Consider an arbitary positive semidefinite operator ρ, acting on ℂ<sup>A</sup> ⊗ ℂ<sup>B</sup> ⊗ ℂ<sup>C</sup>, for A,B,C finite. Also, let P be an orthogonal projector on ℂ<sup>B</sup> ⊗ ℂ<sup>C</sup> . For the sake of concision, I will write R = 1<sub>ℂ<sup>A</sup></sub> ⊗ P ; this of course is also an orthogonal projector. Consider the completely positive transformation > M(ρ) = (1 − R) ρ (1 − R) + R ρ R . As R is an orthogonal projector, it is easy to show that || M(ρ) ||<sub>2</sub> ≤ || ρ ||<sub>2</sub> . This is because we may represent ρ as matrix in a basis consisting of the eigenvectors of R; if we divide ρ into block according to rows/columns representing vectors in the image or the kernel of R, the effect of the map M is to set the non-diagonal blocks to zero. I am interested in how the map M may similarly affect the operator 2 norm of reduced operators on ℂ<sup>A</sup> ⊗ ℂ<sup>B</sup>. So I would like to know: Is it also true that || tr<sub>C</sub>( M(ρ) ) ||<sub>2</sub> ≤ || tr<sub>C</sub>(ρ) ||<sub>2</sub> — where tr<sub>C</sub> is the trace operator acting on ℂ<sup>C</sup>, taken in tensor product with 1<sub>ℂ<sup>A</sup></sub> ⊗ 1<sub>ℂ<sup>B</sup></sub> ?