How do I solve $AB+BA=C$ for $A$?

Question

It might be a silly question... But

how do I solve $$A\,B + B\,A = C$$ for $A$???

Like... Is there a solution of the kind $A = f(C, B^{-1})$ or even $A = a\,C\,B^{-1} + b\,B^{-1}\,C$ for some $a$ and $b$?

Thanks!

Answer 1

Assume that for every eigenvalue of $B$, say $\lambda$, one has $Re(\lambda)>0$. Then the equation $AB+BA=C$ has a unique solution

$A=\int_0^{+\infty}e^{-tB}Ce^{-tB}dt$.

EDIT. More generally.

$\textbf{Proposition}$. Let $A,B,C$ be complex $n\times n$ matrices, $(\lambda_i)_i=spectrum(A),(\mu_i)_i=spectrum(B)$. Assume that , for every $i,j$, $Re(\lambda_i+\mu_j)>0$.

Then the equation $AX+XB=C$ has the unique solution

$X=\int_0^{+\infty}e^{-tA}Ce^{-tB}dt$.

$\textbf{Proof}$. We stack the matrices into vectors, row by row.

i) Notice that $f=A\otimes I+I\otimes B^T:X\mapsto AX+XB$ is one to one. Indeed, $0\notin spectrum(f)=(\lambda_i+\mu_j)_{i,j}$. Moreover, $-f$ is asymptotically stable (its eigenvalues have a negative real part).

ii) Let $g_t=e^{-tA}\otimes e^{-tB^T}:C\mapsto e^{-tA}Ce^{-tB}$; then $g_t=e^{-tf}$ and

$\int_0^{\infty} g_t(C)dt=\int_0^{\infty}g_tdt C=f^{-1}C=X$.