class quasinilpotent operator $B$ on the original separable $H$ to get ${tr}_{H} (B) = 0$ . Step 5.1 then gives ${tr}_{H} (T) = {tr}_{M} (T_{M})$ , and step 4.1 identifies this with the absolutely convergent eigenvalue sum. If there are no nonzero eigenvalues then $M = 0$ , $B = T$ , and the same argument yields the empty sum $0$ ; this includes $T = 0$ . If $H = C$ and $T = q I$ , then for $q \neq 0$ the sole eigenvalue is $q$ with multiplicity one, $M = H$ , and the two traces are $q$ and $0$ ; for $q = 0$ the eigenvalue list and $M$ are empty/zero and all traces vanish. There is no endpoint parameter, and the statement is not an equivalence, so both iff directions are inapplicable. AC is explicit in [A1] and propagates to AC $_{ω}$ through [A2] for the trace, Weyl, projection and basis suppliers. [A1, A2, A17, step 4.1, step 5.1, step 6.1] \qed

Depends on

Used by

Spectral product from traces of powers Lemma

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Sources

Kostenko, Trace Ideals with Applications, §3.4.4, Theorem 3.4.7 proof, printed pp. 41–42 (PDF pp. 50–51); comparison only (standard reference, not scraped)
van Neerven, Functional Analysis, Proposition 14.22, printed pp. 574–575 (PDF pp. 585–586), and Theorem 14.33 proof, §14.5.a, printed pp. 583–591 (PDF pp. 594–602); comparison only (standard reference, not scraped)