Given a matrix $F \in \mathbb{C}^{m \times n}$ such that $m > n$ and other (non-symmetric) square matrix $A$ of size $n \times n$, how can one formulate
$$ \arg \min_b \left\|A- {F}^{*} \operatorname{diag} \left( b \right) \, {F} \right\|_{2}$$
where $b \in \mathbb{C}^m$ is some vector and $*$ denotes the conjugate transpose, as a semidefinite program?
I started as follows. Writing the above problem in epigraph form by introducing a variable $x$,
\begin{array}{ll} \text{minimize} & x\\ \text{subject to} & \left\|A- {F}^{*} \operatorname{diag} \left( b \right) \, {F} \right\|_{2} \leq x\end{array}
which is equivalent to
\begin{array}{ll} \text{minimize} & x\\ \text{subject to} & \sigma_{\max}(A- {F}^{*} \operatorname{diag} \left( b \right) \, {F} ) \leq x\end{array}
which is equivalent to
\begin{array}{ll} \text{minimize} & x\\ \text{subject to} & \lambda_{\max}\big((A- {F}^{*} \operatorname{diag} \left( b \right) \, {F} )^*(A- {F}^{*} \operatorname{diag} \left( b \right) \, {F} ) \big) \leq x^2\end{array}
Can anybody tell me how I can proceed with this?
There are two ways to approach this. One is to recognize that $$\sigma_\max(X)\leq y \quad\Longleftrightarrow\quad \begin{bmatrix} yI & X \\ X^T & yI \end{bmatrix} \succeq 0$$ So the constraint becomes $$\begin{bmatrix} x I & A - F^T\mathop{\textrm{diag}}(b) F \\ A - F^T\mathop{\textrm{diag}}(b) F & x I \end{bmatrix} \succeq 0$$ Another way is to recognize that, for a symmetric matrix, $$\sigma_\max(X) = \max\{-\lambda_\min(X),\lambda_\max(X)\}$$ And with that, we could do $$ -x I \preceq A - F^T\mathop{\textrm{diag}}(b)F \preceq x I$$ The latter will be preferred because a pair of LMIs is more performant than one twice the size.