Minimizing $(1-\lambda)||X-HA||_F^2+\lambda||Y-HB||_F^2$ subject to $H \succeq 0$ I have two matrices, $X\in R^{m \times n}, Y \in R^{m \times p}$ and I also have another two matrices, $A \in R^{r \times n}, B \in R^{r \times p}$. Now I want to a matrix $H \in R^{m \times r}$ to satisfy the following objective function. Here H is a common matrix.
The objective function is list as follows:
$f=arg\min_{H}(1-\lambda)||X-HA||_F^2+\lambda||Y-HB||_F^2$, s.t. $H\succeq0$
where X, Y, A, B, and ($0 \leq \lambda \le1$) are known, what is the best way to get H? Can you tell me this kinds of objective function belongs to which problem? matrix factorization? but it is two matrices X and Y. For the constrain of H, how to deal with this? I understand that $\partial f/\partial H = 0$ can deal with this, but I still do not know how to handle H constrain.
Is this H matrix called correlation matrix between X and Y? If X and Y are similarity matrices, can I use the objective function to get H as the correlation matrix to indicate the correlation between X and Y?
Can you help me to get the solution of H?
Any help is appreciated!
 A: Given $\lambda \in [0,1]$, we have a convex quadratic program (QP) in $\mathrm X \in \mathbb R^{m \times r}$
$$\begin{array}{ll} \text{minimize} & (1-\lambda) \, \| \mathrm X \mathrm A_1 - \mathrm B_1 \|_{\rm F}^2 + \lambda \, \| \mathrm X \mathrm A_2 - \mathrm B_2 \|_{\rm F}^2\\ \text{subject to} & \mathrm X \geq \mathrm O_{m \times r}\end{array}$$
Vectorizing, we obtain
$$\| \mathrm X \mathrm A_i - \mathrm B_i \|_{\rm F}^2 = \| \mbox{vec} (\mathrm X \mathrm A_i - \mathrm B_i) \|_2^2 = \| \left(\mathrm A_i^{\top} \otimes \mathrm I_m\right) \mbox{vec} (\mathrm X) - \mbox{vec} (\mathrm B_i) \|_2^2$$
Thus, we can write the QP in a more standard form
$$\boxed{\begin{array}{ll} \text{minimize} & (1-\lambda) \, \| \left(\mathrm A_1^{\top} \otimes \mathrm I_m\right) \mbox{vec} (\mathrm X) - \mbox{vec} (\mathrm B_1) \|_2^2 + \lambda \, \| \left(\mathrm A_2^{\top} \otimes \mathrm I_m\right) \mbox{vec} (\mathrm X) - \mbox{vec} (\mathrm B_2) \|_2^2\\ \text{subject to} & \mbox{vec} (\mathrm X) \geq 0_{mr}\end{array}}$$
