Farkas' 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).[6]

Here, the notation ${\displaystyle \mathbf {x} \geq 0}$ means that all components of the vector ${\displaystyle \mathbf {x} }$ are nonnegative.

Let m, n = 2, ${\displaystyle \mathbf {A} ={\begin{bmatrix}6&4\\3&0\end{bmatrix}},}$ and ${\displaystyle \mathbf {b} ={\begin{bmatrix}b_{1}\\b_{2}\end{bmatrix}}.}$ The lemma says that exactly one of the following two statements must be true (depending on b₁ and b₂):

1. There exist x₁ ≥ 0, x₂ ≥ 0 such that 6x₁ + 4x₂ = b₁ and 3x₁ = b₂, or
2. 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 ${\displaystyle x_{1}={\tfrac {b_{2}}{3}}}$ and ${\displaystyle x_{2}={\tfrac {b_{1}-2b_{2}}{4}}.}$ Option 2 is false, since
  $${\displaystyle b_{1}y_{1}+b_{2}y_{2}\geq b_{2}(2y_{1}+y_{2})=b_{2}{\frac {6y_{1}+3y_{2}}{3}},}$$

  
  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:
    $${\displaystyle b_{1}y_{1}+b_{2}y_{2}=2b_{2}y_{1}+b_{2}y_{2}-By_{1}=b_{2}{\frac {6y_{1}+3y_{2}}{3}}-By_{1}.}$$

    
    Thus we can take, for example, y₁ = 1, y₂ = −2.

Consider the closed convex cone ${\displaystyle C(\mathbf {A} )}$ spanned by the columns of A; that is,

${\displaystyle C(\mathbf {A} )=\{\mathbf {A} \mathbf {x} \mid \mathbf {x} \geq 0\}.}$

Observe that ${\displaystyle C(\mathbf {A} )}$ 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 ${\displaystyle C(\mathbf {A} ).}$ The lemma follows from the observation that b belongs to ${\displaystyle C(\mathbf {A} )}$ if and only if there is no hyperplane that separates it from ${\displaystyle C(\mathbf {A} ).}$

More precisely, let ${\displaystyle \mathbf {a} _{1},\dots ,\mathbf {a} _{n}\in \mathbb {R} ^{m}}$ denote the columns of A. In terms of these vectors, Farkas' lemma states that exactly one of the following two statements is true:

1. There exist non-negative coefficients ${\displaystyle x_{1},\dots ,x_{n}\in \mathbb {R} }$ such that ${\displaystyle \mathbf {b} =x_{1}\mathbf {a} _{1}+\dots +x_{n}\mathbf {a} _{n}.}$
2. There exists a vector ${\displaystyle \mathbf {y} \in \mathbb {R} ^{m}}$ such that ${\displaystyle \mathbf {a} _{i}^{\top }\mathbf {y} \geq 0}$ for ${\displaystyle i=1,\dots ,n,}$ and ${\displaystyle \mathbf {b} ^{\top }\mathbf {y} <0.}$

The sums ${\displaystyle x_{1}\mathbf {a} _{1}+\dots +x_{n}\mathbf {a} _{n}}$ with nonnegative coefficients ${\displaystyle x_{1},\dots ,x_{n}}$ form the cone spanned by the columns of A. Therefore, the first statement tells that b belongs to ${\displaystyle C(\mathbf {A} ).}$

The second statement tells that there exists a vector y such that the angle of y with the vectors a_i 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 a_i on one side and the vector b on the other side. Hence, this hyperplane separates the cone spanned by ${\displaystyle \mathbf {a} _{1},\dots ,\mathbf {a} _{n}}$ 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.

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 ${\displaystyle \mathbf {Ax} \leq \mathbf {b} }$ is unsolvable then ${\displaystyle \mathbf {y} ^{\top }\mathbf {A} =0,}$ ${\displaystyle \mathbf {y} ^{\top }\mathbf {b} =-1,}$ ${\displaystyle \mathbf {y} \geq 0}$ has a solution.[7] Note that ${\displaystyle \mathbf {y} ^{\top }\mathbf {A} }$ is a combination of the left-hand sides, ${\displaystyle \mathbf {y} ^{\top }\mathbf {b} }$ 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: ${\displaystyle \mathbf {Ax} \leq \mathbf {b} }$ 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.[8]^: 92–94

The Farkas Lemma has several variants with different sign constraints (the first one is the original version):[8]^: 92

• Either the system ${\displaystyle \mathbf {Ax} =\mathbf {b} }$ has a solution with ${\displaystyle \mathbf {x} \geq 0,}$ or the system ${\displaystyle \mathbf {A} ^{\top }\mathbf {y} \geq 0}$ has a solution with ${\displaystyle \mathbf {b} ^{\top }\mathbf {y} <0.}$
• Either the system ${\displaystyle \mathbf {Ax} \leq \mathbf {b} }$ has a solution with ${\displaystyle \mathbf {x} \geq 0,}$ or the system ${\displaystyle \mathbf {A} ^{\top }\mathbf {y} \geq 0}$ has a solution with ${\displaystyle \mathbf {b} ^{\top }\mathbf {y} <0}$ and ${\displaystyle \mathbf {y} \geq 0.}$
• Either the system ${\displaystyle \mathbf {Ax} \leq \mathbf {b} }$ has a solution with ${\displaystyle \mathbf {x} \in \mathbb {R} ^{n},}$ or the system ${\displaystyle \mathbf {A} ^{\top }\mathbf {y} =0}$ has a solution with ${\displaystyle \mathbf {b} ^{\top }\mathbf {y} <0}$ and ${\displaystyle \mathbf {y} \geq 0.}$
• Either the system ${\displaystyle \mathbf {Ax} =\mathbf {b} }$ has a solution with ${\displaystyle \mathbf {x} \in \mathbb {R} ^{n},}$ or the system ${\displaystyle \mathbf {A} ^{\top }\mathbf {y} =0}$ has a solution with ${\displaystyle \mathbf {b} ^{\top }\mathbf {y} \neq 0.}$

The latter variant is mentioned for completeness; it is not actually a "Farkas lemma" since it contains only equalities. Its proof is an exercise in linear algebra.

Generalized Farkas' lemma can be interpreted geometrically as follows: either a vector is in a given closed convex cone, or there exists a hyperplane separating the vector from the cone; there are no other possibilities. The closedness condition is necessary, see Separation theorem I in Hyperplane separation theorem. For original Farkas' lemma, ${\displaystyle \mathbf {S} }$ is the nonnegative orthant ${\displaystyle \mathbb {R} _{+}^{n},}$ hence the closedness condition holds automatically. Indeed, for polyhedral convex cone, i.e., there exists a ${\displaystyle \mathbf {B} \in \mathbb {R} ^{n\times k}}$ such that ${\displaystyle \mathbf {S} =\{\mathbf {B} \mathbf {x} \mid \mathbf {x} \in \mathbb {R} _{+}^{k}\},}$ the closedness condition holds automatically. In convex optimization, various kinds of constraint qualification, e.g. Slater's condition, are responsible for closedness of the underlying convex cone ${\displaystyle C(\mathbf {A} ).}$

By setting ${\displaystyle \mathbf {S} =\mathbb {R} ^{n}}$ and ${\displaystyle \mathbf {S} ^{*}=\{0\}}$ in generalized Farkas' lemma, we obtain the following corollary about the solvability for a finite system of linear equalities:

Farkas's lemma can be varied to many further theorems of alternative by simple modifications,[5] such as Gordan's theorem: Either ${\displaystyle \mathbf {Ax} <0}$ has a solution x, or ${\displaystyle \mathbf {A} ^{\top }\mathbf {y} =0}$ has a nonzero solution y with y ≥ 0.

Common applications of Farkas' lemma include proving the strong duality theorem associated with linear programming and the Karush–Kuhn–Tucker conditions. An extension of Farkas' lemma can be used to analyze the strong duality conditions for and construct the dual of a semidefinite program. It is sufficient to prove the existence of the Karush–Kuhn–Tucker conditions using the Fredholm alternative but for the condition to be necessary, one must apply von Neumann's minimax theorem to show the equations derived by Cauchy are not violated.

• Dual linear program
• Fourier–Motzkin elimination – can be used to prove Farkas' lemma.

• Goldman, A. J.; Tucker, A. W. (1956). "Polyhedral Convex Cones". In Kuhn, H. W.; Tucker, A. W. (eds.). Linear Inequalities and Related Systems. Princeton: Princeton University Press. pp. 19–40. ISBN 0691079994.
• Rockafellar, R. T. (1979). Convex Analysis. Princeton University Press. p. 200.
• Kutateladze S.S. The Farkas lemma revisited. Siberian Mathematical Journal, 2010, Vol. 51, No. 1, 78–87.