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The Galerkin coarse correction is an energy-orthogonal projector

Statement

Let Ah∈Rn×n be symmetric positive definite and let P∈Rn×m have full column rank, where 1≤m≤n. Put AH=PTAhP. Then AH is symmetric positive definite, hence invertible, and the map C=In−PAH−1PTAh of Galerkin coarse matrices and the exact coarse-correction error map satisfies

C2=C,CP=0,PTAhC=0,CTAh=AhC.

Consequently Ran⁡(C)=Ran⁡(P)⊥Ah; C is the orthogonal projector onto that complement in the energy inner product ⟨x,y⟩Ah:=xTAhy.

Facts & Assumptions

Given: A symmetric positive-definite Ah∈Rn×n and a full-column-rank P∈Rn×m, where 1≤m≤n.

[F1]

Galerkin coarse matrices and the exact coarse-correction error map: AH=PTAhP, and when AH is invertible, C=In−PAH−1PTAh.

[F2]

For Hermitian positive-definite A, the exact solution of Ax=b is the unique minimizer of the quadratic energy: a real symmetric positive-definite matrix has a unique solution to AHz=b for every right-hand side b, so it is invertible.

[F3]

The energy inner product ⟨u,v⟩A=⟨Au,v⟩ and energy norm ∥u∥A for a Hermitian positive-definite matrix: ⟨x,y⟩Ah=xTAhy is an inner product and ∥x∥Ah2=xTAhx.

[F4]

Transpose is linear and involutive, and (AB)T=BTAT: transposition is involutive and reverses matrix products, (XY)T=YTXT.

Proof

1.1givenF1F2F4algebra

For every nonzero q∈Rm, full column rank gives Pq≠0, so qTAHq=(Pq)TAh(Pq)>0. Also AHT=PTAhTP=AH by [F4]. Thus AH is symmetric positive definite and [F2] makes it invertible; [F1] therefore defines C.

2.1F1step 1.1algebra

From PTAhP=AH, direct multiplication gives CP=P−PAH−1AH=0 and PTAhC=PTAh−AHAH−1PTAh=0. Moreover, C2=In−2PAH−1PTAh+PAH−1(PTAhP)AH−1PTAh=C.

3.1F1F3F4step 2.1algebra

Transposing AHAH−1=I=AH−1AH and using [F4] shows that (AH−1)T is an inverse of AHT=AH. If B and D are two-sided inverses of AH, then B=B(AHD)=(BAH)D=D, so the inverse is unique and (AH−1)T=AH−1. Thus [F4] yields CTAh=Ah−AhPAH−1PTAh=AhC, so C is self-adjoint in [F3]. Also PTAhC=0 says Ran⁡(C)⊆Ran⁡(P)⊥Ah. If w is in that complement, then PTAhw=0, so [F1] gives Cw=w; therefore every such w is in Ran⁡(C) and the ranges are equal.

4.1step 2.1step 3.1∎

The identities in step 2.1 and the range equality in step 3.1 prove all stated projection properties.

Depends on

Used by

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Sources