Is there a closed form solution for the following quadratic program with inequality constraint? Let $P \in \mathbb{S}^n_{++}$ be a symmetric positive definite matrix, and $B$, $F \in \mathbb{R}^{k \times n}$ two non-square matrices for $k \geq n$. $F \geq 0$ is non-negative. Also let $R \in V_k(F_n)$ be the stiefel matrix such that $R^T R = Id_k$. We consider the following quadratic program in $X$ and $Y$. It should be really easy to solve, but I am interested in the closed form which is easier for me to analyze.
$$\min_{X \in \mathbb{R}^{n^2}, Y\in \mathbb{R}^{k \times n}} ||P^{\frac{1}{4}}X||^2_F+tr(B^TX)$$
$$Y+F \geq 0$$ $$RY = X$$
The lagrangian is $$L(X,Y, \Gamma, U) = ||P^{\frac{1}{4}}X||^2_F+tr(B^TX) +tr(\Gamma^T(Y+F))+tr(U^T(RY-X)),$$
with the gradient
$$\nabla_{X, Y}L(X,Y, \Gamma, U) = \begin{pmatrix} 2P^{\frac{1}{2}}X + B -U \\ \Gamma + R^TU \\ \end{pmatrix}$$
Thus, set $X = -\frac{1}{2} P^{-\frac{1}{2}}V$, $V = B-U$, $Y = 0$, and use the fact that $\frac{1}{2}tr(V^TP^{-\frac{1}{2}}V) = \frac{1}{2}||P^{-\frac{1}{4}}V||^2_F$, we derive the dual problem
$$\max_{U, V \in \mathbb{R}^{n^2}, \Gamma \in \mathbb{R}^{k \times n}} -\frac{1}{4}||P^{-\frac{1}{4}}V||^2_F+tr(\Gamma^TF) $$
$$\Gamma \geq 0$$ $$U-B = V$$ $$\Gamma + R^TU = 0$$
I am wondering if this is possible to just look at the KKT condition for the primal and dual and derive a reasonable closed form solution. The KKT conditions are
- $Y+F \geq 0$
- $RY = X$
- $\Gamma \geq 0$
- $U-B = V$
- $\Gamma + R^TU = 0$
- $2P^{\frac{1}{2}}X + B -U = 0$
- $tr(\Gamma^T (Y+F)) = 0$