# minimize frobenius norm

My question is the following:

Suppose $M$ is an $n \times n$ symmetric real matrix. I want to find an $n \times n$ symmetric real matrix X such that $|| X -M||_F$ is minimized with the constraint that $U^T X U \succeq 0$ (positive semidefinite), where $U$ is an $n \times d$ matrix, $U^TU = I_d$ and $n > d$. $|| \cdot ||_F$ is the frobenius norm of matrix. In short: $$X^* = \displaystyle \text{argmin}_{X:U^TXU\succeq 0}||X-M ||_F, X, M \in \mathcal{S_n}$$

Some useful information may be: $U$ can be obtained from eigenvalue decomposition of $BB^T$ or QR decomposition of $B$ in the sense that there exists $Q = [U,V]$ such that $Q^TQ = QQ^T = I_n, U^TB = 0$.

I would like to find a formula or construction based on eigenvalue decomposition or singular value decomposition for M. For example, if the constraint is $X \succeq 0$ instead of $U^T X U \succeq 0$, we can first apply eigenvalue decomposition to $M$ such that $M = Q \text{diag}(\lambda_1,\cdots, \lambda_n) Q^T$, then $X^* =Q \text{diag}(\max\{\lambda_1,0\},\cdots, \max\{\lambda_n,0\}) Q^T$ would minimize $|| X-M ||_F$.

I used CVX package to compute $X^*$ for some examples and I found some interesting facts: $(X^*-M) B = 0$ where $U^TB=0$ and $X^* -M$ is always positive semidefinite while $M$ and $X^*$ may not be positive semidefinite. Any help or suggestion of what method or tools I should use for these kind of problems would be much appreciated! Thanks~

Consider $Q$ as you defined it. Let $Y = Q^TXQ$ and $N = Q^TMQ$. Let $$E = \pmatrix{I_d\\0}$$ We can rewrite your optimization problem as $X^* = Q^TY^*Q$, where $$Y^* = \arg\min_{Y:E^TYE \geq 0} \| Y - N\|_F$$ Note, however, that $E^TYE$ is simply the upper-left $d \times d$ principle submatrix of $Y$. Denote this submatrix as $\tilde Y$. Let $\tilde N$ denote the corresponding submatrix of $N$.
It therefore suffices to set all other entries of $Y$ to those of $N$ and setting $\tilde Y$ to solve $$\tilde Y = \arg\min_{Z \geq 0} \|Z - \tilde N\|_F$$ Now, find an orthogonal $W$ such that $W^T\tilde N W$ is diagonal. Make the substitution $\tilde A = W^T \tilde Y W$. We then find $$\tilde A = \arg\min_{A \geq 0} \|A - D\|_F$$ This has an obvious and unique solution: take $\tilde A$ to be diagonal matrix. Set $a_{ii} = \max\{d_{ii},0\}$.
With that, we have enough information to get the unique solution $X$.