Orthogonal Projection onto Half Space of Matrices

Question

Given a matrix $\boldsymbol{Y} \in \mathbb{R}^{m \times n}$ how can one solve the orthogonal projection:

$$\begin{align} \arg \min_{\boldsymbol{X}} \quad & \frac{1}{2} {\left\| \boldsymbol{X} - \boldsymbol{Y} \right\|}_{F}^{2} \\ \text{subject to} \quad & \begin{aligned} \boldsymbol{X} \boldsymbol{a} & \leq \boldsymbol{b} \\ \end{aligned} \end{align}$$

Where $\boldsymbol{a} \in \mathbb{R}^{n}$ is a given vector.

This is similar to Orthogonal Projection onto a Half Space yet the argument is a matrix.

Answer 1

The way to solve this is using the Vectorization Operator.
Using the operator:

$$ \boldsymbol{X} \boldsymbol{a} \leq \boldsymbol{b} \iff \operatorname{Vec} \left( \boldsymbol{X} \boldsymbol{a} \right) \leq \operatorname{Vec} \left( \boldsymbol{b} \right) = \boldsymbol{b} $$

By the identity of the operator:

$$ \operatorname{Vec} \left( \boldsymbol{X} \boldsymbol{a} \right) = \left( \boldsymbol{a}^{T} \otimes \boldsymbol{I}_{m} \right) \operatorname{Vec} \left( \boldsymbol{X} \right) $$

Where $\otimes$ is the Kronecker Product and $\boldsymbol{I}_{m}$ is the $m \times m$ identity matrix.

By setting $\boldsymbol{A} = \boldsymbol{a}^{T} \otimes \boldsymbol{I}_{m} $ and $\boldsymbol{x} = \operatorname{Vec} \left( \boldsymbol{X} \right)$ the problem becomes:

$$ \begin{align} \arg \min_{\boldsymbol{x}} \quad & \frac{1}{2} {\left\| \boldsymbol{x} - \boldsymbol{y} \right\|}_{2}^{2} \\ \text{subject to} \quad & \begin{aligned} \boldsymbol{A} \boldsymbol{x} & \leq \boldsymbol{b} \\ \end{aligned} \end{align} $$

Which is equivalent to Orthogonal Projection onto a Polyhedron (Matrix Inequality).