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Trace Norm Unitary Invariance And Positive Formula

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. If $U : K \to K$ and $V : H \to H$ are unitary, then $∥ U A V ∥_{1} = ∥ A ∥_{1}$ . If $P : H \to H$ is non-negative (positive semidefinite), then

$∥ P ∥_{1} = {tr}_{H} (P) .$

Here non-negative means self-adjoint with $⟨ P v, v ⟩ \geq 0$ for all $v$ , as in Non-negative and positive operators.

Facts & Assumptions

Given: The finite-dimensional spaces and maps in the Statement.

[F1]

An SVD writes $A z = \sum_{i < r} s_{i} ⟨ z, e_{i} ⟩ f_{i}$ , and the trace norm is the sum of those singular values; singular values are the eigenvalues of $\sqrt{A^{*} A}$ and are well-defined (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition, Trace Norm, 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).

[F2]

A unitary endomorphism preserves orthonormal bases and has $V^{- 1} = V^{*}$ (Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces, For an endomorphism in finite dimension, preserving lengths, preserving inner products, carrying orthonormal bases to orthonormal bases, and $T^{*} T = I$ are equivalent).

[F3]

The adjoint identity is $⟨ T v, w ⟩ = ⟨ v, T^{*} w ⟩$ , 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]

Non-negative means self-adjoint with nonnegative quadratic form; such an endomorphism has an orthonormal eigenbasis with nonnegative real eigenvalues (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).

[F5]

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

[F6]

The trace of an endomorphism is the sum of diagonal entries in any basis, and is $0$ on the zero space (The basis-independent trace of an endomorphism of a finite-dimensional vector space).

Proof

1.1F1F2F3choosealgebra

Choose an SVD of $A$ as in [F1], with domain basis $(e_{j})$ and codomain basis $(f_{i})$ . By [F2], $(V^{*} e_{j})$ and $(U f_{i})$ are orthonormal bases. For every $z \in H$ , [F3] gives $U A V z = \sum_{j < r} s_{j} ⟨ V z, e_{j} ⟩ U f_{j} = \sum_{j < r} s_{j} ⟨ z, V^{*} e_{j} ⟩ U f_{j} .$ This is an SVD of $U A V$ with the same positive singular values, so [F1] gives $∥ U A V ∥_{1} = ∥ A ∥_{1}$ .

1.2F1F4F5choosealgebra

Let $P$ be non-negative. By [F4], choose an orthonormal eigenbasis $(e_{j})$ with $P e_{j} = λ_{j} e_{j}$ and $λ_{j} \geq 0$ . Since $P^{*} = P$ , $P^{*} P = P^{2}$ . The operator $P$ is a non-negative square root of $P^{2}$ , so uniqueness [F5] gives $\sqrt{P^{*} P} = P$ . Its singular values are therefore the $λ_{j}$ ; by [F1] the trace norm is $\sum_{j} λ_{j}$ .

2.1F1F6step 1.2∎

Computing the trace in the same basis by [F6] gives ${tr}_{H} (P) = \sum_{j} λ_{j} = ∥ P ∥_{1}$ . If $H = 0$ , both sums are empty and equal $0$ . This proves both assertions.

Depends on

Used by

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Sources

John Watrous, The Theory of Quantum Information (2018 author-hosted draft), Chapter 1, §1.1.3 (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)