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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 an integer linear polyhedron $P=\{x\in {\mathbb{R}}^n:Ax\ge b\}$ with coefficients bounded by $U\ge 1$, put $D=[(n+1)U{]}^{n+1}$ and $0<\epsilon <1/D$. If $x_0\in P$, every $h$ with $\Vert h{\Vert }_2\le \epsilon /(nU)$ satisfies $A_{i}h\ge -\sqrt{n}U\Vert h{\Vert }_2\ge -\epsilon$. Thus $P_{\epsilon }$ contains a positive-radius Euclidean ball and is a full-dimensional linear polyhedron. If $P$ is empty, the normalized integer Farkas infeasibility gap gives $\lambda \ge 0$, $A^T\lambda =0$, $1^T\lambda =1$ and $b^T\lambda \ge 1/D$. A point of $P_{\epsilon }$ would give $0={\lambda }^{T}Ax\ge b^T\lambda -\epsilon >0$, a contradiction. Consequently this relaxation preserves emptiness and supplies the inner-volume bound needed by the ellipsoid method, even when $P$ has empty interior.

Let $x$ be an extreme point and consider its active constraints. Their row vectors span ${\mathbb{R}}^n$. Otherwise a nonzero vector $h$ is orthogonal to every active row. All inactive inequalities have strictly positive slack; since there are finitely many, a sufficiently small $\epsilon >0$ makes both $x+\epsilon h$ and $x-\epsilon h$ feasible. Their midpoint is $x$, contradicting extremality.

Choose $n$ independent active rows and write the corresponding equalities as $Bx=d$. The matrix $B$ has integer entries, so $\det B$ is a nonzero integer and $|\det B|\ge 1$. By Cramer's rule,where $B_j$ replaces column $j$ by $d$. Every entry of $B_j$ has absolute value at most $U$, even when $b$ is real. The determinant expansion has $n!$ terms, each with absolute value at most $U^n$. ThereforeThis proves the integer-matrix vertex coordinate bound; only the denominator determinant requires integrality.

For an exact finite-input ellipsoid method, the numerical data must have a specified encoding. In the standard model the input is the dimensions, the integer matrix and a rational right-hand side in binary, together with bounds derived from their encoding lengths. The vertex estimate above remains valid for arbitrary real $b$, but arbitrary unencoded reals do not by themselves specify an exact computational decision problem. Clear the rational row denominators to obtain an equivalent integer system, and let $V\ge 1$ bound the resulting integer coefficients and right-hand sides.

Two precautions make a feasibility account valid even for unbounded or lower-dimensional polyhedra. First, nonempty polyhedra need not have extreme points. Nevertheless they have a feasible point in the box $[-R,R{]}^n$ with $R=n!V^n$. To prove this feasible-point box bound for a polyhedron with lines, intersect the polyhedron with an orthant containing one feasible point. Minimize the sum of the signed coordinates in that orthant. Its nonempty sublevel sets are closed and bounded, so the minimum is attained. The compact minimum face has an extreme point, which is also extreme in the orthant intersection. Its active rows come from the integer system and signed coordinate inequalities, still bounded by $V$. Applying the proved vertex estimate gives the box bound. This is an existence argument; the algorithm need not enumerate all orthants.

Second, a nonempty feasible set can have zero ambient volume. Use a small uniform relaxation with a certified rational infeasibility gap, rather than stopping on volume alone. For the integer system $\tilde{A}x\ge \tilde{b}$, putThe normalized integer Farkas infeasibility gap says that if the system is empty there is a multiplier $y\ge 0$ with ${\tilde{A}}^{T}y=0$, $1^{T}y=1$ and ${\tilde{b}}^{T}y\ge 1/D$. Its bound follows by maximizing the certificate's value in the normalized multiplier polytope: a vertex has at most $n+1$ positive entries, and a determinant denominator bounded by $D$. Any point satisfying the relaxed inequalities would then give the contradictionThus relaxation preserves emptiness. If the original system is nonempty, take its point $x_0\in [-R,R{]}^n$. The bounded relaxed setcontains a Euclidean ball around $x_0$ of radius $\rho =\min (1/2,\eta /(nV))$: each row changes by at most $\sqrt{n}V\rho \le \eta$, and the expanded box leaves room for the ball. Hence $Q$ is either empty or has an explicit positive inner-volume bound. This is the full-dimensional relaxation of integer inequalities principle.

Start with a ball, viewed as an ellipsoid, centered at zero with radius $n(R+1)$, which contains the search box. At each iteration test its center against all inequalities defining $Q$. If the center is feasible, the preserved-emptiness result answers that the original problem is nonempty. Otherwise a violated row supplies a separation oracle cut. Move that cut parallel to itself through the current center if necessary, and compute a smaller containing ellipsoid for the surviving half. Every feasible point remains in the new ellipsoid.

The central-cut ellipsoid volume bound gives a shrinkage factor at most $e^{-1/(2(n+1))}$ for $n\ge 2$. After $O(n^2\log (n(R+1)/\rho ))$ cuts, the volume is below that of a radius-$\rho$ ball. If no feasible center has been found, $Q$ and therefore the original system are empty. Dimension one uses interval bisection. The logarithms of $R$, $D$ and $1/\rho$ are polynomial in the encoded input size. Standard outward-controlled finite-precision ellipsoid updates preserve enclosure with polynomial bit complexity.

The coordinate bound therefore supplies the initial bounded search region; the rational gap supplies a valid stopping threshold. Neither unboundedness nor zero volume of the original polyhedron can be ignored when deciding emptiness.

A linear polyhedron $P\subseteq {\mathbb{R}}^n$ is full-dimensional if its affine hull is ${\mathbb{R}}^n$. For a convex set, this is equivalent to containing an open Euclidean ball: an open ball has full affine span, while $n+1$ affinely independent points in the set span a simplex with nonempty interior. In particular, a nonempty linear polyhedron need not be full-dimensional; a singleton or a hyperplane is an example.

Assume $U\ge 1$, increasing the coefficient bound to one if necessary. If $x_0\in P$ and $A_i$ denotes the $i$th row, then $\Vert A_i{\Vert }_2\le \sqrt{n}U$. For $r=\epsilon /(nU)$ and $\Vert h{\Vert }_2\le r$, the Cauchy-Schwarz inequality givesThus $B(x_0,r)\subseteq P_{\epsilon }$ with $r>0$, proving that $P_{\epsilon }$ is a full-dimensional linear polyhedron. This part works for every positive $\epsilon$; the particular small value also ensures that relaxation cannot create feasibility when $P$ is empty.

Here is that exact-feasibility point. Put $D=[(n+1)U{]}^{n+1}$. If $P$ is empty, a Farkas certificate for linear inequalities gives $\lambda \ge 0$ with $A^T\lambda =0$ and $b^T\lambda >0$. Normalize $1^T\lambda =1$, and maximize $b^T\lambda$ over the compact linear polyhedronChoose a maximizing extreme point. Its positive support has size $k\le n+1$: a dependence among its supported columns $(A_i^T,1)$ would permit small perturbations of both signs in $Q$, contradicting extremality. Select $k$ independent equations for those $k$ coordinates. Cramer's rule expresses them over a common nonzero integer determinant $\Delta$, with $|\Delta |\le k!U^k\le D$. Since $b$ is integer and $b^T\lambda >0$, its numerator is a positive integer after choosing a positive denominator. Hence $b^T\lambda \ge 1/D$. This is the normalized integer Farkas infeasibility gap. A point in $P_{\epsilon }$ would implya contradiction. Consequently $P=\varnothing$ if and only if $P_{\epsilon }=\varnothing$.

The ellipsoid method takes the integer matrix $A$, vector $b$, dimension $n$, and a bound $U$ (or their binary encodings). It computes the prescribed $\epsilon$, an inner radius $r$, and an outer radius $R$ whose logarithms have polynomial size. For an explicit outer bound, let $M=(nU{)}^n$. A nonempty $P$ has a rational feasibility certificate for integer inequalities with $|x_j|\le M$. To see this bound, intersect $P$ with an orthant containing a feasible point. This intersection is a nonempty linear polyhedron containing no whole nontrivial affine line, and it has an extreme point. Indeed, if its active rows do not yet span the ambient space, move along a nonzero direction annihilating them. The orthant prevents motion in both directions from being feasible forever, so in at least one direction a new independent row becomes active. Repeating at most $n$ times produces an extreme point. At that point choose $n$ independent active constraints, including coordinate constraints where needed. Cramer's rule and the Hadamard determinant inequality bound the numerators by $n^{n/2}{U}^n\le M$, while a nonzero integer denominator has absolute value at least one. Thus the point has norm at most $nM$. Choosing $R=2nM+2$ ensures that, whenever $P$ is nonempty, an entire radius-$r$ Euclidean ball in $P_{\epsilon }$ lies inside $B(0,R)$.

Start with the ellipsoid $E_0=B(0,R)$ and retain $K=P_{\epsilon }\cap B(0,R)$. At each iteration test the centre $c$ against every relaxed inequality. If it passes all tests, $P_{\epsilon }$ is nonempty and hence so is $P$. Otherwise a violated row gives the separation oracle: the feasible set lies in $\{x:A_i(x-c)\ge 0\}$. Cut the current ellipsoid by this half-space through its centre and replace it by the standard containing ellipsoid. Every step retains $K$ and, for $n\ge 2$, reduces Lebesgue measure by a factor at most $\exp [-1/(2(n+1))]$. In dimension one the corresponding step bisects an interval.

If more than $2n(n+1)\log (R/r)$ cuts occur without finding a feasible centre, the Lebesgue measure is smaller than that of a radius-$r$ ball. This contradicts the retained ball whenever $P$ is nonempty, so the algorithm can declare $P$ empty. Both $\log R$ and $\log (1/r)$ are polynomial in $n$ and $\log U$, and each inequality test uses $O(mn)$ arithmetic operations for $m$ rows. The bit implementation uses rational rounding with outward enlargement of the containing ellipsoids; the encoding bounds allow polynomial precision while keeping a constant fraction of the volume decrease. Detailed update and rounding formulae are not needed for this account. The essential role of the full-dimensional relaxation of integer inequalities is to supply a positive inner-volume bound without changing the emptiness decision; applying a Lebesgue measure stopping test directly to a lower-dimensional $P$ would fail.