An exact penalty approach for optimization with nonnegative orthogonality constraints

Abstract

Optimization with nonnegative orthogonality constraints has wide applications in machine learning and data sciences. It is NP-hard due to some combinatorial properties of the constraints. We first propose an equivalent optimization formulation with nonnegative and multiple spherical constraints and an additional single nonlinear constraint. Various constraint qualifications, the first- and second-order optimality conditions of the equivalent formulation are discussed. By establishing a local error bound of the feasible set, we design a class of (smooth) exact penalty models via keeping the nonnegative and multiple spherical constraints. The penalty models are exact if the penalty parameter is sufficiently large but finite. A practical penalty algorithm with postprocessing is then developed to approximately solve a series of subproblems with nonnegative and multiple spherical constraints. We study the asymptotic convergence and establish that any limit point is a weakly stationary point of the original problem and becomes a stationary point under some additional mild conditions. Extensive numerical results on the problem of computing the orthogonal projection onto nonnegative orthogonality constraints, the orthogonal nonnegative matrix factorization problems and the K-indicators model show the effectiveness of our proposed approach.

A Construction of problem (4.1) with unique solution

Proposition 1

Choose $X^* \in {\mathcal {S}}^{n,k}_+$ and $L \in {\mathbb {R}}^{k \times k}$ with positive diagonal elements satisfying $L_{ii}L_{jj} > \max \{L_{ij}, L_{ji}, 0\}^2\ \forall i, j \in [k], i \ne j.$ Then the optimal solution of (4.1) with $C = X^*L^{\top }$ is unique and equals to $X^*$.

Proof

For simplicity of notation, we use $\sum _i$ to denote $\sum _{i \in [k]}$ in the proof. Since problem (4.1) is equivalent to $\max _{X \in {\mathcal {S}}^{n,k}_+} \, \langle C, X \rangle$, we only need to show that $\langle C,Y\rangle <\langle C,X^*\rangle =\sum _{i}L_{ii}$, $\forall \ {\mathcal {S}}^{n,k}_+\ni Y \ne X^*$. Let $Z = \mathsf {sgn}(Y)$ and $P = \varPi _{{\mathbb {R}}_+^n}(L)$. We have

$$\begin{aligned} \langle C,Y \rangle = \mathrm {tr}(L(X^*)^{\top }Y) = \sum \nolimits _{i} \sum \nolimits _j L_{ji} \mathbf {y}_i^{\top } \mathbf {x}^*_j \le \sum \nolimits _{i} \sum \nolimits _j P_{ji}\mathbf {y}_i^T(\mathbf {x}^*_j\circ \mathbf {z}_i).\nonumber \\ \end{aligned}$$

(A. 1)

Define $w_{ji} = \Vert \mathbf {x}_j^* \circ \mathbf {z}_i\Vert ^2$. With $X^* \in {\mathcal {S}}^{n,k}_+$, we have $\Vert \sum _{j} P_{ji} (\mathbf {x}_j^* \circ \mathbf {z}_i)\Vert$ = $(\sum _{i} P_{ji}^2 w_{ji})^{1/2}$. Using the Cauchy-Schwarz inequality, $\Vert \mathbf {y}_i\Vert =1$ and the requirements on L, we have

$$\begin{aligned} \sum \nolimits _{j} P_{ji}\mathbf {y}_i^T(\mathbf {x}^*_j\circ \mathbf {z}_i) \le \Big (\sum \nolimits _{j} P^2_{ji}w_{ji}\Big )^{\frac{1}{2}} \le P_{ii} \Big (\sum \nolimits _{j} \frac{P_{jj}}{P_{ii}}w_{ji}\Big )^{\frac{1}{2}}. \end{aligned}$$

(A. 2)

With (A. 1) and $\langle C, X^*\rangle = \sum _{i}L_{ii} = \sum _{i} P_{ii}$, we further have

$$\begin{aligned} \langle C, Y \rangle \le \sum \nolimits _{i} P_{ii} \Big (\sum \nolimits _{j} \frac{P_{jj}}{P_{ii}}w_{ji}\Big )^{\frac{1}{2}} {\le } \Big (\sum \nolimits _{i} P_{ii} \Big )^{\frac{1}{2}} \Big ( \sum \nolimits _{i} \sum \nolimits _{j} P_{jj} w_{ji}\Big )^{\frac{1}{2}} {\le } \langle C, X^* \rangle ,\nonumber \\ \end{aligned}$$

(A. 3)

where the second inequality uses the fact that $\sum _{i} a_i x_i^{1/2} \le (\sum _{i} a_i)^{1/2}(\sum _{i} a_i x_i)^{1/2}$ for $a_i>0$ and $x_i\ge 0$, and the third inequality uses $\sum _{i} w_{ji} \le 1$. Obviously, the equalities in (A. 2) and (A. 3) hold if and only if $Y=X^*$. The proof is completed. $\square$