Does a closed form solution exist for $\beta$?

Question

I am having trouble trying to solve this equation for an unknown matrix $\beta$.$$K = \sum_{i=1}^n \tau_i \sum_{j=1}^p (A_j + A_j^T)\beta(C_j + C_j^T)$$ where $\tau_i$ is a scalar constant, and $A_j$, $C_j$, K are matrices of constants.

Answer 1

I don't believe that this can be solved using only typical matrix operations (e.g. additions multiplications and transposes). However, it can be solved using vectorization.

Notation: Let $\operatorname{vec}(A)$ denote the (column-major) vectorization of $A$. Let $\otimes$ denote the Kronecker product. Denote $B_j = A_j + A_j^T$, $D_j = C_j + C_j^T$; note that these matrices are symmetric.

We can vectorize both sides of the equation and simplify as follows. $$ K = \sum_i \tau_i \sum_j B_j \beta D_j\\ \operatorname{vec}(K) = \operatorname{vec}\left[\sum_i \tau_i \sum_j B_j \beta D_j \right]\\ \operatorname{vec}(K) = \sum_i \tau_i \sum_j \operatorname{vec}(B_j \beta D_j)\\ \operatorname{vec}(K) = \sum_i \tau_i \sum_j (D_j \otimes B_j)\operatorname{vec}(\beta). $$ We can solve this for $\operatorname{vec}(\beta)$ to find that $$ \operatorname{vec}(\beta) = \left[\sum_i \tau_i \sum_j (D_j \otimes B_j)\right]^{-1}\operatorname{vec}(K) = \frac{\left[\sum_j (D_j \otimes B_j)\right]^{-1}\operatorname{vec}(K)}{\sum_i \tau_i }. $$ Obtain $\beta$ by "unvectorizing" $\operatorname{vec}(\beta)$.