LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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A rank-one Hermitian perturbation changes signature by at most two

Statement

Let $A$ and $B$ be Hermitian forms on the same finite-dimensional complex vector space, and suppose that the linear map representing $A - B$ has rank at most one. Then $∣ sign (A) - sign (B) ∣ \leq 2$ , where the signature is the number of positive eigenvalues minus the number of negative eigenvalues, with multiplicity and with zero eigenvalues omitted.

Statement references

Signature and codimension-one restriction

Facts & Assumptions

Given: Hermitian forms $A$ and $B$ on the same finite-dimensional complex vector space, with $rank (A - B) \leq 1$ .

[F1]

For a Hermitian form $Q$ , its positive index $n_{+} (Q)$ is the largest dimension of a subspace on which $Q (v, v) > 0$ for every nonzero $v$ ; its negative index $n_{-} (Q)$ is defined using $Q (v, v) < 0$ . Its signature is $n_{+} (Q) - n_{-} (Q)$ .

Proof

1.1givenF1algebra

Let $W = \ker (A - B)$ . Rank-nullity and the rank hypothesis give $codim W \leq 1$ . If $P$ is a positive subspace for $A$ of dimension $n_{+} (A)$ , then $\dim (P \cap W) \geq n_{+} (A) - 1$ . On $P \cap W$ the forms agree, so this intersection is positive for $B$ . Thus $n_{+} (B) \geq n_{+} (A) - 1$ . Interchanging $A$ and $B$ proves $∣ n_{+} (A) - n_{+} (B) ∣ \leq 1$ .

2.1F1step 1.1algebra∎

Apply the same argument to $- A$ and $- B$ to obtain $∣ n_{-} (A) - n_{-} (B) ∣ \leq 1$ . Therefore $∣ sign (A) - sign (B) ∣ \leq ∣ n_{+} (A) - n_{+} (B) ∣ + ∣ n_{-} (A) - n_{-} (B) ∣ \leq 2$ .

Used by

The classical signature bounds unknotting number Theorem

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Ozsváth, Stipsicz, and Szabó, Grid Homology for Knots and Links (standard reference, not scraped)