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For smooth equality constraints $h_j=0$ and inequality constraints $g_i\ge 0$, this condition requires linear independence of all equality gradients and active inequality gradients at the feasible point. It ensures existence of Lagrange multipliers at a local optimum and uniqueness of their values for a fixed objective gradient. It implies the Mangasarian–Fromovitz constraint qualification by solving the independent gradient equations with zero equality derivatives and positive active inequality derivatives.

For smooth inequalities $c_j\ge 0$, the displayed intersection of half-spaces contains the contingent tangent cone. This follows by differentiating each active feasibility inequality along a feasible tangent sequence. The linear independence constraint qualification, and more generally the Mangasarian–Fromovitz constraint qualification, ensure equality in the inequality-only setting. Without a qualification the inclusion can be strict: for $c(x)=-x^2\ge 0$, the feasible set is $\{0\}$ and its tangent cone is $\{0\}$, while $c^{\prime }(0)=0$ makes the linearized cone the whole real line.

Write $x_1=t$, $x_2=t^2$. The two inactive constraints are strict exactly for $-2<t<2$. Complementary slackness gives ${\lambda }_1={\lambda }_2=0$. The KKT conditions reduce toIf $t=0$ these give ${\lambda }_3=0$. Otherwise ${\lambda }_3=2t$ and $t=-3/2$, which gives a negative multiplier and is inadmissible. Thereforeis the only KKT point with this active set. Its active gradient $(0,1)$ is nonzero, so the linear independence constraint qualification, and hence the Mangasarian–Fromovitz constraint qualification, holds.

It is not a local maximizer: $(0,\epsilon )$ is feasible for small $\epsilon >0$ and has objective ${\epsilon }^2>0=F(0,0)$. The feasible boundary path $(t,t^2)$ for small negative $t$ has objective $2t^3+t^4<0$, so it is not a local minimizer either.

All three active constraints force $x_1=2$ and $x_2=4$, which also satisfy $x_1^2+x_2^2=20$. For maximization use $\mathcal{L}=F+{\lambda }^{T}g$, $\lambda \ge 0$. At $x^{\ast }=(2,4)$,Stationarity gives the complete familyIn particular $({\lambda }_1,{\lambda }_2,{\lambda }_3)=(2,3/2,2)$ is strictly positive. Three gradients in two dimensions cannot be linearly independent, so the linear independence constraint qualification fails. But $d=(-1,-1)$ gives $\nabla g_i\cdot d=(1,12,3)>0$, so the Mangasarian–Fromovitz constraint qualification holds.

For a nonzero linearized feasible direction $d$, each $\nabla g_i\cdot d\ge 0$, andThe inequality is strict because equality would require all three inner products to vanish, and the first and third gradients already span ${\mathbb{R}}^2$. This proves a strict local maximum: otherwise a sequence of nearby feasible points with no smaller objective would, after normalizing their differences from $x^{\ast }$ and taking a convergent subsequence, give a unit linearized feasible direction with nonnegative objective derivative, contradicting the displayed inequality. Thus

The equality gradients are independent everywhere. At an interior point of the ellipsoid, there is no active inequality, so the Mangasarian–Fromovitz constraint qualification holds. At a boundary point $x=x_0+z$, a qualifying direction must satisfy $Av=0$ and $2x^{T}Qv<0$.

If ${\rho }^2<1$, choose $v=x_0-x=-z$. At a boundary point $z^{T}Qz=1-{\rho }^2>0$, andThus the qualification holds at every feasible point. If ${\rho }^2=1$, the only feasible point is $x_0$; for every $v\in \ker A$, $x_0^{T}Qv=0$, so no strict direction exists. This provesThe equivalence also identifies this case with strict feasibility of the inequality on the affine equality space.

The active constraint set is $\boxed{\mathcal{A}(x)=\{j:c_j(x)=0\}}$, and the linearized tangent cone isDifferentiating each active feasibility inequality along a tangent sequence gives $T_{\Omega }(x)\subseteq F(x)$. A sufficient condition for equality is the linear independence constraint qualification: the active constraint gradients are linearly independent. The weaker Mangasarian–Fromovitz constraint qualification also suffices in this inequality-only setting.

Under the stated independence condition, the necessary first-order conditions become the Karush-Kuhn-Tucker conditions: there are multipliers ${\lambda }_j$ withUse the allowed span assertion to write $\nabla f(x_{\ast })={\sum }_{j\in \mathcal{A}}{\lambda }_j\nabla c_j(x_{\ast })$. Independence gives vectors ${\phi }_i$ with $\nabla c_j(x_{\ast }{)}^T{\phi }_i={\delta }_{ji}$ for the active indices. Each ${\phi }_i$ lies in $F=T_{\Omega }$, so part (a) gives ${\lambda }_i=\nabla f(x_{\ast }{)}^T{\phi }_i\ge 0$. Assign zero multipliers to inactive constraints. This proves all displayed conditions, including complementarity. Conversely, their stationarity and sign conditions imply $\nabla f(x_{\ast }{)}^{T}d={\sum }_{j\in \mathcal{A}}{\lambda }_j\nabla c_j(x_{\ast }{)}^{T}d\ge 0$ for every $d\in F$. Thus they reformulate the first-order inequality; they do not alone prove a minimum for a nonconvex problem.