# Nonnegative matrix factorization

(Difference between revisions)
 Revision as of 13:44, 24 November 2011 (edit)← Previous diff Current revision (17:40, 30 October 2016) (edit) (undo) Line 9: Line 9: by solving by solving - $\begin{array}{cl}\mbox{find}_{A\in\mathbb{S}^3,\,B\in\mathbb{S}^3,\,W\in\mathbb{R}^{3\times2},\,H\in\mathbb{R}^{2\times3}}&W\,,\,H\\ + [itex]\begin{array}{cl}{\text find}_{A\in\mathbb{S}^3,\,B\in\mathbb{S}^3,\,W\in\mathbb{R}^{3\times2},\,H\in\mathbb{R}^{2\times3}}&W\,,\,H\\ - \mbox{subject to}&Z=\left[\begin{array}{ccc}I&W^{\rm T}&H\\W&A&X\\H^{\rm T}&X^{\rm T}&B\end{array}\right]\succeq0\\ + {\text subject to}&Z=\left[\begin{array}{ccc}I&W^{\rm T}&H\\W&A&X \\H^{\rm T}&X^{\rm T}&B\end{array}\right]\succeq0\\ &W\geq0\\ &W\geq0\\ &H\geq0\\ &H\geq0\\ - &\mbox{rank}\,Z\leq2\end{array}$ + &{\text rank}\,Z\leq2\end{array}[/itex] which follows from the fact, at optimality, which follows from the fact, at optimality, Line 30: Line 30: $\begin{array}{cl}\mbox{minimize}_{A\in\mathbb{S}^3,\,B\in\mathbb{S}^3,\,W\in\mathbb{R}^{3\times2},\,H\in\mathbb{R}^{2\times3}}&\langle Z\,,Y\rangle\\ [itex]\begin{array}{cl}\mbox{minimize}_{A\in\mathbb{S}^3,\,B\in\mathbb{S}^3,\,W\in\mathbb{R}^{3\times2},\,H\in\mathbb{R}^{2\times3}}&\langle Z\,,Y\rangle\\ - \mbox{subject to}&Z=\left[\begin{array}{ccc}I&W^{\rm T}&H\\W&A&X\\H^{\rm T}&X^{\rm T}&B\end{array}\right]\succeq0\\ + \mbox{subject to}&Z=\left[\begin{array}{ccc}I&W^{\rm T}&H\\W&A&X \\H^{\rm T}&X^{\rm T}&B\end{array}\right]\succeq0\\ &W\geq0\\ &W\geq0\\ &H\geq0\end{array}$ &H\geq0\end{array}[/itex]

## Current revision

Exercise from Convex Optimization & Euclidean Distance Geometry, ch.4:

Given rank-2 nonnegative matrix $LaTeX: X=\!\left[\!\begin{array}{ccc}17&28&42\\

 16&47&51\\

17&82&72\end{array}\!\right],$

find a nonnegative factorization $LaTeX: X=WH\,$ by solving

$LaTeX: \begin{array}{cl}{\text find}_{A\in\mathbb{S}^3,\,B\in\mathbb{S}^3,\,W\in\mathbb{R}^{3\times2},\,H\in\mathbb{R}^{2\times3}}&W\,,\,H\\ {\text subject to}&Z=\left[\begin{array}{ccc}I&W^{\rm T}&H\\W&A&X \\H^{\rm T}&X^{\rm T}&B\end{array}\right]\succeq0\\ &W\geq0\\ &H\geq0\\ &{\text rank}\,Z\leq2\end{array}$

which follows from the fact, at optimality,

$LaTeX: Z^*=\left[\!\begin{array}{c}I\\W\\H^{\rm T}\end{array}\!\right]\begin{array}{c}\textbf{[}\,I~~W^{\rm T}~H\,\textbf{]} \end{array}$

Use the known closed-form solution for a direction vector $LaTeX: Y\,$ to regulate rank (rank constraint is replaced) by Convex Iteration;

set $LaTeX: _{}Z^*\!=Q\Lambda Q^{\rm T}\!\in\mathbb{S}^{\mathbf{8}}$ to a nonincreasingly ordered diagonalization and $LaTeX: _{}U^*\!=_{\!}Q(:\,,_{^{}}3:8)\!\in_{\!}\mathbb{R}^{\mathbf{8}\times\mathbf{6}}$, then $LaTeX: Y\!=U^* U^{*\rm T}.$

In summary, initialize $LaTeX: Y=I\,$ then alternate solution of

$LaTeX: \begin{array}{cl}\mbox{minimize}_{A\in\mathbb{S}^3,\,B\in\mathbb{S}^3,\,W\in\mathbb{R}^{3\times2},\,H\in\mathbb{R}^{2\times3}}&\langle Z\,,Y\rangle\\ \mbox{subject to}&Z=\left[\begin{array}{ccc}I&W^{\rm T}&H\\W&A&X \\H^{\rm T}&X^{\rm T}&B\end{array}\right]\succeq0\\ &W\geq0\\ &H\geq0\end{array}$

with

$LaTeX: Y\!=U^* U^{*\rm T}.$ Global convergence occurs, in this example, in only a few iterations.