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LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Positive Gaussian observation noise makes the observed covariance positive definite

Statement

Let n≥1, let K∈Rn×n be symmetric positive semidefinite, let σ2≥0, and put C=K+σ2I. If σ2>0, then C is symmetric positive definite, even when K is singular. If σ2=0, then C is positive definite exactly when K is positive definite; zero noise is not guaranteed to make the solve available.

Facts & Assumptions

Given: A symmetric matrix K with vTKv≥0 for all v∈Rn, and a real σ2≥0.

[F1]

A real matrix is positive definite when it is symmetric and vTAv>0 for every nonzero v, as in Hermitian positive-definite matrices and Cholesky factorisation A = LL* with positive diagonal.

Proof

1.1givenF1algebra

The matrix C is symmetric because K and I are symmetric. For every nonzero v, vTCv=vTKv+σ2∥v∥22. If σ2>0, the first term is nonnegative and the second is strictly positive, so vTCv>0; hence C is positive definite by [F1].

2.1givenF1algebra∎

If σ2=0, then C=K, so C is positive definite if and only if K is. This permits zero-noise solves exactly in the positive-definite observed-Gram cases and makes no pseudoinverse assertion in the singular cases.

Depends on

Used by

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Sources