Entrywise exponential of a PSD matrix is PSD
Part of the Schur-Polya-Loewner theory of entrywise functions preserving PSD. The proof uses the Schur product theorem iteratively: exp_⊙(A) = ∑ A^{⊙k}/k!, each Hadamard power is PSD, and the convergent series of PSD ma…
Overview
Entrywise exponential of a PSD matrix is PSD
posSemidef_map_exp — a formalization challenge from the lean-eval benchmark.
Notes
Part of the Schur-Polya-Loewner theory of entrywise functions preserving PSD. The proof uses the Schur product theorem iteratively: exp_⊙(A) = ∑ A^{⊙k}/k!, each Hadamard power is PSD, and the convergent series of PSD matrices is PSD.
Formal statement
theorem posSemidef_map_exp
{n : Type*} [Fintype n] [DecidableEq n]
{A : Matrix n n ℝ} (hA : A.PosSemidef) :
(A.map Real.exp).PosSemidef := by
sorry
Informal solution sketch
Write exp(a_{ij}) as the convergent series ∑ (a_{ij})^k / k!. The matrix with entries (a_{ij})^k is the k-fold Hadamard product A^{⊙k}, which is PSD by iterated Schur product. The partial sums are nonneg combinations of PSD matrices, hence PSD. PSD is a closed condition, so the limit is PSD.
Source
I.J. Schoenberg, Positive definite functions on spheres, 1942.
How to submit
Challenge.lean and Solution.lean are part of the trusted benchmark and must not be modified.
Write your proof in Submission.lean (plus any local modules under Submission/).
Mathlib may be used freely; anything not in Mathlib has to be inlined into the submission.
Comparator configuration
- Solution module:
Solution - Theorems checked:
posSemidef_map_exp - Permitted axioms:
propext,Quot.sound,Classical.choice
Submitted by Kim Morrison.
Problems
1 problemFiles
View on GitHub- Submission
- Helpers.lean53 B
- Challenge.lean208 B
- config.json233 B
- holes.json428 B
- lakefile.toml451 B
- lean-toolchain25 B
- README.md1.4 KB
- Solution.lean259 B
- Submission.lean272 B
- WorkspaceTest.lean1.6 KB