# Farkas' lemma

(Difference between revisions)
 Revision as of 17:59, 29 May 2010 (edit)← Previous diff Revision as of 18:01, 29 May 2010 (edit) (undo)Next diff → Line 13: Line 13: == Proof == == Proof == - '''('''[https://ccrma.stanford.edu/~dattorro/mybook.html Dattorro, ch.2.13]''')''' Define a convex cone + '''('''[http://ccrma.stanford.edu/~dattorro/mybook.html Dattorro, ch.2.13]''')''' Define a convex cone * $\mathcal{K}=\{y~|~A^Ty\succeq0\}\quad$ * $\mathcal{K}=\{y~|~A^Ty\succeq0\}\quad$ whose dual cone is whose dual cone is

## Revision as of 18:01, 29 May 2010

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, ch.2.13) 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 $LaTeX: \,\mathcal{K}^*\!=\{b~|~b^{\rm T}y\!\geq\!0~~\forall~y\!\in_{}\!\mathcal{K}\}$ we get

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

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.

# Extended Farkas' lemma

For any closed convex cone $LaTeX: \mathcal J$ in the Hilbert space $LaTeX: (\mathbb H,\langle\cdot,\cdot\rangle)$ $LaTeX: (\,$in particular we can have $LaTeX: \mathbb H=\mathbb R^n)$, denote by $LaTeX: \mathcal J^\circ$ the polar cone of $LaTeX: \mathcal J.$

Let $LaTeX: \mathcal K$ be an arbitrary closed convex cone in $LaTeX: \mathbb H.$

Then, the extended Farkas' lemma asserts that $LaTeX: \mathcal K^{\circ\circ}=\mathcal K.$

Hence, denoting $LaTeX: \mathcal L=\mathcal K^\circ,$ it follows that $LaTeX: \mathcal L^\circ=\mathcal K.$

Therefore, the cones $LaTeX: \mathcal K$ and $LaTeX: \mathcal L$ are called mutually polar pair of cones.

### notes

For definition of convex cone see in finite dimension see Convex cones, Wikimization.

For definition of polar cone see Moreau's theorem.

## Proof of extended Farkas' lemma

(Sándor Zoltán Németh) Let $LaTeX: z\in\mathbb H$ be arbitrary. Then, by Moreau's theorem we have

$LaTeX: z=P_{\mathcal K}z+P_{\mathcal K^\circ}z$

and

$LaTeX: z=P_{\mathcal K^\circ}z+P_{\mathcal K^{\circ\circ}}z.$

Therefore,

$LaTeX: P_{\mathcal K}z=P_{\mathcal K^{\circ\circ}}z=z-P_{\mathcal K^\circ}z.$

In particular, for any $LaTeX: z\in K$ we have $LaTeX: z=P_{\mathcal K}z=P_{\mathcal K^{\circ\circ}}z\in\mathcal K^{\circ\circ}.$ Hence, $LaTeX: \mathcal K \subset \mathcal K^{\circ\circ}.$ Similarly, for any $LaTeX: z\in K^{\circ\circ}$ we have $LaTeX: z=P_{\mathcal K^{\circ\circ}}z= P_{\mathcal K}z\in\mathcal K.$ Hence, $LaTeX: \mathcal K^{\circ\circ}\subset\mathcal K.$ Therefore, $LaTeX: \mathcal K^{\circ\circ}=\mathcal K.$