Rewriting a quadratic Matrix equation as a quadratic vector equation

Question

Consider the set of $N \times N$ matrices $\{W_i\}_{i=1}^{i=L}$, set of $N \times 1$ vectors $\{g_i\}_{i=1}^{i=L}$ and $\{h_i\}_{i=1}^{i=L}$. Now consider the following sum \begin{align} S=\sum_{i}\sum_{j}g_i^HW_ih_ih_j^{H}W_j^{H}g_j \end{align} where the summation runs through all $L$ for all $i,j$. Clearly, this equation is quadratic in the matrix variables $\{W_i\}_{i=1}^{i=L}$. Now define the column vector \begin{align} w=\begin{bmatrix} \operatorname{vec}(W_1) \\ \operatorname{vec}(W_2) \\ \vdots \\ \operatorname{vec}(W_L) \end{bmatrix} \end{align} where for a matrix $A$, $\operatorname{vec}(A)$ denotes the column vector containing the columns of $A$ starting from the first column. The question is, can we write \begin{align} S=w^{H}Qw \end{align} where $Q$ is a matrix which is a function of $\{g_i\}_{i=1}^{i=L}$ and $\{h_i\}_{i=1}^{i=L}$. If so, what is the structure of $Q$?

Answer 1

\begin{align} S&=\sum_{i}\sum_{j}g_i^HW_ih_ih_j^HW_j^Hg_j\\ &=\sum_{i}\sum_{j}\operatorname{trace}(g_jg_i^HW_ih_ih_j^HW_j^H)\\ &=\sum_{i}\sum_{j}\operatorname{vec}(W_j)^H \operatorname{vec}(g_jg_i^HW_ih_ih_j^H)\\ &\qquad\qquad\qquad\{\because \operatorname{trace}(AB^{H})=\operatorname{trace}(A^{H}B)=\operatorname{vec}(A)^{H}\operatorname{vec}(B)\}\\ & \\ &=\sum_{i}\sum_{j}\operatorname{vec}(W_j)^H \left((h_ih_j^H)^T\otimes(g_jg_i^H)\right)\operatorname{vec}(W_i)\\ &\qquad\qquad\qquad\{\because \operatorname{vec}(ABC)=(C^{T}\otimes A)\operatorname{vec}(B)\}\\ & \\ &=\sum_{i}\sum_{j}\operatorname{vec}(W_j)^H \left((\bar{h}_jh_i^T)\otimes(g_jg_i^H)\right)\operatorname{vec}(W_i)\\ &=w^H Qw, \end{align} where $Q$ is a block matrix whose $(j,i)$-th (note: not $(i,j)$-th) subblock is $(\bar{h}_jh_i^T)\otimes(g_jg_i^H)$.