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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.

Assume $n\ge 1$ and the integer coefficient bound $U\ge 1$. There are $m(n+1)$ integers to record. Each signed integer in $[-U,U]$ can be encoded in $\lceil {\log }_2(2U+1)\rceil$ bits, so a fixed-width dense encoding needsbits, including the dimensions and bound. Its size is $\boxed{O(mn\log (U+1))}$. This is the bit-input model used for polynomial time; an arithmetic-operation count alone would not suffice.

We first construct a rational feasibility certificate for integer inequalities. Choose a feasible point and an orthant containing it, adding coordinate inequalities of the form $x_i\ge 0$ or $-x_i\ge 0$. The intersection $Q$ is nonempty. Starting at a point of $Q$, if the active constraint normals have rank less than $n$, choose a nonzero direction $d$ annihilating them. A sufficiently short move in either direction remains feasible, since every inactive constraint has positive slack. At least one direction has a finite endpoint: a nonzero coordinate of $d$ can be oriented toward the orthant boundary. Move to the first constraint met along such a direction. Its normal does not annihilate $d$, so it is independent of the previously active normals. Their rank strictly increases. In at most $n$ moves we obtain a point with $n$ independent active constraints.

At that point $Bx=d_0$ for an invertible integer $n\times n$ matrix $B$. Its rows are original constraint rows or signed coordinate rows, and $d_0$ has entries from $b$ or zero. All entries have absolute value at most $U$. By Cramer's rule,where $B_i$ replaces column $i$ by $d_0$. The Hadamard determinant inequality bounds both determinants by $(\sqrt{n}U{)}^n\le (nU{)}^n$. The denominator is a nonzero integer. Hence each coordinate has the claimed representation with numerator and denominator bounded in absolute value by $R=(nU{)}^n$. The orthant argument is needed because $P$ itself may contain lines and have no vertex. If $U=0$ is allowed literally, all coefficients and right sides vanish, and $x=0$ is feasible; the printed fraction bound then needs $U$ replaced by $\max (1,U)$, since a nonzero denominator cannot have absolute value at most zero. The bound just proved uses the customary positive coefficient bound.

Recording $n$ such fractions takes $O(n^2\log (nU))$ bits. A verifier checks that denominators are nonzero, changes their signs if needed, and verifies all inequalities by exact integer arithmetic after multiplying by positive denominators. The intermediate integers still have polynomial bit length. Thus a yes-instance has a polynomial-size, polynomial-time verifiable witness, and $\boxed{\mathrm{FP}\in \mathbf{NP}}$.

For optimization, let $c$ be rational with a finite binary encoding; clear its denominators to make it integer, which multiplies the objective by a positive constant and preserves its minimizers. All bit bounds below include the encoding length of $c$. Let $V=\max (1,{\max }_i|c_i|)$ and $B_0=nVR$.

A useful refinement of the preceding vertex argument is that if an objective is bounded below, a feasible point can be moved to an orthant vertex without increasing its objective. Choose the annihilating direction to have $c^{T}d\le 0$. If the inequality is strict, the forward feasible interval must have a finite endpoint, since otherwise the objective would be unbounded below. If $c^{T}d=0$, choose either finite endpoint supplied by the orthant. Rank increases as before. There are only finitely many such vertices across the finitely many orthants. Every feasible point therefore has one of these vertices with no larger objective; taking the least vertex value proves that a finite optimum is attained at one of them.

The Cramer representations at this optimum have one common denominator $\det B$, of magnitude at most $R$. Therefore its objective value $z^{\ast }$ has a representation with denominator at most $R$ and numerator at most $B_0$. In particular $|z^{\ast }|\le B_0$. No enumeration of orthants or vertices is part of the algorithm; they provide only polynomial-bit bounds.

Here is an exact linear optimization from a feasibility oracle. First call the assumed ellipsoid method feasibility routine on $P$; if it is empty, report infeasibility. Otherwise query feasibility with the extra inequality $c^{T}x\le -B_0-1$. A yes answer proves unboundedness below, because any finite optimum would be at least $-B_0$; an unbounded objective necessarily gives a yes answer. Thus the infeasible, unbounded and finite-optimum cases are distinguished using two feasibility calls.

In the finite case, maintain an infeasible lower threshold $-B_0-1$ and a feasible upper threshold $B_0$. Bisect the interval, querying $c^{T}x\le t$ each time. After polynomially many calls its length is less than $1/(2R^2)$, and $z^{\ast }$ lies in the resulting interval. Distinct reduced fractions with positive denominators at most $R$ differ by at least $1/R^2$, so there is only one possible value. It can be recovered in polynomial time by continued fractions: for the midpoint $t$, $|z^{\ast }-t|<1/(4R^2)\le 1/(2q^2)$, where $q\le R$ is the denominator of $z^{\ast }$. Thus $z^{\ast }$ occurs among the continued-fraction convergents of $t$. Compute those by the Euclidean algorithm and choose the unique fraction with denominator at most $R$ in the interval. If $t=z^{\ast }$, use the final exact convergent. The Euclidean algorithm and all thresholds have polynomial bit length.

Even if the feasibility routine returns only a decision, an optimal point can be recovered without assuming a witness-producing oracle. Add $c^{T}x=z^{\ast }$ and $-R\le x_i\le R$ for all $i$. This bounded optimal convex polytope is nonempty, since the optimal vertex above lies in it. Clear the denominator of $z^{\ast }$ and setAll vertex coordinates of this fixed convex polytope have denominators at most $D_0$ by the same determinant argument. Minimize $x_1$, recover its exact minimum by feasibility bisection and rational reconstruction, then fix it; do the same for $x_2$, and so on. Each restriction is an optimal face of the same original bounded convex polytope, so its vertices are original vertices. The single denominator bound $D_0$ therefore works at every step; repeated new precision bounds do not grow exponentially. After $n$ steps all coordinates are fixed to a feasible optimizer. There are polynomially many calls on polynomial-length rational instances, converted to integer inequalities by clearing denominators. Consequently linear optimization has a polynomial-time algorithm whenever linear feasibility does, with exact optimal value and optimizer, and correct reports of infeasibility or unboundedness.