Summary: Complete Dictionary Recovery over the Sphere

The data matrix is generated as \(\mathbf{Y} = \mathbf{A}_0 \mathbf{X}_0\), where \(\mathbf{Y} \in \mathbb{R}^{n \times p}\), and the following conditions are assumed.

The dictionary \(\mathbf{A} \in \mathbb{R}^{n \times n}\) is complete, namely square and invertible.

The elements in \(\mathbf{X}_0\) is independently drawn from Bernoulli-Gaussian (BG) model. \(X_{0{[ij]}} = \Omega_{ij} V_{ij}\), where \(\Omega_{ij} \sim \text{Ber}(\theta)\) and \(V_{ij} \sim N(0,1)\).

The goal is to recover \(\mathbf{A}_0\) and \(\mathbf{X}_0\) from \(\mathbf{Y}\).

The main result is that when \(\theta \in (0, 1/3)\), \(\mathbf{A}_0\) and \(\mathbf{X}_0\) can be exactly recovered with a polynomial time algorithm with high probability, if number of samples \(p\) is no less than a polynomial of \((n, 1/\theta, \kappa(\mathbf{A}_0), 1/\mu)\), where \(\kappa(\mathbf{A}_0)\) is the conditional number of the dictionary.

Since \(\mathbf{A_0}\) is complete, the row space of \(\mathbf{Y}\) is equivalent to the row space of \(\mathbf{X}_0\). By the result from (Spielman 2013), under the generative model, rows of \(\mathbf{X}_0\) corresponds to the \(n\) sparsest vectors in the row space. Once \(\mathbf{X}_0\) is recovered, \(\mathbf{A}_0\) could be solved from least square \(\mathbf{A}_0 = \mathbf{Y} \mathbf{X}_0 (\mathbf{X}_0 \mathbf{X}_0^T)^{-1}\).

A straightforward formulation for recovering one row of \(\mathbf{X}_0\) is \[\min \| \mathbf{q}^{T} \mathbf{Y} \|_0 \quad s.t.\ \mathbf{q} \neq \mathbf{0}.\] \(\mathbf{X}_0\) can b

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