Nonnegative matrix factorization

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Exercise from [http://meboo.convexoptimization.com/Meboo.html Convex Optimization & Euclidean Distance Geometry], ch.4:
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Given rank-2 nonnegative matrix
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<math>X=\!\left[\!\begin{array}{ccc}17&28&42\\
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16&47&51\\
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17&82&72\end{array}\!\right],</math>
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find a nonnegative factorization
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<math> X=WH\,</math>
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by solving
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<math>\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\\
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\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\\
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&W\geq0\\
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&H\geq0\\
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&\mbox{rank}\,Z\leq2\end{array}</math>
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which follows from the fact, at optimality,
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<math> Z^\star=\left[\!\begin{array}{c}I\\W\\H^{\rm T}\end{array}\!\right]\begin{array}{c}\textbf{[}\,I~~W^{\rm T}~H\,\textbf{]}
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\end{array}</math>
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Use the known closed-form solution for a direction vector <math>Y\,</math> to regulate rank (rank constraint is replaced) by [[Convex Iteration]];
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set <math>_{}Z^\star\!=Q\Lambda Q^{\rm T}\!\in\mathbb{S}^\mathbf{8}</math> to a nonincreasingly ordered diagonalization and
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<math>_{}U^\star\!=_{\!}Q(:\,,_{^{}}3\!:\!8)\!\in_{\!}\reals^{\mathbf{8}\times\mathbf{6}}</math>,
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then <math>Y\!=U^\star U^{\star\rm T}.</math>
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<br>
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In summary, initialize <math>Y=I\,</math> then alternate solution of
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<math>\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\\
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\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\\
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&W\geq0\\
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&H\geq0\end{array}</math>
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with
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<math>Y\!=U^\star U^{\star\rm T}.</math>
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Global convergence occurs, in this example, in only a few iterations.

Revision as of 05:24, 17 February 2010

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

Given rank-2 nonnegative matrix LaTeX: X=\!\left[\!\begin{array}{ccc}17&28&42\\ </p> <pre> 16&47&51\\ 17&82&72\end{array}\!\right],

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

LaTeX: \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\\ \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\\ &\mbox{rank}\,Z\leq2\end{array}

which follows from the fact, at optimality,

LaTeX: Z^\star=\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^\star\!=Q\Lambda Q^{\rm T}\!\in\mathbb{S}^\mathbf{8} to a nonincreasingly ordered diagonalization and LaTeX: _{}U^\star\!=_{\!}Q(:\,,_{^{}}3\!:\!8)\!\in_{\!}\reals^{\mathbf{8}\times\mathbf{6}}, then LaTeX: Y\!=U^\star U^{\star\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^\star U^{\star\rm T}. Global convergence occurs, in this example, in only a few iterations.

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