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A wide Gaussian matrix has full row rank almost surely

Statement

Assume Countable Choice for the finite-dimensional Gaussian probability law. Let k≥1, ℓ>k, and let G∈Fk×ℓ have iid standard real Gaussian or normalized complex Gaussian entries. Then rank⁡(G)=k almost surely.

Facts & Assumptions

Given: Countable Choice for the finite Gaussian law, and a finite matrix with the density and independence specified in Standard real and normalized complex Gaussian matrix laws.

[F1]

Matrix entries have the stated product density, so each row is absolutely continuous with respect to finite-dimensional real-coordinate Lebesgue measure (Standard real and normalized complex Gaussian matrix laws).

[F2]

Orthogonal or unitary changes of coordinates preserve this Gaussian law (Orthogonal and unitary invariance of Gaussian matrix blocks).

[F3]

For a nonnegative product-measurable function, the product integral equals either iterated integral (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[A1]

Countable Choice supplies the Lebesgue measure used for the density (The Axiom of Countable Choice (ACω)).

Proof

1.1F1F2F4A1choose

A fixed proper real or complex linear subspace L⊊Fℓ has probability zero for one Gaussian row: choose an orthogonal/unitary coordinate change sending L into {z1=0}, use [F2], and integrate the density over this coordinate subspace. It is Lebesgue-null: inside each bounded box [−M,M]d it lies in a coordinate hyperplane coverable by a box of width 2ε and bounded remaining sides, whose volume tends to zero with ε; [F4] makes that hyperplane null, and countable subadditivity covers the whole hyperplane.

2.1step 1.1F3choose

Induct on the number j≤k of revealed rows. The first row is nonzero almost surely by step 1.1 applied to the zero subspace. The dependence event for j rows is measurable because it is the common zero set of the finitely many j×j minors. If the first j−1 rows are independent, their span is proper since j−1<k<ℓ; for each fixed such tuple, step 1.1 makes the section of the event that row j lies in their span null. The exceptional dependent tuples are null by induction. Applying [F3] to the nonnegative indicator of dependence gives probability zero for the first j rows to be dependent. The induction is finite and concludes at j=k.

3.1step 2.1algebra∎

Outside this null event, the k rows are linearly independent, hence the row rank equals k.

Depends on

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