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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:

  1. Why is $S_\infty >0$ unique?
  2. Which condition causes $S_\infty >0$ unique?

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

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    $\begingroup$ I would suspect the uniqueness has more to do with $A$ being stable. $\endgroup$ – copper.hat May 19 '15 at 1:27
  • $\begingroup$ I adapt my question. $\endgroup$ – sleeve chen May 19 '15 at 1:45
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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.

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  • $\begingroup$ Easier to understand. $\endgroup$ – sleeve chen May 19 '15 at 6:28
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    $\begingroup$ @sleevechen Thank you. Also I believe that the observability of $(A,\sqrt{Q})$ is needed to ensure the positive definiteness of $S_{\infty}$. $\endgroup$ – RTJ May 19 '15 at 6:43
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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.

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  • $\begingroup$ How to claim "The eigenvalues of $L$ are $\lambda_1 \lambda_j -1$"? I know that $A^TSA-S = A^T(U^TDU)A - U^TDU = (UA)^TD(UA) - U^TDU $ So you suppose $1$ is one eigenvalue of $S$? $\endgroup$ – sleeve chen May 19 '15 at 5:11
  • $\begingroup$ It is straightforward if $A$ has a full set of eigenvectors. Use denseness to complete the proof. There is an algebraic proof as well, but I prefer the eigenvector approach. $\endgroup$ – copper.hat May 19 '15 at 5:13
  • $\begingroup$ So you assume $1$ is an eigenvalue of $S$ for convenience? In general, it will become $\lambda_1 b\lambda_j -b$ with $b >0$ and we can obtain the same conclusion. true? $\endgroup$ – sleeve chen May 19 '15 at 5:22
  • $\begingroup$ No. $S$ is a dummy variable used to define $L$. All eigenvalues of the operator $S \mapsto -S$ (or $\Xi \mapsto -\Xi$) are $-1$. $\endgroup$ – copper.hat May 19 '15 at 5:23
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    $\begingroup$ @sleevechen: Yes, it is exactly like $Ax=b$. You could take a basis $E_{ij}$ and write it as an operator $\mathbb{C}^{n \times n} \to \mathbb{C}^{n \times n}$ to see this. (In fact, you can use this to show that the eigenvalues are $\lambda_i \lambda_j -1$.) $\endgroup$ – copper.hat May 19 '15 at 15:51

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