Manuscript VII-E@1 · VII-E.tex · lemma · b713cf3c

Lemma: Intertwining forced by sharp pullback

Record compatibility, residuals, and composition › Sharp-sector intertwining · lines 1381–1421 · proof included or linked

VII-E@1#lem:intertwining, VII-E@1#eq:kraus-intertwining, VII-E@1#eq:corner-identity

If $T^{*} (E_{w}) = Π_{w}$ , then every Kraus operator $K_{α}$ in any Kraus representation of $T$ satisfies

E_{w} K_{α} = K_{α} Π_{w} .

Consequently,

E_{w} T (X) E_{w} = T (Π_{w} X Π_{w}) (X \in L (H_{f})) .

Proof. For every vector $v$ ,

α \sum ∥(I - E_{w}) K_{α} Π_{w} v ∥^{2} = ⟨ v, Π_{w} T^{*} (I - E_{w}) Π_{w} v ⟩ = ⟨ v, Π_{w} (I - Π_{w}) Π_{w} v ⟩ = 0.

Every summand is nonnegative. Hence $(I - E_{w}) K_{α} Π_{w} = 0$ for every $α$ . Similarly,

α \sum ∥ E_{w} K_{α} (I - Π_{w}) v ∥^{2} = ⟨ v, (I - Π_{w}) T^{*} (E_{w}) (I - Π_{w}) v ⟩ = 0,

so $E_{w} K_{α} (I - Π_{w}) = 0$ . These two identities give

E_{w} K_{α} = E_{w} K_{α} Π_{w} = K_{α} Π_{w} .

Substitution into the Kraus representation proves eq:corner-identity.

Displayed formulas in this passage

VII-E@1#eq:kraus-intertwining · line 1385
$E_{w} K_{α} = K_{α} Π_{w} .$
VII-E@1#eq:corner-identity · line 1390
$E_{w} T (X) E_{w} = T (Π_{w} X Π_{w}) (X \in L (H_{f})) .$
unlabeled · line 1399
$α \sum ∥(I - E_{w}) K_{α} Π_{w} v ∥^{2} = ⟨ v, Π_{w} T^{*} (I - E_{w}) Π_{w} v ⟩ = ⟨ v, Π_{w} (I - Π_{w}) Π_{w} v ⟩ = 0.$
unlabeled · line 1408
$α \sum ∥ E_{w} K_{α} (I - Π_{w}) v ∥^{2} = ⟨ v, (I - Π_{w}) T^{*} (E_{w}) (I - Π_{w}) v ⟩ = 0,$
unlabeled · line 1416
$E_{w} K_{α} = E_{w} K_{α} Π_{w} = K_{α} Π_{w} .$