LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Entrywise products of finite positive semidefinite matrices are positive semidefinite

Statement

If $A, B \in R^{n \times n}$ are positive semidefinite, then their entrywise product $A \circ B$ , defined by $(A \circ B)_{j k} = A_{j k} B_{j k}$ , is positive semidefinite.

Facts & Assumptions

Given: Two finite real symmetric positive semidefinite matrices $A, B$ of the same size.

[F1]

Real and complex inner product spaces, with the inner product linear in the first argument: the standard real inner product on $R^{n}$ is positive definite and bilinear.

[F2]

A non-negative operator is equivalently self-adjoint with nonnegative eigenvalues, a positive semidefinite matrix in an orthonormal basis, or an operator of the form S^*S: a finite nonnegative real symmetric matrix has an orthonormal eigenbasis and nonnegative eigenvalues.

Proof

1.1givenF2chooseconstruct

By [F2], choose real vectors $a_{1}, \dots, a_{n}$ with $A_{j k} = a_{j}^{T} a_{k}$ and real vectors $b_{1}, \dots, b_{n}$ with $B_{j k} = b_{j}^{T} b_{k}$ : diagonalize each matrix orthogonally and multiply each eigenvector coordinate by the square root of its nonnegative eigenvalue.

2.1givenF1step 1.1algebra∎

For each $j$ , let $c_{j} \in R^{n^{2}}$ have $(p, q)$ -coordinate $(a_{j})_{p} (b_{j})_{q}$ , and give $R^{n^{2}}$ its standard dot product. Then $c_{j}^{T} c_{k} = \sum_{p, q} (a_{j})_{p} (b_{j})_{q} (a_{k})_{p} (b_{k})_{q} = (a_{j}^{T} a_{k}) (b_{j}^{T} b_{k}) = A_{j k} B_{j k}$ . Thus $A \circ B$ is the Gram matrix of the $c_{j}$ . For every real vector $z$ , $z^{T} (A \circ B) z = {∥ \sum_{j} z_{j} c_{j} ∥}^{2} \geq 0$ , so $A \circ B$ is positive semidefinite.

Depends on

Used by

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Troy Lee, Rajat Mittal, Ben Reichardt, Robert Špalek, and Mario Szegedy, Quantum Query Complexity of State Conversion, Appendix A.1 (standard reference, not scraped)
Andrew M. Childs, Lecture Notes on Quantum Algorithms, Section 23.1 (standard reference, not scraped)