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Given $A$ and $B$, are there some $C$ and $D$ such that $AXB = CX + XD$ is true for all $X$. ($A, B, C, D$ and $X$ are matrices)

Américo Tavares
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Jose
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1 Answers1

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The answer is "no" in general. The reason is obvious if you know what "tensor" is, but we can also solve the problem in an elementary way. If $AXB=CX+XD$ for all $X$, then in particular it is true for every matrix $E_{ij}$ in the standard basis of the matrix space (i.e. $E_{ij}$ is the matrix with a $1$ at the $(i,j)$-th entry and zero elsewhere). By comparing coefficients, it can be shown that either $A$ or $B$ must be a scalar multiple of the identity matrix $I$. So, if both $A$ and $B$ are not multiples of $I$, the required $C$ and $D$ do not exist.

Without proving the previous assertion, you may also construct a counterexample as follows. On one hand, if $AXB=CX+XD$ for all $X$, then for $X=I$, we have $AB=C+D$. On the other hand, $AXB=CX+XD$ implies that $\operatorname{trace}(AXB)=\operatorname{trace}(CX+XD)$. Using the identity $\operatorname{trace}(MN)\equiv\operatorname{trace}(NM)$, the above simplifies to $\operatorname{trace}(X(BA-C-D))=0$ for all $X$. Therefore $BA-C-D=0$, or $BA=C+D$. In other words, for $AXB=CX+XD$ to hold for all $X$, a necessary condition is $AB=BA$. So, every pair of non-commuting matrices $A$ and $B$ will produce a counterexample to the existence of $C$ and $D$.

user1551
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