Alphabeta Math
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Blockwise Frobenius errors propagate by disjoint block sums

Statement

For any finite matrix partition into disjoint blocks, if each local residual satisfies ∥Eb∥F≤δb, then the assembled residual satisfies ∥E∥F≤(∑bδb2)1/2. In particular, let A,B be conformable finite real or complex matrices, and let A~,B~ be approximations with ∥A−A~∥F≤ϵA and ∥B−B~∥F≤ϵB. If a successful block addition or block product from Tolerance-controlled H-matrix block addition or Tolerance-controlled H-matrix block product returns S^ or C^, with disjoint output blocks b and local Frobenius errors eb≤δb, put δ=(∑bδb2)1/2. Then

∥A+B−S^∥F≤ϵA+ϵB+δ,

and

∥AB−C^∥F≤ϵA∥B∥F+∥A∥FϵB+ϵAϵB+δ.

Facts & Assumptions

Given: The target matrices, their H-matrix approximations, successful block-operation outputs, and local block-error budgets.

[F1]

Tolerance-controlled H-matrix block addition and Tolerance-controlled H-matrix block product: near blocks are computed exactly; on success, each far output block has a certified Frobenius error bound eb no larger than its allocated tolerance.

[F2]

Bottom-up factor-sum recompression has the reported rank and Frobenius error bound: for each successful factor-sum tree, the returned block approximation error is at most the sum of its nodewise core-SVD discarded-tail norms, which is bounded by the sum of its node tolerances.

[F3]

Block-cluster leaves partition the index-pair set: every recursively generated terminal block partition is pairwise disjoint and covers the output index-pair set.

[F4]

Finite tensor arrays and the Frobenius inner product: the Frobenius norm is the Euclidean norm of the finite matrix-entry list.

[F5]

Proof

1.1F1F2F3F4algebra

Let Δb be the actual residual in output block b. For near blocks it is zero; for far blocks [F1] and [F2] give ∥Δb∥F≤eb≤δb. By [F3] their entry supports are disjoint, so [F4] gives ∥Δ∥F2=∑b∥Δb∥F2≤∑bδb2. Hence ∥Δ∥F≤δ.

2.1F4F5step 1.1algebra

For addition, write S^=A~+B~+Δ. Expanding the squared Frobenius norm as the finite entry inner product and applying [F5] gives the triangle inequality. Therefore ∥A+B−S^∥F≤∥A−A~∥F+∥B−B~∥F+∥Δ∥F≤ϵA+ϵB+δ.

3.1F4F5step 1.1algebra∎

For multiplication, put EA=A~−A and EB=B~−B. Since C^=A~B~+Δ, expansion gives AB−C^=−(EAB+AEB+EAEB+Δ). For any conformable X,Y, entrywise Cauchy–Schwarz gives ∣(XY)ij∣2≤(∑ℓ∣Xiℓ∣2)(∑ℓ∣Yℓj∣2); summing over i,j gives ∥XY∥F2≤∥X∥F2∥Y∥F2. Thus ∥EAB∥F≤ϵA∥B∥F, ∥AEB∥F≤∥A∥FϵB, and ∥EAEB∥F≤ϵAϵB. The triangle inequality from [F4,F5] and step 1.1 give the stated product bound.

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