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PUE singular-space invariant decomposition

Statement

Let A:HR→HL be the PUE map of Projected-unitary encoding and singular vectors, with positive singular pairs (σj,vj,uj). For every 0<σj<1, set

wj=(I−ΠR)U∗uj1−σj2,w~j=(I−ΠL)Uvj1−σj2.

The vectors wj form an orthonormal family in ker⁡ΠR and the vectors w~j form an orthonormal family in ker⁡ΠL. In the input basis (vj,wj) and output basis (uj,w~j),

U∣span⁡(vj,wj)=(σj1−σj21−σj2−σj).

For σj=1, Uvj=uj and the complementary vector is absent. A right zero-singular vector v∈ker⁡A∩HR satisfies Uv∈ker⁡ΠL; a left zero-singular vector u∈ker⁡A∗∩HL satisfies U∗u∈ker⁡ΠR. Let MR be the orthogonal sum of all positive-singular input planes or lines, ker⁡A, and U∗(ker⁡A∗); let ML be the corresponding sum of positive-singular output planes or lines, U(ker⁡A), and ker⁡A∗. Then U(MR)=ML. The residual ambient subspaces MR⊥ and ML⊥ lie in ker⁡ΠR and ker⁡ΠL, and U(MR⊥)=ML⊥. Thus each named block is invariant under its corresponding projector, with no implicit pairing of the two kernels.

Facts & Assumptions

Given: A PUE (U,ΠR,ΠL) and the SVD data from Projected-unitary encoding and singular vectors.

[F1]

Projected-unitary encoding and singular vectors: A=ΠLU∣HR, A∗=ΠRU∗∣HL, the positive singular pairs satisfy Avj=σjuj and A∗uj=σjvj, the trailing bases span ker⁡A and ker⁡A∗, and U∗ is unitary.

[F2]

Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition: every finite-dimensional linear map has complete orthonormal singular bases and a positive-term SVD formula.

[F5]

The orthogonal projection PWv is the W-component in V=W⊕W⊥: a vector is the sum of its projection into a subspace and an orthogonal component.

[F6]

Pythagoras and finite orthogonal sums: orthogonal components have squared norms that add.

Proof

1.1givenF1F3F4F5F6algebra

For 0<σi,σj<1, [F1] gives ΠRU∗ui=σivi. Since U∗ preserves inner products by [F1], orthogonal projection and [F6] give ⟨(I−ΠR)U∗ui,(I−ΠR)U∗uj⟩=⟨U∗ui,U∗uj⟩−⟨ΠRU∗ui,ΠRU∗uj⟩=δij(1−σi2). Dividing by the positive square roots shows that the wi are orthonormal and lie in ker⁡ΠR. The same calculation, using ΠLUvi=σiui, proves that the w~i are orthonormal in ker⁡ΠL.

1.2F1F3F5F6algebra

If σi=1, then ΠRU∗ui=vi has norm one, as does U∗ui. By the orthogonal decomposition in [F5] and Pythagoras [F6], its component in ker⁡ΠR is zero, so U∗ui=vi and Uvi=ui.

2.1F1F3F4step 1.1algebra

Decompose Uvi=ΠLUvi+(I−ΠL)Uvi=σiui+1−σi2 w~i. Also 1−σi2 Uwi=U(I−ΠR)U∗ui=ui−UΠRU∗ui=ui−σiUvi, where the middle equality uses UU∗=I and the last uses [F1]. Substitution of the first equation yields Uwi=1−σi2 ui−σiw~i, which is the displayed matrix.

3.1F1F2F3F4F5step 1.1step 1.2algebra∎

If v∈ker⁡A∩HR, then ΠLUv=Av=0, so Uv∈ker⁡ΠL. If u∈ker⁡A∗∩HL, then ΠRU∗u=A∗u=0, so U∗u∈ker⁡ΠR. Every right singular vector lies in HR and is orthogonal to each wi and each U∗u from the left kernel, since those vectors lie in ker⁡ΠR. For a positive singular vector and a left-kernel vector, ⟨(I−ΠR)U∗ui,U∗uk⟩=⟨U∗ui,U∗uk⟩−⟨ΠRU∗ui,ΠRU∗uk⟩=0−0=0, because ui⊥uk and ΠRU∗uk=0. Thus the listed source blocks are mutually orthogonal. Their orthogonal complement is contained in ker⁡ΠR, because the right singular vectors, including a basis of ker⁡A, span HR. Applying U preserves orthogonality. The listed output blocks are also mutually orthogonal: the left singular vectors, including a basis of ker⁡A∗, span HL; the vectors w~i are orthonormal by step 1.1; and for v∈ker⁡A, ⟨w~i,Uv⟩=⟨Uvi,Uv⟩−⟨ΠLUvi,ΠLUv⟩=0, since vi⊥v and Av=0. The remaining named output vectors Uv lie in ker⁡ΠL, so they are orthogonal to HL. The output complement is therefore contained in ker⁡ΠL. Since U maps each listed source block onto its corresponding output block, it maps the source complement onto the output complement. These are ambient-space decompositions; the residual blocks are not part of HR or HL. Each projector phase is scalar on the residual source and output blocks, so no residual component mixes into a projected singular-space block.

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