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Trace Norm Duality And Norm Properties

Statement

Let $A : H \to K$ be a linear map (Linear map between vector spaces over the same field) between finite-dimensional complex inner-product spaces. Then

$∥ A ∥_{1} = \max {∣ {tr}_{H} (B^{*} A) ∣ : B : H \to K linear and ∥ B ∥ \leq 1} .$

Consequently, on the vector space of linear maps $H \to K$ , the trace norm is nonnegative, absolutely homogeneous, definite, and satisfies the triangle inequality.

Facts & Assumptions

Given: Finite-dimensional complex inner-product spaces $H, K$ and linear maps $A, D : H \to K$ ; where quantified, $B : H \to K$ is also a map $H \to K$ .

[F1]

The trace norm is the sum of the singular values, including zeros (Trace Norm).

[F2]

A finite map has an SVD $A z = \sum_{i < r} s_{i} ⟨ z, e_{i} ⟩ f_{i}$ , with orthonormal bases in its domain and codomain and positive singular values $s_{i}$ (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition, The singular values of a linear map as the eigenvalues of the positive square root of T^*T, Singular values are well defined because the positive square root of T^*T is unique).

[F3]

The adjoint is characterized by $⟨ T v, w ⟩ = ⟨ v, T^{*} w ⟩$ , and finite-dimensional adjoints exist uniquely (The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ , Every linear map between finite-dimensional inner product spaces has a unique adjoint).

[F4]

In an ordered basis, the trace of an endomorphism is the sum of its diagonal entries; it is basis-independent and is $0$ on the zero space (The basis-independent trace of an endomorphism of a finite-dimensional vector space).

[F5]

The operator norm is $0$ on the zero domain and otherwise is defined as the maximum of $∥ C z ∥$ over unit vectors $z$ (The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||).

[F6]

Cauchy–Schwarz gives $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[F7]

An orthonormal basis gives finite vector expansions, including the empty basis case (Bessel's inequality for a finite orthonormal list and Parseval's identity for an orthonormal basis).

[F8]

Complex modulus is nonnegative, multiplicative, and subadditive (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[F9]

In a fixed ordered basis, a linear map's $j$ -th matrix column is the coordinate column of the image of the $j$ -th basis vector (Coordinate columns $[v]_{B}$ and matrices $[T]_{B}^{C}$ of linear maps relative to ordered bases).

[F10]

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

Proof

1.1F4F9algebra

Fix an ordered basis of $H$ . For endomorphisms $R, S$ of $H$ and a scalar $α$ , [F9] shows that the diagonal entries of $α R + S$ are $α R_{j j} + S_{j j}$ : the $j$ -th coordinate column is obtained from $(α R + S) e_{j} = α R e_{j} + S e_{j}$ . Summing these entries by [F4] gives ${tr}_{H} (α R + S) = α {tr}_{H} (R) + {tr}_{H} (S)$ . Thus trace homogeneity and additivity used below follow from the diagonal-sum definition. The same calculation covers $H = 0$ by empty sums.

1.2F2F5F7algebra

Let $C : H \to K$ be any finite-dimensional linear map. If $H = 0$ , its operator norm is $0$ by [F5]. If $H \neq 0$ , choose an SVD as in [F2] with positive singular values $t_{1} \geq \dots \geq t_{q} > 0$ and orthonormal domain vectors $u_{1}, \dots, u_{\dim H}$ . If $q > 0$ , Bessel's inequality [F7] gives, for every unit vector $z$ , $∥ C z ∥^{2} = \sum_{j \leq q} t_{j}^{2} ∣ ⟨ z, u_{j} ⟩ ∣^{2} \leq t_{1}^{2}$ , with equality at $z = u_{1}$ . If $q = 0$ , the SVD says $C = 0$ , so every unit vector attains the value $0$ . Thus the maximum in [F5] exists for every map used below, including the zero-rank case.

1.3givenF1F2choose

Choose an SVD as in [F2], with $r$ positive singular values. Then $∥ A ∥_{1} = \sum_{i < r} s_{i}$ by [F1]. If $r = 0$ , [F2] gives $A = 0$ and this sum is empty.

1.4F2F3F4F7F10algebra

Write the SVD domain basis as $(e_{j})_{j < n}$ and its positive singular values as $s_{i}$ for $i < r$ . By [F4], [F7], and the fact that the $j$ -th diagonal entry is the $e_{j}$ -coordinate of the image of $e_{j}$ , ${tr}_{H} (B^{*} A) = \sum_{j < n} ⟨ B^{*} A e_{j}, e_{j} ⟩_{H} = \sum_{i < r} s_{i} ⟨ B^{*} f_{i}, e_{i} ⟩_{H} = \sum_{i < r} s_{i} ⟨ f_{i}, B e_{i} ⟩_{K} .$ For the last equality, [F3] gives $⟨ B e_{i}, f_{i} ⟩_{K} = ⟨ e_{i}, B^{*} f_{i} ⟩_{H}$ , and conjugate symmetry [F10] gives the displayed order. This is a direct endomorphism-trace calculation and uses no cyclic trace identity for maps.

2.1F2F3F7F10step 1.2construct

Suppose $r > 0$ and define $B_{0} : H \to K$ by $B_{0} z = \sum_{i < r} ⟨ z, e_{i} ⟩ f_{i}$ . By [F10], this is linear; it sends $e_{i}$ to $f_{i}$ for $i < r$ and sends the remaining basis vectors to $0$ . By [F7], $∥ B_{0} z ∥^{2} = \sum_{i < r} ∣ ⟨ z, e_{i} ⟩ ∣^{2} \leq ∥ z ∥^{2}$ , with equality at $z = e_{0}$ . The maximum property established in step 1.2 therefore gives $∥ B_{0} ∥ = 1$ . Define $C w = \sum_{i < r} ⟨ w, f_{i} ⟩ e_{i}$ . Expanding in the finite orthonormal bases using [F7] verifies $⟨ B_{0} z, w ⟩_{K} = ⟨ z, C w ⟩_{H}$ , so [F3] gives $C = B_{0}^{*}$ . Hence $B_{0}^{*} A e_{i} = s_{i} e_{i}$ for $i < r$ and $B_{0}^{*} A e_{j} = 0$ for $j \geq r$ . If $r = 0$ , set $B_{0} = 0$ .

2.2F5F6F8step 1.2step 1.3step 1.4algebra

Let $B : H \to K$ satisfy $∥ B ∥ \leq 1$ . By [F5] and the maximum property established in step 1.2, $∥ B e_{i} ∥ \leq 1$ for every unit vector $e_{i}$ ; [F6] then gives $∣ ⟨ f_{i}, B e_{i} ⟩ ∣ \leq 1$ . Applying [F8] to the sum in step 1.4 yields $∣ {tr}_{H} (B^{*} A) ∣ \leq \sum_{i < r} s_{i} = ∥ A ∥_{1}$ . The same bound is immediate when $r = 0$ .

3.1F1F2F4F7step 1.3step 2.2step 2.1

By [F4], [F7], and step 1.3, ${tr}_{H} (B_{0}^{*} A) = \sum_{i < r} s_{i} = ∥ A ∥_{1}$ ; when $r = 0$ , both sides are $0$ . Together with step 2.2 this proves the stated maximum, including zero-dimensional spaces.

4.1F1F2F8step 1.1step 3.1algebra

For any scalar $α$ , apply the maximum formula to $α A$ . By step 1.1, ${tr}_{H} (B^{*} (α A)) = α {tr}_{H} (B^{*} A)$ , so taking absolute values and maxima gives $∥ α A ∥_{1} = ∣ α ∣ ∥ A ∥_{1}$ . The definition and [F2] give $∥ A ∥_{1} \geq 0$ , with equality exactly when all singular values vanish, which by the SVD is exactly $A = 0$ .

5.1F8step 1.1step 3.1algebra∎

For maps $A, D : H \to K$ , every contraction $B$ satisfies $∣ {tr}_{H} (B^{*} (A + D)) ∣ \leq ∣ {tr}_{H} (B^{*} A) ∣ + ∣ {tr}_{H} (B^{*} D) ∣ \leq ∥ A ∥_{1} + ∥ D ∥_{1}$ by the additivity from step 1.1, [F8], and step 3.1. Taking the maximum over $B$ proves $∥ A + D ∥_{1} \leq ∥ A ∥_{1} + ∥ D ∥_{1}$ . Thus the trace norm has all stated norm properties.

Depends on

Used by

Dimension-dependent quantum entropy continuity and Holevo control Lemma
Fidelity classical measurement minimum Lemma
Finite braid compilation and adiabatic error budget Lemma
Finite cq wiretap achievability with strong Holevo secrecy Lemma
Finite square-root measurement packing bound for cq states Lemma
Finite unitary averaging and a decoupling second-moment bound Lemma
Replacing Bob's key by a copy of Alice's costs the correctness error Lemma
Trace Holder Inequality Proposition
Channel Composition Hybrid Bound Theorem
Channel Diamond Norm One Theorem
Channel Tensor Error Bound Theorem
Circuit-to-adiabatic simulation Theorem
Diamond Norm Ancilla Dimension Bound Theorem
Fidelity Purification Characterization Theorem
Fidelity range equality and products Theorem
Full Rank Pair No Broadcasting Theorem
Hybrid bound for a composed channel circuit Theorem
Operator error bounds measured output variation Theorem
QKD composable secrecy is bounded by correctness plus secrecy Theorem
Tensor-product state errors add under trace distance Theorem

Dependency tree · two levels

44 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John Watrous, The Theory of Quantum Information (2018 author-hosted draft), Chapter 1, §1.1.3, Eqs. (1.173)–(1.174) (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, arXiv:1907.09415v6 (2026-08-27), Appendices A.3, A.5 and A.7 (standard reference, not scraped)