Solving the matrix equation $X-AXB=C$

Question

Let $A\in\mathbb{C}^{p\times p}, B\in\mathbb{C}^{q\times q}, C\in\mathbb{C}^{p\times q}$, if $\|A^n\|\cdot\|B^n\|<1$, then the matrix equation $X-AXB=C$ have a unique solution $X\in\mathbb{C}^{p\times q}$, where $\|\cdot\|$ represents the Frobenius norm.

I don't know how to prove the proposition. When $n=1$ situation, by letting $F(x)=AXB+C$ and applying Banach's contraction mapping theorem, I can finish the proof, but the method does not seem to work for $n\neq 1$.

Answer 1

Starting with $n=1$ is the correct approach. Since you don't summarize your solution in that case, I will start by providing one here for future readers.

Note that $X-AXB=C$ iff $AXB+C=X$; that is, the function $F_{A,B,C}(X)=AXB+C$ has a fixed point. So we are to show that $F$ is a contraction. Well, \begin{align*} \|(AXB+C)-(AYB+C)\|&=\|A(X-Y)B\|\\ &\leq\|A\|\cdot\|X-Y\|\cdot\|B\|\\ &\leq(1-\epsilon)\|X-Y\| \end{align*} since $\|A\|\cdot\|B\|<1$ is assumed. Thus $F_{A,B,C}$ is a contraction, and by Banach's theorem has a unique fixed point.

I claim that we can reduce the general case to the $n=1$ case. To see this, note that $$F_{A,B,C}^2(X)=A(AXB+C)B+C=A^2XB^2+ACB+C=F_{A^2,B^2,ACB+C}(X)$$ In general, $F_{A,B,C}^n=F_{A^n,B^n,P(A,B,C)}$, where $P(A,B,C)$ is some (noncommutative) polynomial in those three variables. By the above argument, then, $F_{A,B,C}^n$ has a unique fixed point; all that remains is to connect the fixed points of $F_{A,B,C}^n$ to those of $F_{A,B,C}$.

Any fixed point of $F_{A,B,C}$ is necessarily a fixed point of $F_{A,B,C}^n$, so there is at most one fixed point of $F_{A,B,C}$. Moreover, a fixed point of $F_{A,B,C}^n$ must necessarily have finite orbit under $F_{A,B,C}$. But every point of that orbit must also be a fixed point of $F_{A,B,C}^n$! So the unique fixed point of $F_{A,B,C}^n$ is the unique fixed point of $F_{A,B,C}$, and vice versa.

Edit: the OP has mentioned that they are not familiar with dynamical-systems based terminology I use in the last paragraph. Here's a more elementary phrasing (I drop the subscripts, since they no longer vary):

Let $x$ be the unique fixed point of $F^n$ and consider the set $\mathcal{O}=\{F^j(x):j\in\{0,1,\dots\}\}$. Since $F^n(x)=x$, we have, for each $j$, $$F^j(x)=F^j(F^n(x))=F^{j+n}(x)=F^n(F^j(x))$$ Thus (1) $\mathcal{O}$ is equal to the finite set $\{F^j(x):k\in\{0,1,\dots,n-1\}\}$ and (2) each $F^j(x)\in\mathcal{O}$ is a fixed point of $F^n$. Since $F^n$ has a unique fixed point, $\mathcal{O}$ contains exactly one element, whence the claim.