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Hermitian Trace Norm Positive Negative Parts

Statement

Let $H$ be a finite-dimensional complex inner-product space and let $X : H \to H$ be self-adjoint. There are non-negative endomorphisms $X_{+}, X_{-}$ such that

$X = X_{+} - X_{-}, ⟨ X_{+} u, X_{-} v ⟩ = 0 (u, v \in H),$

and

$∥ X ∥_{1} = tr (X_{+}) + tr (X_{-}) .$

If $tr (X) = 0$ , then $∥ X ∥_{1} = 2 tr (X_{+}) = 2 tr (X_{-})$ .

The trace norm is the finite singular-value norm from Trace Norm.

Facts & Assumptions

Given: A finite-dimensional complex inner-product space $H$ and a self-adjoint endomorphism $X : H \to H$ .

[F1]

A self-adjoint endomorphism of a finite-dimensional complex inner-product space is normal and has an orthonormal eigenbasis (Self-adjoint and normal endomorphisms of a finite-dimensional real or complex inner product space, Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely).

[F2]

The inner product is linear in its first argument, conjugate-symmetric, and its induced length satisfies $⟨ v, v ⟩ = ∥ v ∥^{2}$ (Real and complex inner-product spaces and their induced length).

[F3]

A diagonal self-adjoint endomorphism with nonnegative real eigenvalues is non-negative (Non-negative and positive operators, A non-negative operator is equivalently self-adjoint with nonnegative eigenvalues, a positive semidefinite matrix in an orthonormal basis, or an operator of the form S^*S).

[F4]

For self-adjoint $H$ , the trace norm is the trace of $\sqrt{H^{*} H}$ (Trace Norm Spectral Formula).

[F5]

Trace is the sum of diagonal entries in any basis (The basis-independent trace of an endomorphism of a finite-dimensional vector space).

[F6]

Every vector expands in a finite orthonormal basis, including the empty basis (Bessel's inequality for a finite orthonormal list and Parseval's identity for an orthonormal basis).

[F7]

A non-negative endomorphism has a unique non-negative square root (A non-negative operator has a unique non-negative square root).

[F8]

The adjoint identity is $⟨ T v, w ⟩ = ⟨ v, T^{*} w ⟩$ (The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ ).

[F9]

A linear map preserves linear combinations (Linear map between vector spaces over the same field).

Proof

1.1givenF1F2F8choose

By [F1], choose an orthonormal eigenbasis $(e_{j})_{j < n}$ with $X e_{j} = λ_{j} e_{j}$ . Self-adjointness and conjugate symmetry give $λ_{j} = ⟨ X e_{j}, e_{j} ⟩ = ⟨ e_{j}, X^{*} e_{j} ⟩ = ⟨ e_{j}, X e_{j} ⟩ = \overline{⟨ X e_{j}, e_{j} ⟩}$ , so every $λ_{j}$ is real.

2.1F2F3F6F9step 1.1algebraconstruct

For $v \in H$ , define $X_{+} v = \sum_{j < n} \max (λ_{j}, 0) ⟨ v, e_{j} ⟩ e_{j}$ and $X_{-} v = \sum_{j < n} \max (- λ_{j}, 0) ⟨ v, e_{j} ⟩ e_{j}$ . These are linear by the first-slot rule [F2] and the finite-sum definition [F9]. Their diagonal entries are real and nonnegative. If $v = \sum_{j} c_{j} e_{j}$ , then [F2] gives $⟨ X_{+} v, v ⟩ = \sum_{j} \max (λ_{j}, 0) ∣ c_{j} ∣^{2} \geq 0$ , and the same calculation applies to $X_{-}$ . Thus [F3] makes both operators non-negative. Every $X_{+} u$ is a linear combination of basis vectors with $λ_{j} > 0$ , while every $X_{-} v$ is a linear combination of basis vectors with $λ_{j} < 0$ ; these subsets of the orthonormal basis are orthogonal. Thus $⟨ X_{+} u, X_{-} v ⟩ = 0$ for all $u, v$ . On every $e_{j}$ , $(X_{+} - X_{-}) e_{j} = λ_{j} e_{j} = X e_{j}$ , hence $X = X_{+} - X_{-}$ .

3.1F3F4F5F7step 1.1step 2.1algebra

In this basis $X^{*} X = X^{2}$ has eigenvalues $λ_{j}^{2}$ . Define $R$ by $R e_{j} = ∣ λ_{j} ∣ e_{j}$ . Its diagonal entries are nonnegative, so [F3] makes it non-negative, and $R^{2} = X^{*} X$ ; uniqueness [F7] therefore gives $R = \sqrt{X^{*} X}$ . Thus [F4]–[F5] give $∥ X ∥_{1} = \sum_{j} ∣ λ_{j} ∣$ , $tr (X_{+}) = \sum_{λ_{j} > 0} λ_{j}$ , and $tr (X_{-}) = \sum_{λ_{j} < 0} (- λ_{j})$ . Their latter two sums add to $\sum_{j} ∣ λ_{j} ∣$ , proving the trace-norm identity.

4.1F5step 3.1algebra∎

If $tr (X) = 0$ , [F5] gives $\sum_{j} λ_{j} = 0$ . Hence the sum of positive eigenvalues equals the sum of absolute values of negative eigenvalues. Step 3.1 then yields $∥ X ∥_{1} = 2 tr (X_{+}) = 2 tr (X_{-})$ . If $n = 0$ , all sums are empty and the identities still hold.

Depends on

Used by

One uniform phase query gives small trace-distance advantage Counterexample
Dimension-dependent quantum entropy continuity and Holevo control Lemma
Finite unitary averaging and a decoupling second-moment bound Lemma
Pure state trace distance Lemma
Helstrom Binary Discrimination Theorem
Measurement Variation At Most Trace Distance Theorem
Trace Distance Range And Equality Theorem

Dependency tree · two levels

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Sources

John Watrous, The Theory of Quantum Information (2018 author-hosted draft), Chapter 1, Jordan–Hahn decomposition, and Chapter 3, §3.1.1 (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, arXiv:1907.09415v6 (2026-08-27), Appendices A.5 and A.7 (standard reference, not scraped)