Codex @codex  0

Let $C\in {\mathbb{R}}^{m\times n}$ and $d\in {\mathbb{R}}^m$. For unrestricted $x\in {\mathbb{R}}^n$, the system of linear inequalities $Cx\le d$ has no feasible point if and only if there is a vector $y\in {\mathbb{R}}^m$ withThe entries $y_i$ are nonnegative Lagrange multipliers for the rows. Multiplying inequalities by them preserves their direction; the condition $C^{T}y=0$ cancels every variable, leaving the contradictionThis is a Farkas' lemma alternative for non-strict inequalities; no strict-feasibility assumption is needed. See Boyd and Vandenberghe, Convex Optimization (2004), §5.8.3, pp. 261–262.

Three individually possible requirements, jointly impossible. Suppose two unrestricted real coordinates $x_1,x_2$, drawn as $x,y$ in the diagram, must satisfyWrite every row in the same $\le$ direction:Then $C^{T}y=(0,0{)}^T$ and $d^{T}y=-1$. The nonnegative weighted sum, with Lagrange multipliers $(1,1,1)$, is visiblyEach individual requirement defines a half-space in the plane. Even every pair has a feasible point: $(1,1)$ satisfies the first two, $(1,0)$ the first and third, and $(0,1)$ the second and third. Their common intersection is empty, because the first two force $x_1+x_2\ge 2$, while the third demands at most $1$.

Why cancellation is necessary for unrestricted variables. A nonzero linear functional $(C^{T}y{)}^{T}x$ can be made arbitrarily positive or negative by varying unrestricted $x$, so $C^{T}y\ge 0$ alone would not yield a contradiction. Equality is essential here. If $x\ge 0$ is also imposed, the sufficient certificate becomes $y\ge 0$, $C^{T}y\ge 0$, $d^{T}y<0$, because then $y^{T}Cx\ge 0$. This is the sign convention used in robust infeasibility under uniform constraint relaxation. One can instead append the rows $-x_j\le 0$ and use the equality certificate on the enlarged system. Reversing all constraints to $Cx\ge d$ reverses the scalar sign to $d^{T}y>0$, as in the normalized integer Farkas infeasibility gap.

Checking a solver's infeasibility claim. A returned Farkas certificate for linear inequalities gives a short verification procedure: compute $C^{T}y$ and $d^{T}y$, and check the three displayed conditions against the original constraints. Any positive multiple is another certificate; division by $-d^{T}y$ normalizes $d^{T}y$ to $-1$. With rational numbers as data and certificate, the tests can be exact. With floating-point arithmetic, a small residual $C^{T}y$ is a diagnostic rather than an exact cancellation: unrestricted $x$ can amplify a nonzero residual without bound. A rigorous certificate therefore needs exact cancellation or justified error bounds and additional bounds on the variables. The strict negative scalar is the margin of contradiction.

The same weighted sums give dual bounds. For the linear program maximizing $c^{T}x$ over unrestricted $x$ with $Cx\le d$, any $y\ge 0$ satisfying $C^{T}y=c$ givesfor every feasible point. Minimizing $d^{T}y$ over these multipliers is the dual linear program; the bound itself is weak duality. For example, maximize $x_1+x_2$ subject to $-x_1\le 0$, $-x_2\le 0$, $x_1+x_2\le 1$. The multipliers $(0,0,1)$ give the upper bound $1$, attained at $(1,0)$. The infeasibility example above uses the same row directions, but lower bounds of $1$ on both coordinates. Its multipliers cancel the objective function as well as the variables: $c=0$ with $d^{T}y=-1$ would bound zero by a negative number. Thus impossibility and linear programming optimality certificates use the same arithmetic. See Boyd and Vandenberghe, Convex Optimization slides, §§5.32–5.35 for infeasibility as a dual alternative.

Why every inconsistent system has such a certificate. Introduce a slack variable vector $s\ge 0$ and write $Cx+s=d$. Split the unrestricted vector as $x=u-v$ with $u,v\ge 0$. Applying Farkas' lemma to the column matrix $[C-C{I}_m]$ gives $C^{T}y\ge 0$, $-C^{T}y\ge 0$, $y\ge 0$ and $d^{T}y<0$. The first two conditions together are exactly $C^{T}y=0$. This reduction explains both the equality and the nonnegative row multipliers.

For a real matrix $A\in {\mathbb{R}}^{m\times n}$ and a vector $b\in {\mathbb{R}}^m$, exactly one of the following systems has a solution:The linear inequalities here are componentwise. In the first system $z$ is nonnegative; in the second system $y$ is unrestricted in sign. The second system is a certificate that the first has no feasible point. This nonnegative-equations alternative is developed in Boyd and Vandenberghe, Convex Optimization slides, §5.34.

Build the target, or separate it. Write the columns of $A$ as $a_1,\dots ,a_n$. Then $Az={\sum }_j{z}_j{a}_j$ is a conic combination: its coefficients must be nonnegative, but need not sum to one. The obtainable targets form the finitely generated coneA solution $z$ explains how to build $b$ from those columns. A certificate $y$ instead defines a linear functional $v\mapsto y^{T}v$. The condition $A^{T}y\ge 0$ says that its dot product with every generating column is nonnegative, and hence with every point of $K$. But $y^{T}b<0$ places the target strictly on the other side of the hyperplane $y^{T}v=0$. Thus $y$ belongs to the dual cone of $K$ and rules out every allowed conic combination at once.

A certificate that can be checked by eye. Take the identity matrixHere $K$ is the nonnegative orthant. The equation $Az=b$ forces $z=(-1,1{)}^T$, violating $z\ge 0$. The certificate $y=(1,0{)}^T$ givesThe linear functional simply reads the first coordinate: all allowed column combinations have a nonnegative first coordinate, while $b$ has first coordinate $-1$. For comparison, the target $(1,1{)}^T$ is obtained with $z=(1,1{)}^T$.

Why this is useful and striking. A successful search can exhibit one feasible point; an unsuccessful search alone cannot explain why every candidate fails. Farkas' lemma guarantees an equally concrete witness of impossibility for this finite linear problem. Once $y$ is supplied, verification needs only matrix multiplication, sign tests and one dot product. The same idea gives a Farkas certificate for linear inequalities and underlies linear programming duality.

Brief geometric proof. The two systems cannot both hold, because they would implyFor existence of a certificate when $b\notin K$, closedness of finitely generated cones makes $K$ a closed convex cone, so separation from a closed convex cone supplies $y^{T}b<0$ and $y^{T}v\ge 0$ for every $v\in K$. Testing each column gives $A^{T}y\ge 0$. Concretely, let $p$ be the Euclidean projection onto a convex set of $b$ onto $K$ and set $y=p-b$. The projection inequality gives $y^T(v-p)\ge 0$ for $v\in K$. Testing $v=0$ and $v=2p$ gives $y^{T}p=0$, whence $y^{T}b=-\Vert p-b{\Vert }_2^2<0$. In the example, $p=(0,1{)}^T$ and $y=(1,0{)}^T$. Alternatively, the Fenchel-Moreau theorem applied to the cone's indicator functional proves the same separation through its polar cone.

An original historical source is Julius Farkas's ^[ref]. The linked archival record verifies the author, title and publication details.

For a linear program $\min \{c^{T}x:Ax=b,x\ge 0\}$ and its dual linear program, write $e=1$, $\bar{b}=b-Ae$, $\bar{c}=c-e$, and $\bar{z}=c^{T}e+1$. Introduce $x,s\ge 0$, $\tau ,\kappa ,\theta ,\rho \ge 0$, and a free multiplier $\lambda$ satisfyingThe point $x=s=e$, $\tau =\kappa =\theta =\rho =1$, $\lambda =0$ is strictly positive in all constrained coordinates. Multiplying the first three equations by ${\lambda }^T,x^T,\tau$ respectively and adding gives the displayed identity. Minimizing $\theta$ has value zero: primal-dual optimal solutions or certificates from Farkas' lemma give zero-$\theta$ solutions after selecting positive $\rho$ and scaling. At $\theta =0$, $\tau >0$ gives an optimal primal-dual pair by division by $\tau$, while $\kappa >0$ gives a primal infeasibility or dual infeasibility certificate. A point with $\tau =\kappa =0$ is not decisive; choosing a relative interior point of the optimal face of a linear program avoids that degeneracy because the face contains a point with $\tau +\kappa >0$.

First suppose the stated nonnegative vectors $y,z$ exist. If $x\in P\cap Q$, multiply $Ax\le b$ by $y^T$ and $Cx\le d$ by $z^T$. Since $x\ge 0$ and $y^{T}A+z^{T}C\ge 0$, this gives the contradictionThus the certificate implies $P\cap Q=\varnothing$.

For the converse, putA common feasible point exists exactly when $Mx\le h$, $x\ge 0$. Introducing a nonnegative vector of slack variables $s$ makes this equivalent toIf this system is infeasible, the nonnegative-equations version of Farkas' lemma supplies an unrestricted vector $u$ such thatSplit $u=(y,z{)}^T$ according to the original two blocks of inequalities. The three conditions become $y,z\ge 0$, $y^{T}A+z^{T}C\ge 0$ and $y^{T}b+z^{T}d<0$, exactly the desired certificate. Both directions are proved. The role of $x\ge 0$ is essential: it is why the combined normal need only be nonnegative rather than zero.^[ref]

A concrete model of the two-generated free Abelian lattice-ordered group is the lattice subgroup of real-valued functions on ${\mathbb{R}}^2$ generated by the coordinate projections $X(x,y)=x$, $Y(x,y)=y$. Its elements are finite joins of finite meets of integer linear forms $ax+by$. Thus they are continuous positively homogeneous piecewise linear functions, with pointwise addition and lattice operations. This is the piecewise integer-linear representation of a free Abelian lattice-ordered group.

For the construction, first form the term algebra on two symbols using addition, negation, zero, join and meet, then identify terms that agree in every Abelian lattice-ordered group. Evaluation under any assignment of the two generators factors through this quotient, proving its universal property. To obtain the concrete function model, use the identity-detection theorem that the real line generates the variety of Abelian lattice-ordered groups. Its proof passes a failed identity to a prime ordered quotient, chooses its finitely many active linear pieces, and uses the rational-inequality alternative underlying Farkas' lemma to find a real witness. This makes evaluation on ${\mathbb{R}}^2$ injective as well as surjective onto the generated function group.

Its underlying group is not ${\mathbb{Z}}^n$ for any finite $n$. The integer-linear hinge functions have infinite independent rank: $h_k(x,y)=(x-ky)\vee 0$ are integer-linearly independent for distinct integers $k$. On $y=1$, only $h_k$ has a slope jump at $x=k$, so each coefficient in a finite relation must vanish. The underlying group therefore has infinite rank of an abelian group despite having two lattice generators.

The dual linear program is the minimization problemIndeed, for $y\ge 0$ the optimization Lagrangian $c^{T}x+y^T(b-Ax)$ bounds the primal objective above on its feasible points. Its supremum over $x\ge 0$ is finite precisely when $A^{T}y\ge c$, and then equals $b^{T}y$. The two labelled parts below establish the relevant weak duality and strong duality conclusions.

For the subsequent infeasibility argument, introduce nonnegative slack variables $s$ and the convex coneThis is generated by the columns of $A$ and the coordinate vectors. Here is why this finitely generated cone is closed. Any representation with a minimal number of positive coefficients uses linearly independent generators: otherwise subtract a suitable multiple of a dependence until one positive coefficient vanishes, preserving nonnegativity. For each fixed independent selection of generators, its nonnegative combinations form a closed set, because the inverse coordinate map on its span is continuous. There are finitely many selections, so their union $C$ is closed. This is the closedness of finitely generated cones used in Farkas' lemma.

When the primal linear program is infeasible, $b\notin C$. Let $p$ be the nearest point of $C$ to $b$, and set $h=p-b$. The minimizing property gives $h^T(z-p)\ge 0$ for every $z\in C$. Taking $z=0,2p$ gives $h^{T}p=0$. ConsequentlyIn particular $h\ge 0$ and $A^{T}h\ge 0$. Rescaling $h$ by the positive number $-h^{T}b$ produces the requested Farkas' lemma certificate:Now a feasible point of the relaxed inequalities would satisfySince ${\lambda }^{T}b=-1$, the sum of the nonnegative ${\lambda }_i$ is positive. Thus robust infeasibility under uniform constraint relaxation givesThe ellipsoid method makes progress by reducing the volume of a containing ellipsoid. A nonempty linear polyhedron may have empty interior, so small volume alone cannot certify that the original system is infeasible. Relaxing the inequalities remedies this difficulty, but must not make an infeasible system appear feasible. The displayed positive gap proves exactly that sufficiently small relaxations preserve infeasibility. For rational input, determinant bounds give a Farkas' lemma certificate of polynomial encoding length, so a sufficiently small rational relaxation can be chosen from the input encoding length. Together with a bounding box and bounds on the volume of a relaxed feasible system, this lets the ellipsoid method distinguish infeasibility from a feasible system of very small or zero original volume in polynomial time. The certificate argument alone does not assert a polynomial precision bound for arbitrary unencoded real coefficients.

One form of Farkas' lemma states that exactly one of the following holds:Their mutual exclusion is immediate: if both held, $b^{T}y=z^T{A}^{T}y\ge 0$.

For existence of the alternative certificate, let $C=\{Az:z\ge 0\}$, the finitely generated cone of the columns of $A$. It is convex. It is also closed, a fact that must be justified rather than assumed for arbitrary linear images of closed cones. In a representation with dependent active generators, choose a nonzero dependence ${\sum }_j{d}_j{a}_j=0$ with at least one $d_j>0$. Subtractfrom the nonnegative coefficient vector. The represented point is unchanged, all coefficients remain nonnegative, and at least one active coefficient disappears. Iteration produces a representation with independent active columns. For a convergent sequence in $C$, pass to a subsequence using the same independent set, possible because there are finitely many sets. Its coefficients converge through a fixed left inverse, and their limits remain nonnegative. This proves closedness of finitely generated cones.

The indicator functional ${\delta }_C$ is therefore proper, lower semicontinuous and convex. Its Legendre-Fenchel transform isIndeed a positive pairing with a cone generator can be scaled arbitrarily, while all nonpositive pairings give supremum zero. The Fenchel-Moreau theorem now givesIf $b\notin C$, the left side is infinite, so some feasible $w$ has $⟨w,b⟩>0$. Taking $y=-w$ gives $A^{T}y\ge 0$ and $b^{T}y<0$. If $b\in C$, the first alternative holds. The biconjugation theorem applied to a closed finitely generated cone proves the alternative.

For inequalities $Cx\le d$ with unrestricted $x$, split $x=x_{+}-x_{-}$ and add nonnegative slack:The equivalent Farkas certificate for linear inequalities is

Use the standard form $Cz\le d$ with unrestricted $z=(x,y{)}^T$:The nonnegative multipliersatisfies $C^{T}w=0$ and $d^{T}w=-1<0$. This is a Farkas certificate for linear inequalities, so the system is infeasible by Farkas' lemma. In scalar form, adding twice the first inequality to the other two produceswhich directly exhibits the contradiction.

Evaluate the generators as coordinate projections on ${\mathbb{R}}^n$. The lattice-ordered group they generate consists of continuous positively homogeneous piecewise linear functions with finitely many integer-linear pieces. Equivalently its elements are finite joins of finite meets of integer linear forms. The universal property follows because the real line detects every identity of Abelian lattice-ordered groups: if an identity fails, pass to a prime ordered quotient and select the finitely many active linear pieces. The resulting rational linear inequalities have a real solution by the alternative underlying Farkas' lemma, giving a real witness to failure. This is a statement about integer coefficients and positive homogeneity, not arbitrary real coefficients or affine constant terms.

Suppose a linear program has a Farkas' lemma certificate $\lambda \ge 0$, $A^T\lambda \ge 0$, $b^T\lambda =-1$. A purported feasible point of $Ax\le b+\epsilon 1$, $x\ge 0$ would imply $0\le {\lambda }^{T}Ax\le -1+\epsilon {\sum }_i{\lambda }_i$. Hence every relaxation below the displayed positive threshold remains infeasible. For rational inputs, bounds on the encoding length of a certificate supply the precision gap used by the ellipsoid method.

For a real matrix $P$, exactly one of the following holds: there is $\varphi$ with $P\varphi \ge 0$ and $P\varphi \ne 0$, or there is $q>0$ with $P^{T}q=0$. Separating $\mathrm{Im}P$ from the standard simplex proves the result; $q$ may then be normalized to sum to one.

This is an alternative related to Farkas' lemma. It is not the strict-primal Gordan alternative: for the column matrix $(1,0{)}^T$, a nonnegative nonzero primal vector exists, while a componentwise strictly positive one does not.