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The LP/QP primal-dual Newton matrix is nonsingular from a strict positive start

Statement

Let $Q ⪰ 0$ be symmetric, let $A \in R^{p \times n}$ have full row rank, and let $x, s \in R^{n}$ have strictly positive coordinates. Then the LP/QP Newton matrix $M = [\begin{array}{ccc} Q & - A^{T} & - I \\ A & 0 & 0 \\ S & 0 & X \end{array}], X = diag (x), S = diag (s),$ is nonsingular.

Facts & Assumptions

Given: The matrix and residual convention of Log-barrier central paths and primal-dual Newton residuals, with $Q ⪰ 0$ , full-row-rank $A$ , and $x, s > 0$ .

[F1]

The Newton equations have first block $Q Δ x - A^{T} Δ y - Δ s$ , second block $A Δ x$ , and complementarity block $S Δ x + X Δ s$ (Log-barrier central paths and primal-dual Newton residuals).

[F2]

A full-row-rank matrix has injective transpose, and a symmetric positive-semidefinite $Q$ satisfies $u^{T} Q u \geq 0$ (Conic, LP, and convex QP primal-dual models).

Proof

1.1F1givenalgebra

Let $(u, v, w)$ lie in the kernel of $M$ . The last block equation is $S u + X w = 0$ . Since $X$ is invertible, $w = - X^{- 1} S u = - D u$ , where $D = diag (s_{i} / x_{i}) ≻ 0$ . Substituting into the first block gives $(Q + D) u - A^{T} v = 0$ , and the second block gives $A u = 0$ .

2.1F2step 1.1algebra∎

Take the inner product of $(Q + D) u - A^{T} v = 0$ with $u$ . Since $A u = 0$ , this yields $u^{T} Q u + u^{T} D u = 0$ . Both terms are nonnegative by [F2], and $D ≻ 0$ , so $u = 0$ . The first block then gives $A^{T} v = 0$ , hence $v = 0$ by [F2], and the last block gives $w = 0$ . The square matrix $M$ has trivial kernel and is therefore nonsingular.

Depends on

Used by

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Sources

Stephen Boyd and Lieven Vandenberghe, Convex Optimization (standard reference, not scraped)
Hamza Fawzi, Topics in Convex Optimisation, Lecture 16 (standard reference, not scraped)