Linear matrix equation involving $\sum_i A_i X B_i$

linear algebramatrix equations

I am dealing with a non-standard linear matrix equation:
\begin{equation}
X \quad+\quad \sum_{i=1}^r \Big( (A_i X B_i) + (A_i X B_i)^T + (B_i X A_i) + (B_i X A_i)^T \Big) \quad=\quad C
\end{equation}
The symbols in capital letters, i.e. $X, A_i, B_i, C$, denote $n \times n$ real matrices. $X$ is the unknown that we wish to solve for. Additional considerations:

  1. The number of terms $r$ is large.
  2. $C$ is symmetric
  3. $A_i$'s are all rank one matrices (not sure if this information is helpful for expressing $X$ though).

Is there an elegant way to express the solution form $X$?

Thanks for your help!

Golabi

Best Answer

Each summand in the summation sign is symmetrised and $C$ is symmetric. Hence $X$ must be symmetric and the equation is equivalent to $$ X+\sum_i\left(A_iXB_i+B_i^TXA_i^T+B_iXA_i+A_i^TXB_i^T\right)=C, $$ which can be rewritten as (see Wikipedia) $$ \left[I+\sum_i\left(B_i^T\otimes A_i+A_i\otimes B_i^T+A_i^T\otimes B_i+B_i\otimes A_i^T\right)\right]\operatorname{vec}(X)=\operatorname{vec}(C). $$ Call the matrix inside the pair of square brackets $M$. The equation $M\operatorname{vec}(X)=\operatorname{vec}(C)$ is solvable if and only if $MM^+\operatorname{vec}(C)=\operatorname{vec}(C)$, where $M^+$ denotes the Moore-Penrose pseudoinverse of $M$. In case it is solvable, $\operatorname{vec}(X)=M^+\operatorname{vec}(C)$ is always a solution.

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