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Embedding Ran(A) alone does not control the sketch-and-solve residual

Statement refuted

False claim: If S is an ε-subspace embedding for Ran⁡(A), then every exact sketch least-squares minimizer x^ satisfies

∥Ax^−b∥22≤1+ε1−εmin⁡x∥Ax−b∥22.

Facts & Assumptions

Given: The claim above and the matrices in the Counterexample below.

[F1]

An ε-embedding preserves every squared norm on its stated fixed subspace within factors 1±ε (Oblivious squared-norm subspace embeddings).

[F2]

Sketch-and-solve minimizes ∥S(Ax−b)∥2 on its exact converged branch (Rank-aware sketch-and-solve least squares).

Counterexample

1.1F1algebra

In R2, set A=e1=(1,0)T, b=e1+e2=(1,1)T, and S=(11). For every x∈R, ∥S(Ax)∥2=∣x∣=∥Ax∥2, so [F1] holds on Ran⁡(A)=span⁡(e1) with ε=1/4 (indeed with zero distortion).

1.2F2algebra

The sketched residual is S(Ax−b)=x−2, so its unique minimizer is x^=2 by [F2]. The original squared residual is ∥Ax−b∥22=(x−1)2+1, whose minimum is 1 at x∗=1; at x^=2 it equals 2.

2.1step 1.1step 1.2algebra∎

At ε=1/4, the claimed factor is (1+ε)/(1−ε)=5/3, but 2>5/3⋅1. Thus preserving Ran⁡(A) alone does not imply the residual guarantee; the lost direction is the component of b orthogonal to Ran⁡(A).

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