Why algebraic Lyapunov equation has an unique solution? In the following text book (p.47): Optimal Control (Lewis 2nd edition) 
There is a theorem: (Zero input case)

If $A$ is stable, and $(A,\sqrt Q )$ is observable, then $S_\infty= \sum_{i=0}^\infty(A^T)^iQA^i >0$ is the unique solution to the algebraic Lyapunov equation: $S = A^TSA+Q$.My question is:  



*

*Why is $S_\infty >0$ unique?  

*Which condition causes $S_\infty >0$ unique? 


I try to come up with an example, but I can not find one to explain this.

I think this answers both questions:
Let $L(S) = A^T S A - S$. The eigenvalues of $L$ are $\lambda_1 \lambda_j -1$,
where the $ \lambda_k$ are eigenvalues of $A$. Since $A$ is stable, we
have $|\lambda_k| <1$, hence $\lambda_1 \lambda_j -1 \neq 0$, and so $L$
is invertible. Hence $L(S) = -Q$ has a unique solution.
