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Block LDL congruence adds the Hermitian pivot inertias

Statement

Let k≥1, m≥0, and let H=(DB∗BC)∈C(k+m)×(k+m) be Hermitian, where D=D∗∈Ck×k is nonsingular, B∈Cm×k, and C=C∗∈Cm×m. Define the Schur complement S=C−BD−1B∗. Inertia triples are added coordinatewise, and the 0×0 matrix has inertia (0,0,0). Then

inertia⁡(H)=inertia⁡(D)+inertia⁡(S),

where the triples are added coordinatewise.

Facts & Assumptions

Given: H,D,B,C,S as in the statement.

[F1]

Nonsingular complex congruence preserves Hermitian inertia (Complex Hermitian inertia is invariant under nonsingular congruence).

Proof

1.1givenalgebra

Set L=(I0BD−1I). It is invertible, with inverse (I0−BD−1I). Direct block multiplication gives H=L(D00S)L∗, so H is congruent to diag⁡(D,S).

2.1givenF1F2step 1.1algebra∎

Since D=D∗ and D is invertible, (D−1)∗=(D∗)−1=D−1. Therefore (BD−1B∗)∗=BD−1B∗, and C=C∗ gives S∗=S. By [F1], inertia⁡(H)=inertia⁡(diag⁡(D,S)). If m=0, the trailing block is empty and its inertia is (0,0,0) by the stated convention. If m>0, [F2] diagonalizes the Hermitian matrices D and S separately; concatenating their orthonormal eigenbases diagonalizes the block diagonal matrix, so its positive, negative and zero counts are the sums of the corresponding counts for D and S.

Depends on

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Sources