Consider the following optimization problem $$\mathop{\arg\min}_{X\text{ is positive semi-definite}} \|X+A\|_F^2$$ where $A\in\mathbb R^{n\times n}$ is symmetric and $\|\cdot\|_F^2$ is the Frobenius norm.
I am wondering what's the solution $X$ of the above problem and how to derive the solution. The only thing I know so far is that it seems to be a convex optimization problem...
Thanks in advance.
Let $H=\frac12(A+A^T)$ and $K=\frac12(A-A^T)$. By orthogonally diagonalising $H$, we may assume that $$ H=\pmatrix{-D_1&0\\ 0&D_3}\ \text{ and }\ X=\pmatrix{X_1&X_2^T\\ X_2&X_3} $$ where $D_1$ is a nonnegative diagonal matrix and $D_3$ is a positive diagonal matrix. Since $X$ is positive semidefinite, so are $X_1$ and $X_3$. Therefore $X_3+D_3\succeq D_3\succeq0$ in positive semidefinite partial ordering, and $$ \|X_3+D_3\|_F^2=\sum_i\lambda_i(X_3+D_3)^2\ge\sum_i\lambda_i(D_3)^2=\|D_3\|_F^2. $$ It follows that \begin{aligned} \|X+A\|_F^2 &=\|X+H\|_F^2+\|K\|_F^2\\ &\ge\|X+H\|_F^2+\|K\|_F^2\\ &=\|X_1-D_1\|_F^2+\|X_3+D_3\|_F^2+\|D_1X_2^T\|_F^2+\|D_3X_2\|_F^2+\|K\|_F^2\\ &\ge\|D_3\|_F^2+\|K\|_F^2. \end{aligned} Clearly, $\|X+H\|_F^2=\|D_3\|_F^2$ when $X=\pmatrix{D_1&0\\ 0&0}$. Therefore this $X$ is the global minimiser. In terms of $H$, this means $$ X=\frac{(H^2)^{1/2}-H}{2}. $$ (Note: since $H$ is not necessarily positive semidefinite, $(H^2)^{1/2}$ in general is not equal to $H$.)