Farkas' lemma

From Wikimization

(Difference between revisions)
Jump to: navigation, search
(Geometrical Interpretation)
Line 32: Line 32:
== Geometrical Interpretation ==
== Geometrical Interpretation ==
-
Farkas' lemma simply states that
+
Farkas' lemma simply states that either vector <math>\,b</math> belongs to convex cone <math>\mathcal{K}^*</math> or it does not.
-
either vector <math>\,b</math> belongs to convex cone <math>\mathcal{K}^*</math>
+
 
-
or it does not.
+
When <math>b\notin\mathcal{K}^*</math>, then there is a vector <math>\,y\!\in\!\mathcal{K}</math>
 +
normal to a hyperplane separating point <math>\,b</math> from cone <math>\mathcal{K}^*</math>.
== 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 12:42, 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.

http://dz-srv1.sub.uni-goettingen.de/sub/digbib/loader?ht=VIEW&did=D261364

Personal tools