Given $A$ a matrix with spectral radius smaller than 1 and a matrix $C$ symmetric.
It can be shown that $U=\sum\_{k=0}^\infty (A^T)^k C A^k$ converges, is symmetric and is the solution of the equation above.


Is it possible to show that if $C$ is non-negative also $U$ is non-negative?



non negative solution of the matrix equation $A^T U A = U - C$ if C is non-negative
Here is an alternative approach to solving the original equation.


The desired equation subject to the constraint $U \ge 0$ (elementwise nonnegativity) can be written as


\begin{equation\*}
 U = C + A^TUA,\qquad U \ge 0.
\end{equation\*}


Define now the iteration:


\begin{equation\*}
 U\_{k+1} = [C + A^TU\_kA]\_+,\qquad k=0,1,\ldots
\end{equation\*}
where $[\cdot]\_+ \equiv \max\lbrace 0, \cdot\rbrace$ denotes **projection** onto the nonnegative orthant. Assuming the slightly stronger condition that the operator norm $\|A\| < 1$, we can show that the above iteration converges to the desired solution.


$\newcommand{\Gc}{\mathcal{G}}$
**Claim.** Let $\Gc$ denote the map $U\mapsto [C + A^TUA]\_+$. Then, $\Gc$ is a strict contraction in the spectral norm.



> 
> *Proof.* Let $U, V$ be two matrices. Then,
>  \begin{eqnarray\*}
> \|\Gc(U)-\Gc(V)\| &=& \|[C + A^TUA]\_+ - [C + A^TVA]\_+\|\\
> &\le& \|C + A^TUA - C - A^TVA]\|\\
> &\le& \|A^T(U-V)A\|\\
> &\le& \|A^T\| \|A\| \|U-V\| \\
> &\le& \gamma\|U-V\|,\qquad \text{where}\ \gamma=\|A\|^2 < 1.
> \end{eqnarray\*}
>  The first inequality follows as projection is a *non-expansive* operator, the rest follow from obvious properties of the spectral norm.
> 
> 
> 


Thus, $\Gc$ is a strict contraction. So by the Banach contraction theorem the iteration $U\_{k+1}=\Gc(U\_k)$ has a unique solution. This solution is nonnegative by construction and satisfies the desired equation.


