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Given $A$ a matrix with spectral radius smaller than 1 and a symmetric matrix $C$. 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?

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1 Answer 1

For every $x$, $x^*Ux=\sum\limits_{k\geqslant0}x_k^*Cx_k$, where $x_k=A^kx $ for every $k\geqslant0$. If $C$ is nonnegative, then $x_k^*Cx_k\geqslant0$ for each $k\geqslant0$ hence $x^*Ux\geqslant0$. Thus, $U$ is nonnegative.

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Thanks Did. I agree that if $C$ is semidefinite positive, then also $U$ is (this is what you proved). I was asking if we can claim that if $C$ is entrywise non-negative then $U$ is entrywise non-negative. Now I doubt this is true but I haven't found a counter-example. –  Giovanni Aug 28 '12 at 8:59

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