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Complex Hermitian inertia is invariant under nonsingular congruence

Statement

For a Hermitian $H \in C^{n \times n}$ , define its inertia as $(n_{+} (H), n_{-} (H), n_{0} (H))$ , the counts with multiplicity of positive, negative, and zero eigenvalues. If $S \in C^{n \times n}$ is nonsingular, then

inertia (S^{*} H S) = inertia (H) .

Facts & Assumptions

Given: Hermitian $H$ and nonsingular $S$ as stated.

[F1]

Realification of a Hermitian matrix duplicates each real eigenvalue, and realification of $S^{*} H S$ is the real congruence $R (S)^{T} R (H) R (S)$ ; $R (S)$ is nonsingular (Realification preserves Hermitian spectra, norms, and congruence).

[F2]

For real symmetric matrices, nonsingular congruence preserves positive, negative, and zero inertia counts (Sylvester's law of inertia: every real symmetric form is congruent to $diag (I_{p}, - I_{q}, 0_{r})$ , and $(p, q, r)$ is unique).

Proof

1.1givenF1algebra

By [F1], $R (H)$ and $R (S^{*} H S)$ are real symmetric and the latter equals $R (S)^{T} R (H) R (S)$ with $R (S)$ nonsingular.

2.1step 1.1F1F2algebra∎

Apply [F2] to the realification congruence in step 1.1. The real positive, negative, and zero counts agree; by [F1] each is twice the corresponding complex count, so division by $2$ gives the asserted complex inertia equality.

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Sources

Nicholas J. Higham, Numerical Linear Algebra and Matrix Analysis, MIMS EPrint 2015.103, §5.6 (standard reference, not scraped)