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.
 A: With respect to uniqueness:
Consider the $vec(\cdot)$ operator that stacks the columns of a matrix into one long vector and define $\otimes$ for the Kronecker product of two matrices. If we apply the $vec(\cdot)$ operator to the Lyapunov equation we have
$$vec\left(S\right)-vec\left(A^TSA\right)=vec(Q)$$
Equivalently we have
$$[\mathbb{I}-(A^T\otimes A^T)]vec\left(S\right)=vec(Q)$$
and a unique solution exists iff the matrix $\mathbb{I}-(A^T\otimes A^T)$ is invertible. Matrix $A^T\otimes A^T$ has $n^2$ eigenvalues $\lambda_i\lambda_j$ ($1\leq i,j\leq n$) and therefore the eigenvalues of $\mathbb{I}-(A^T\otimes A^T)$ are $1-\lambda_i\lambda_j$. So a necessary and sufficient condition for uniqueness is 
$$\lambda_i\lambda_j\neq 1\qquad \forall i,j=1,\cdots,n$$
A sufficient condition for uniqueness is therefore the stability of $A$ ($|\lambda_i|<1$).
Also if $A$ stable then the above series $S_{\infty}$ converges and one can directly verify by substitution that this is a solution (unique) of the Lyapunov equation. 
A: 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.
