# Farkas' lemma

(Difference between revisions)
 Revision as of 21:12, 10 November 2008 (edit) (→Geometrical Interpretation)← Previous diff Revision as of 13:42, 12 November 2008 (edit) (undo)Next diff → Line 32: Line 32: == Geometrical Interpretation == == Geometrical Interpretation == - Farkas' lemma simply states that + Farkas' lemma simply states that either vector $\,b$ belongs to convex cone $\mathcal{K}^*$ or it does not. - either vector $\,b$ belongs to convex cone $\mathcal{K}^*$ + - or it does not. + When $b\notin\mathcal{K}^*$, then there is a vector $\,y\!\in\!\mathcal{K}$ + normal to a hyperplane separating point $\,b$ from cone $\mathcal{K}^*$. == References == == References == * Gyula Farkas, Über die Theorie der Einfachen Ungleichungen, Journal für die Reine und Angewandte Mathematik, volume 124, pages 1–27, 1902. * Gyula Farkas, Über die Theorie der Einfachen Ungleichungen, Journal für die Reine und Angewandte Mathematik, volume 124, pages 1–27, 1902. [http://dz-srv1.sub.uni-goettingen.de/sub/digbib/loader?ht=VIEW&did=D261364 http://dz-srv1.sub.uni-goettingen.de/sub/digbib/loader?ht=VIEW&did=D261364] [http://dz-srv1.sub.uni-goettingen.de/sub/digbib/loader?ht=VIEW&did=D261364 http://dz-srv1.sub.uni-goettingen.de/sub/digbib/loader?ht=VIEW&did=D261364]

## Revision as of 13:42, 12 November 2008

Farkas' lemma is a result used in the proof of the Karush-Kuhn-Tucker (KKT) theorem from nonlinear programming.

It states that if $LaTeX: \,A\,$ is a matrix and $LaTeX: \,b$ a vector, then exactly one of the following two systems has a solution:

• $LaTeX: A^Ty\succeq0$ for some $LaTeX: y\,$ such that $LaTeX: b^Ty<0~~$

or in the alternative

• $LaTeX: Ax=b\,$ for some $LaTeX: x\succeq0$

where the notation $LaTeX: x\succeq0$ means that all components of the vector $LaTeX: x$ are nonnegative.

The lemma was originally proved by Farkas in 1902. The above formulation is due to Albert W. Tucker in the 1950s.

It is an example of a theorem of the alternative; a theorem stating that of two systems, one or the other has a solution, but not both.

## Proof

(Dattorro) Define a convex cone

• $LaTeX: \mathcal{K}=\{y~|~A^Ty\succeq0\}\quad$

whose dual cone is

• $LaTeX: \quad\mathcal{K}^*=\{A_{}x~|~x\succeq0\}$

From the definition of dual cone,

$LaTeX: y\in\mathcal{K}~\Leftrightarrow~b^Ty\geq0~~\forall~b\in\mathcal{K}^*$

rather,

$LaTeX: A^Ty\succeq0~\Leftrightarrow~b^Ty\geq0~~\forall~b\in\{A_{}x~|~x\succeq0\}$

Given some $LaTeX: {\displaystyle b}$ vector and $LaTeX: y\!\in\!\mathcal{K}~$, then $LaTeX: {\displaystyle b^Ty\!<\!0}$ can only mean $LaTeX: b\notin\mathcal{K}^*$.

An alternative system is therefore simply $LaTeX: b\in\mathcal{K}^*$ and so the stated result follows.

## Geometrical Interpretation

Farkas' lemma simply states that either vector $LaTeX: \,b$ belongs to convex cone $LaTeX: \mathcal{K}^*$ or it does not.

When $LaTeX: b\notin\mathcal{K}^*$, then there is a vector $LaTeX: \,y\!\in\!\mathcal{K}$ normal to a hyperplane separating point $LaTeX: \,b$ from cone $LaTeX: \mathcal{K}^*$.

## References

• Gyula Farkas, Über die Theorie der Einfachen Ungleichungen, Journal für die Reine und Angewandte Mathematik, volume 124, pages 1–27, 1902.