Bordered Hessian Matrix¶

Definition¶

For an equality-constrained problem with Lagrangian $L(\boldsymbol{x},\boldsymbol{\lambda})$ and constraint Jacobian $\boldsymbol{D}\boldsymbol{g}(\boldsymbol{x}^\ast)$ , the bordered Hessian matrix is the block matrix

\begin{pmatrix} \boldsymbol{0} & \boldsymbol{D}\boldsymbol{g}(\boldsymbol{x}^\ast)\\ \boldsymbol{D}\boldsymbol{g}(\boldsymbol{x}^\ast)^{\top} & \boldsymbol{H}_xL(\boldsymbol{x}^\ast,\boldsymbol{\lambda}^\ast) \end{pmatrix},

with the exact block orientation depending on whether gradients are written as rows or columns.

Interpretation¶

The bordered Hessian packages the constraint linearization and the curvature of the Lagrangian into one matrix. Its leading principal minors tell whether $L$ bends down or up along the directions that keep the constraints satisfied to first order.

Block diagram of the bordered Hessian: an m by m zero block in the top left, the constraint gradients Dg(x star) as the border, and the n by n Hessian of the Lagrangian in x in the bottom right.

The figure shows the block layout: an $m\times m$ zero block, the constraint Jacobian $\boldsymbol{D}\boldsymbol{g}(\boldsymbol{x}^\ast)$ ( $m \times$ $m\times n$ $n$ ) and its transpose as the border, and the $n\times n$ Hessian of $L$ with respect to $\boldsymbol{x}$ in the corner, $(m+n)\times(m+n)$ in total. This is the matrix of the bordered matrix test, with $\boldsymbol{B}=\boldsymbol{D}\boldsymbol{g}(\boldsymbol{x}^\ast)$ and $\boldsymbol{A}$ the Hessian of $L$ .

The constraint gradient borders the Hessian block; the determinant of the bordered matrix carries the second-order constrained test.

The Hessian of $L$ , not of the objective, is needed because the bending of the constraint matters too. When $x+y$ is maximized on the circle $x^2+y^2=1$ $1$ , the objective has zero Hessian and says nothing. With $L=x+y+\lambda(1-x^2-y^2)$ $y^{2})$ , the candidate $(1/\sqrt{2},1/\sqrt{2})$ has $\lambda^\ast=1/\sqrt{2}$ $1/ 2$ , the Hessian of $L$ is $-\sqrt{2}\boldsymbol{I}$ , and the bordered Hessian has determinant $4\sqrt{2}>0$ $0$ , the sign $(-1)^{m+1}=+1$ that the Key Results require for a strict local maximum. Writing a constraint as $-g$ instead of $g$ flips the sign of a border row and the matching column together, which leaves every leading minor unchanged.

A common slip is to inspect all leading minors. The first $2m$ do not depend on the Hessian block, so they say nothing about curvature; only the last $n-m$ count, as in the second-order sufficient conditions.

Key Results¶

The $m\times m$ zero block corresponds to the multiplier variables; the $m\times n$ border contains the constraint gradients.
The $n\times n$ Hessian block is taken with respect to the choice variables, so the full matrix has size $(m+n)\times(m+n)$ $n)$ .
Determinants of selected bordered submatrices are used in bordered matrix tests.
For $m<n$ equality constraints with independent gradients at a first-order candidate, the second-order sufficient conditions inspect the last $n-m$ leading principal minors. Alternating signs starting with $(-1)^{m+1}$ give a strict local maximum; signs all equal to $(-1)^m$ give a strict local minimum. These tests check the curvature of $L$ along directions tangent to the constraints.

Examples¶

For one constraint $g(x,y)=0$ $0$ , a common bordered Hessian has first row and column built from $g_x$ and $g_y$ .
For two equality constraints in three variables, the bordered Hessian has a $2$ by $2$ zero block and a $3$ by $3$ Hessian block.

Prerequisites¶

Sources¶

Math IV: Multivariate Calculus · Week 6, Lecture 2

Aliases: bordered Hessian Keywords: constraint bordered Hessian Symbols: lambda, square root

Last updated: 20 October 2026 03:02