Let $A \in \mathbb{C}^{n \times m}$ and $B \in \mathbb{C}^{n \times m}$. Let${\Vert\cdot\Vert}_F$ be the Frobenius norm of a matrix. How can we solve the following optimization problem in $X \in \mathbb{C}^{m \times m}$?
$$\begin{equation}
\begin{aligned}\label{P}
&\min_{X\in \mathbb{C}^{m\times m}} \quad {\Vert AX-B \Vert}_F^2\\
&\begin{array}{r@{\quad}r@{}l@{\quad}l}
\text{s.t.} &{\Vert X \Vert}_F = 1
\end{array}
\end{aligned}
\end{equation}
$$
I'm totally a rookie with complex optimization problems. If this problem can't be solved, please give a detailed explanation. Any help would be appreciated.
Best Answer
A suitable Lagrange functional is $$ L(X,λ)=\frac12\|AX-B\|_F^2+\frac{λ}2(\|X\|_F^2-1) $$ Variation of $X$ in direction $H$ gives at a saddle point \begin{align} 0&={\rm trace}\Bigl((AX-B)^*(AH)\Bigr)+λ\,{\rm trace}(X^*H) \\ &={\rm trace}\Bigl((A^*AX-A^*B+λX)^*H\Bigr) \end{align} which implies $$ X=(A^*A+λI)^{-1}A^*B. $$ The function $\phi(λ)=\|(A^*A+λI)^{-1}A^*B\|_F^2$ tends toward zero for $λ\to\infty$ and has a pole at every singular value of $A$. Thus above the largest singular value is a solution for $\phi(λ)=1$.