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Duplicate inputs make the unquotiented NTK Gram matrix singular

Example

Let X be a set, let k:X×X→R be a symmetric positive-semidefinite kernel, fix x∈X, and use two observations at that same input. Their Gram matrix is K=κ(1111), where κ=k(x,x)≥0. For labels y=(1,−1)T and initial prediction f0=(0,0)T, consider the two-point empirical least-squares kernel flow

f˙(t)=−12K(f(t)−y),f(0)=f0.

The vector v−=(1,−1)T/2 is a zero-eigenvalue mode of K, and the residual component in this mode remains unchanged, so the flow cannot reach y. The unquotiented Gram matrix has rank at most one. In particular, assume AC, fix n0≥1, L≥2, β>0, and a globally Lipschitz nonpolynomial activation σ:R→R, and use the width-normalized fully connected network with fan-in scaling 1/nℓ, bias scale β, scalar output, and independent standard-normal initialization. For any x∈Sn0−1, its sequential hidden-width limit kernel K∞ has K∞(x,x)>0 by the sphere-positivity proposition, but the Gram matrix for two observations at x is still singular. This does not conflict with that proposition, which asserts strict positivity only for finite lists of pairwise distinct sphere points.

Facts & Assumptions

Given: A set X, a symmetric positive-semidefinite scalar kernel k:X×X→R, and two identical input observations at x∈X, so κ=k(x,x)≥0; labels (1,−1)T; initial prediction (0,0)T; and, for the sphere comparison, AC, n0≥1, L≥2, β>0, a globally Lipschitz nonpolynomial activation, the stated width-normalized scalar-output network, and x∈Sn0−1.

[F1]

A kernel Gram matrix on an input list has entries given by evaluating the kernel at each ordered input pair; in particular, repeating the input makes all four entries equal to k(x,x).

Verification

technique · compute the two-by-two Gram eigenvectors and inspect the inconsistent target component
1.1F1givenalgebra

Repeating the input makes every matrix entry equal to κ, so K=κ(1111). Multiplication gives Kv−=0 for v−=(1,−1)T/2. Thus K has rank at most one and the target y=(1,−1)T=2v− lies entirely in its nullspace.

2.1givenstep 1.1algebra∎

Set r−(t)=v−T(f(t)−y). The flow equation, symmetry of K, and Kv−=0 give r˙−(t)=−12v−TK(f(t)−y)=−12(Kv−)T(f(t)−y)=0. Since r−(0)=−2, this residual component stays nonzero for every t≥0, so f(t)≠y at every finite time and the flow cannot converge to y. This is consistent with the sphere proposition because its strictness conclusion requires pairwise distinct inputs.

Source notes

[F2] For the source-normalized limiting NTK flow, least-squares residual components in zero-eigenvalue modes remain constant (Least-squares NTK flow decays mode by mode and fits distinct sphere data); the more general two-point claim here is checked directly above.

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