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In mathematics, Farkas' lemma is a solvability theorem for a finite system of linear inequalities. It was originally proven by the Hungarian mathematician Gyula Farkas. Farkas' lemma is the key result underpinning the linear programming duality and has played a central role in the development of mathematical optimization (alternatively, mathematical programming). It is used amongst other things in the proof of the Karush–Kuhn–Tucker theorem in nonlinear programming. Remarkably, in the area of the foundations of quantum theory, the lemma also underlies the complete set of Bell inequalities in the form of necessary and sufficient conditions for the existence of a local hidden-variable theory, given data from any specific set of measurements.Generalizations of the Farkas' lemma are about the solvability theorem for convex inequalities, i.e., infinite system of linear inequalities. Farkas' lemma belongs to a class of statements called "theorems of the alternative": a theorem stating that exactly one of two systems has a solution. Statement of the lemma There are a number of slightly different (but equivalent) formulations of the lemma in the literature. The one given here is due to Gale, Kuhn and Tucker (1951). Here, the notation x≥0{\displaystyle \mathbf {x} \geq 0} means that all components of the vector x{\displaystyle \mathbf {x} } are nonnegative. Example Let m, n = 2, A=[6430],{\displaystyle \mathbf {A} ={\begin{bmatrix}6&4\\3&0\end{bmatrix}},} and b=[b1b2].{\displaystyle \mathbf {b} ={\begin{bmatrix}b_{1}\\b_{2}\end{bmatrix}}.} The lemma says that exactly one of the following two statements must be true (depending on b1 and b2): There exist x1 ≥ 0, x2 ≥ 0 such that 6x1 + 4x2 = b1 and 3x1 = b2, or There exist y1, y2 such that 6y1 + 3y2 ≥ 0, 4y1 ≥ 0, and b1 y1 + b2 y2 < 0.Here is a proof of the lemma in this special case: If b2 ≥ 0 and b1 − 2b2 ≥ 0, then option 1 is true, since the solution of the linear equations is x1=b23{\displaystyle x_{1}={\tfrac {b_{2}}{3}}} and x2=b1−2b24.{\displaystyle x_{2}={\tfrac {b_{1}-2b_{2}}{4}}.} Option 2 is false, since so if the right-hand side is positive, the left-hand side must be positive too. Otherwise, option 1 is false, since the unique solution of the linear equations is not weakly positive. But in this case, option 2 is true: If b2 < 0, then we can take e.g. y1 = 0 and y2 = 1. If b1 − 2b2 < 0, then, for some number B > 0, b1 = 2b2 − B, so: Thus we can take, for example, y1 = 1, y2 = −2.Geometric interpretation Consider the closed convex cone C(A){\displaystyle C(\mathbf {A} )} spanned by the columns of A; that is, C(A)={Ax∣x≥0}.{\displaystyle C(\mathbf {A} )=\{\mathbf {A} \mathbf {x} \mid \mathbf {x} \geq 0\}.}Observe that C(A){\displaystyle C(\mathbf {A} )} is the set of the vectors b for which the first assertion in the statement of Farkas' lemma holds. On the other hand, the vector y in the second assertion is orthogonal to a hyperplane that separates b and C(A).{\displaystyle C(\mathbf {A} ).} The lemma follows from the observation that b belongs to C(A){\displaystyle C(\mathbf {A} )} if and only if there is no hyperplane that separates it from C(A).{\displaystyle C(\mathbf {A} ).} More precisely, let a1,…,an∈Rm{\displaystyle \mathbf {a} _{1},\dots ,\mathbf {a} _{n}\in \mathbb {R} ^{m}} denote the columns of A. In terms of these vectors, Farkas' lemma states that exactly one of the following two statements is true: There exist non-negative coefficients x1,…,xn∈R{\displaystyle x_{1},\dots ,x_{n}\in \mathbb {R} } such that b=x1a1+⋯+xnan.{\displaystyle \mathbf {b} =x_{1}\mathbf {a} _{1}+\dots +x_{n}\mathbf {a} _{n}.} There exists a vector y∈Rm{\displaystyle \mathbf {y} \in \mathbb {R} ^{m}} such that ai⊤y≥0{\displaystyle \mathbf {a} _{i}^{\top }\mathbf {y} \geq 0} for i=1,…,n,{\displaystyle i=1,\dots ,n,} and b⊤y<0.{\displaystyle \mathbf {b} ^{\top }\mathbf {y} <0.}The sums x1a1+⋯+xnan{\displaystyle x_{1}\mathbf {a} _{1}+\dots +x_{n}\mathbf {a} _{n}} with nonnegative coefficients x1,…,xn{\displaystyle x_{1},\dots ,x_{n}} form the cone spanned by the columns of A. Therefore, the first statement tells that b belongs to C(A).{\displaystyle C(\mathbf {A} ).} The second statement tells that there exists a vector y such that the angle of y with the vectors ai is at most 90°, while the angle of y with the vector b is more than 90°. The hyperplane normal to this vector has the vectors ai on one side and the vector b on the other side. Hence, this hyperplane separates the cone spanned by a1,…,an{\displaystyle \mathbf {a} _{1},\dots ,\mathbf {a} _{n}} from the vector b. For example, let n, m = 2, a1 = (1, 0)T, and a2 = (1, 1)T. The convex cone spanned by a1 and a2 can be seen as a wedge-shaped slice of the first quadrant in the xy plane. Now, suppose b = (0, 1). Certainly, b is not in the convex cone a1x1 + a2x2. Hence, there must be a separating hyperplane. Let y = (1, −1)T. We can see that a1 · y = 1, a2 · y = 0, and b · y = −1. Hence, the hyperplane with normal y indeed separates the convex cone a1x1 + a2x2 from b. Logic interpretation A particularly suggestive and easy-to-remember version is the following: if a set of linear inequalities has no solution, then a contradiction can be produced from it by linear combination with nonnegative coefficients. In formulas: if Ax≤b{\displaystyle \mathbf {Ax} \leq \mathbf {b} } is unsolvable then y⊤A=0,{\displaystyle \mathbf {y} ^{\top }\mathbf {A} =0,} y⊤b=−1,{\displaystyle \mathbf {y} ^{\top }\mathbf {b} =-1,} y≥0{\displaystyle \mathbf {y} \geq 0} has a solution. Note that y⊤A{\displaystyle \mathbf {y} ^{\top }\mathbf {A} } is a combination of the left-hand sides, y⊤b{\displaystyle \mathbf {y} ^{\top }\mathbf {b} } a combination of the right-hand side of the inequalities. Since the positive combination produces a zero vector on the left and a −1 on the right, the contradiction is apparent. Thus, Farkas' lemma can be viewed as a theorem of logical completeness: Ax≤b{\displaystyle \mathbf {Ax} \leq \mathbf {b} } is a set of "axioms", the linear combinations are the "derivation rules", and the lemma says that, if the set of axioms is inconsistent, then it can be refuted using the derivation rules.: 92–94  Implications in complexity theory Farkas' lemma implies that the decision problem "Given a system of linear equations, does it have a non-negative solution?" is in the intersection of NP and co-NP. This is because, according to the lemma, both a "yes" answer and a "no" answer have a proof that can be verified in polynomial time. The problems in the intersection NP∩coNP{\displaystyle NP\cap coNP} are also called well-characterized problems. It is a long-standing open question whether NP∩coNP{\displaystyle NP\cap coNP} is equal to P. In particular, the question of whether a system of linear equations has a non-negative solution was not known to be in P, until it was proved using the ellipsoid method.: 25  Variants The Farkas Lemma has several variants w.... Discover the I J Farkas popular books. Find the top 100 most popular I J Farkas books.

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