Independent Linear Constraint System¶

Definition¶

A linear constraint system

\boldsymbol{A}\boldsymbol{x}=\boldsymbol{b}

with $\boldsymbol{A}\in\mathbb{R}^{m\times n}$ has independent constraints if the rows of $\boldsymbol{A}$ are linearly independent, equivalently

\operatorname{rank}\boldsymbol{A}=m.

For nonlinear equality constraints $g_1(\boldsymbol{x})=0,\ldots,g_m(\boldsymbol{x})=0$ $0$ , the corresponding local condition at $\boldsymbol{a}$ is that the gradients $\boldsymbol{\nabla} g_1(\boldsymbol{a}),\ldots,\boldsymbol{\nabla} g_m(\boldsymbol{a})$ are linearly independent.

Interpretation¶

Independent constraints each remove a genuinely new direction. Dependent constraints repeat information and can make multiplier tests misleading.

Two panels of lines in the plane. Left: two lines with independent normal vectors meet in one point (rank 2). Right: two parallel lines with dependent normal vectors (rank 1).

On the right, the rows of $\boldsymbol{A}$ are dependent and $\operatorname{rank}\boldsymbol{A}=1$ $1$ . The corresponding parallel lines either have no common solution or coincide along a whole line. On the left, the two independent constraints have rank $2$ and meet at one point. The arrows $\boldsymbol{a}_1$ and $\boldsymbol{a}_2$ are the rows of $\boldsymbol{A}$ , each normal to its line; on the right $\boldsymbol{a}_2=c\boldsymbol{a}_1$ for a scalar $c\neq0$ $0$ .

The rank counts independent rows. Equality to $m$ means that no constraint equation is an algebraic consequence of the others; for nonlinear constraints the same check applies to their gradients at the chosen point.

Key Results¶

For $m$ independent equality constraints in $\mathbb{R}^n$ , the linearized feasible directions (the null space of the constraint matrix or Jacobian) form an $(n-m)$ $m)$ -dimensional space, by the rank-nullity theorem.
Independence is a rank condition on the constraint Jacobian.
Nondegenerate constraint qualification is the nonlinear optimization version of this idea.

Examples¶

The constraints $x+y=1$ and $2x+2y=2$ are dependent because the second repeats the first.
The constraints $x+y=1$ and $x-y=0$ are independent because the matrix $\begin{pmatrix}1&1\\1&-1\end{pmatrix}$ has rank $2$ .

Prerequisites¶

Sources¶

Math IV: Multivariate Calculus · Week 3, Lecture 2

Aliases: independent constraints Keywords: independent linear constraints, linearly independent constraints, constraint rank Symbols: element of, nabla gradient, real numbers

Last updated: 27 October 2026 11:52