Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
2  
In your infinite sum, all terms are nonnegative. Isn't this enough? –  Federico Poloni Aug 21 '12 at 10:23
    
Federico Poloni is absolutely right. –  Bazin Aug 21 '12 at 16:10
    
Cleary if all the terms are nonnegative I am done, but the matrix $A$ could have negative entries, then how can I be sure that each term is nonnegative? –  Giovanni Aug 22 '12 at 11:52
1  
Wait, what do you mean by "nonnegative"? Entrywise or positive semidefinite? –  Federico Poloni Aug 25 '12 at 16:06
    
Entrywise. Now I doubt that the result holds, but I have not found a counter-examplet. I agree that if C is semidefinite positive, then U is semidefinite positive. –  Giovanni Aug 28 '12 at 8:57
add comment

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.