Rank-one nonnegative matrix factorization

The DGP atom library has several functions of positive matrices, including the trace, (matrix) product, sum, Perron-Frobenius eigenvalue, and :math:(I - X)^{-1} (eye-minus-inverse). In this notebook, we use some of these atoms to approximate a partially known elementwise positive matrix as the outer product of two positive vectors.

We would like to approximate :math:A as the outer product of two positive vectors :math:x and :math:y, with :math:x normalized so that the product of its entries equals :math:1. Our criterion is the average relative deviation between the entries of :math:A and :math:xy^T, that is,

.. math::

\frac{1}{mn} \sum_{i=1}^{m} \sum_{j=1}^{n} R(A_{ij}, x_iy_j),

where :math:R is the relative deviation of two positive numbers, defined as

.. math::

R(a, b) = \max{a/b, b/a} - 1.

The corresponding optimization problem is

.. math::

\begin{equation} \begin{array}{ll} \mbox{minimize} & \frac{1}{mn} \sum_{i=1}^{m} \sum_{j=1}^{n} R(X_{ij}, x_iy_j) \ \mbox{subject to} & x_1x_2 \cdots x_m = 1 \ & X_{ij} = A_{ij}, \quad \text{for } (i, j) \in \Omega, \end{array} \end{equation}

with variables :math:X \in \mathbf{R}^{m \times n}_{++}, :math:x \in \mathbf{R}^{m}_{++}, and :math:y \in \mathbf{R}^{n}_{++}. We can cast this problem as an equivalent generalized geometric program by discarding the :math:-1 from the relative deviations.

The below code constructs and solves this optimization problem, with specific problem data

.. math::

A = \begin{bmatrix} 1.0 & ? & 1.9 \ ? & 0.8 & ? \ 3.2 & 5.9& ? \end{bmatrix},

.. code:: python

import cvxpy as cp

m = 3
n = 3
X = cp.Variable((m, n), pos=True)
x = cp.Variable((m,), pos=True)
y = cp.Variable((n,), pos=True)

outer_product = cp.vstack([x[i] * y for i in range(m)])
relative_deviations = cp.maximum(
  cp.multiply(X, outer_product ** -1),
  cp.multiply(X ** -1, outer_product))
objective = cp.sum(relative_deviations)
constraints = [
  X[0, 0] == 1.0,
  X[0, 2] == 1.9,
  X[1, 1] == 0.8,
  X[2, 0] == 3.2,
  X[2, 1] == 5.9,
  x[0] * x[1] * x[2] == 1.0,
]
problem = cp.Problem(cp.Minimize(objective), constraints)
problem.solve(gp=True)

print("Optimal value:\n", 1.0/(m * n) * (problem.value - m * n), "\n")
print("Outer product approximation\n", outer_product.value, "\n")
print("x: ", x.value)
print("y: ", y.value)

.. parsed-literal::

Optimal value:
 1.7763568394002505e-14 

Outer product approximation
 [[1.         1.84375    1.9       ]
 [0.43389831 0.8        0.82440678]
 [3.2        5.89999999 6.07999999]] 

x:  [0.89637009 0.38893346 2.86838428]
y:  [1.11561063 2.0569071  2.1196602 ]