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Expected spectral error for real and normalized complex Gaussian range finders

Statement

Assume Countable Choice for the finite-dimensional Gaussian probability law. Let A∈Fm×n be fixed, where F=R or C. Let k≥2, p≥2, ℓ=k+p≤min⁡(m,n), and draw Ω with iid standard real entries if F=R or normalized complex entries if F=C. If Q is an orthonormal basis of Ran⁡(AΩ), then E∥A−QQ∗A∥2≤(1+kp−1)σk+1+ek+pp(∑j>kσj2)1/2. This preserves the normalized-complex convention E∣Ωij∣2=1.

Facts & Assumptions

Given: Countable Choice for the finite Gaussian law, and the fixed finite matrix and the real or normalized-complex Gaussian draw under the stated k,p,ℓ constraints.

[F1]

The real-field statement with the displayed constants, for k≥2, p≥2, and ℓ=k+p≤min⁡(m,n), is proved in Expected spectral error of the real Gaussian range finder.

[F2]

The normalized-complex range finder, with the same fixed-matrix, dimension, variance-one Gaussian, and Countable Choice hypotheses, satisfies the displayed expected-error bound (Direct expected spectral bound for the normalized complex range finder).

[F3]

Standard real and normalized complex Gaussian matrix laws: real entries are standard Gaussian and complex entries are (X+iY)/2, with E∣g∣2=1; the finite-dimensional laws are product-density laws.

[F4]

Borel measurability and range correctness of the finite scan proves that the finite-scan algorithm Exact Gaussian randomized range finder returns a measurable orthogonal projector onto Ran⁡(AΩ). If Q and Q0 are any two orthonormal bases of that same subspace, then QQ∗=Q0Q0∗: for every vector x, both QQ∗x and Q0Q0∗x lie in the subspace, while their residuals lie in its orthogonal complement; uniqueness of this orthogonal decomposition makes the projections equal.

[A1]

Countable Choice is the axiom used for the finite-dimensional Gaussian probability laws (The Axiom of Countable Choice (ACω)).

Proof

1.1givenF1

If F=R, the hypotheses and constants are exactly those of [F1], which proves the asserted expectation bound.

1.2givenF2F3F4A1choosealgebra

If F=C, let Q0 be the basis returned by the measurable finite-scan algorithm in [F4]. The Q in the Statement and Q0 are orthonormal bases of the same sample range, so [F4] gives QQ∗=Q0Q0∗ for every draw. The expected error is therefore the same measurable random variable as for Q0, and [F2] supplies the required bound with the stated normalized-complex convention [F3].

2.1step 1.1step 1.2∎

The cases F=R and F=C exhaust the field choice; steps 1.1 and 1.2 prove the full displayed assertion.

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