$H$ is the space of all real $N \times N$ matrices. Let $X$ denote the $H$ such that the Frobenius distance is at most $\alpha$ from $R$. Find the closest matrix $A \in X$, for any given $B \in H$.
My attempt:
$A \in X$, then $||A-R|| \leq \alpha$
$$\min_{A \in X}||A-B||_F \forall B \in H $$ So, in terms of the Frobenius norm, I can write:
$$\min_{a_{i,j}}\sum_{i,j}(a_{i,j}-b_{i,j})^2 \hspace{3mm}\text{such that} \sum(a_{i,j}-r_{i,j})^2 \leq \alpha^2 $$
I do not know how to proceed with this. Can I get a closed form solution for this?