# 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 $(I - X)^{-1}$ (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 $A$ as the outer product of two positive vectors $x$ and $y$, with $x$ normalized so that the product of its entries equals $1$. Our criterion is the average relative deviation between the entries of $A$ and $xy^T$, that is,

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

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

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

The corresponding optimization problem is

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

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

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

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

In [1]:
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)

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 ]