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It is a well-known result that for any positive-definite matrix $Q$, there exists a unique solution $P$ to the Lyapunov equation

$$ A^T P + PA = Q $$

if and only if all eigenvalues of $A$ have negative real parts.

A constructive proof suggests to choose

$$P:=\int \limits_{0}^{\infty} e^{A^Tt}Qe^{At} dt$$

But, in order for this to work, the integral needs to converge. In particular,

$$ \lim_{t \rightarrow \infty} e^{A t} = 0 $$

Can this result (in particular, convergence of the integral) be proven without resorting to a diagonal/Jordan normal form of $A$?

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    $\begingroup$ I don't understand what is the problem with $\lim_{t \rightarrow \infty} e^{t A } = 0$ for the convergence of the integral at $+\infty$ bound ... $\endgroup$ – Jean Marie Dec 25 '16 at 23:53
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    $\begingroup$ You may possibly appeal to exponential stability, i.e. for the stable $\dot x=Ax$ you have $|x(t)|\le e^{-\alpha t}|x(0)|$, but it is just another way to say that the fundamental matrix $e^{At}$ goes to zero exponentially fast. All proofs I have seen use the Jordan normal form of $A$. $\endgroup$ – A.Γ. Dec 26 '16 at 0:15
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    $\begingroup$ It is sufficient to use Schur upper triangularization (or upper-triangularization of any kind), but I'm not sure if that's preferable to Jordan form. $\endgroup$ – Ben Grossmann Dec 26 '16 at 0:36
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    $\begingroup$ @ValerySaharov who said anything about computing eigenvectors? Once you know enough about Jordan form, you can simply assert that when $A$ has eigenvalues with negative real part, $e^{At} \to 0$ regardless of what the eigenvectors are. Also, it's possible to compute the Jordan form of a matrix without ever actually finding its eigenvectors/generalized eigenvectors. $\endgroup$ – Ben Grossmann Dec 26 '16 at 15:41
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    $\begingroup$ @ValerySaharov the computability has nothing to do with proving what you want to prove, since there is no need to actually compute the Jordan form. The point is that we know enough about the matrix to deduce the Jordan form $J$ (whatever it happens to be) satisfies $e^{tJ} \to 0$. From there, we can deduce that $e^{tA} \to 0$. $\endgroup$ – Ben Grossmann Dec 26 '16 at 16:04
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We know that $A$ is asymptotically stable, which is to say that there exists a $\mu \in \Bbb R$ with $\operatorname{Re}\lambda_i \leq \mu < 0$ for all $i$. We know that $A$ can be brought to normal form (since every matrix can be brought to Jordan form), which is to say that there exists an invertible $S$ such that $A = SJS^{-1}$, where $J$ is in Jordan form.

Let $\|\cdot\|$ denote the spectral norm (any matrix norm works, though) It can be shown that there exists a constant $C$ such that for all $t$, $\|e^{tJ}\| \leq C e^{\mu t}$. Moreover, we have $e^{tA} = Se^{tJ}S^{-1}$, so that $$ \left\|e^{tA} \right\| \leq \|S\|\cdot \left\|e^{tJ}\right\| \cdot \|S^{-1}\| = \kappa(S) \|e^{tJ}\| $$ where $\kappa(S)$ is the condition number. All together, we have $\|e^{tA}\| \leq C \kappa(S) e^{\mu t}$ (which is enough to state that $e^{tA} \to 0$). Let $D_1 = C\kappa(S)$.

Similarly, there is a $D_2$ such that $\|e^{tA^T}\| \leq D_2 e^{\mu t}$ From there, we have $$ \left\| \int_0^\infty e^{A^Tt}Qe^{At} dt \right\| \leq \int_0^\infty \left\| e^{A^Tt}Qe^{At} \right\|dt\\ \leq \int_0^\infty \left\| e^{A^Tt}\right\| \cdot \|Q\| \cdot \left\|e^{At} \right\|dt \\ \leq D_1D_2 \|Q\| \int_0^\infty e^{2 \mu t}\,dt $$ which converges.

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    $\begingroup$ @ValerySaharov note that I never actually had to compute a Jordan form for $A$. Are you specifically looking for a constructive proof? One in which the constants (which suffice here since they are independent of $t$) depend only on known quantities? $\endgroup$ – Ben Grossmann Dec 26 '16 at 16:36
  • $\begingroup$ @ValerySaharov Now that I understand what you mean by "uncomputable steps", I don't understand why you're so keen to avoid them. I think your question would get more traction here if there were a motivating objective. For example: given a matrix $\tilde A$ and its Jordan normal form, I want to find a bound for how quickly the integral in terms of $A$ converges. $\endgroup$ – Ben Grossmann Dec 26 '16 at 16:45
  • $\begingroup$ @ValerySaharov I wouldn't, because doing so wouldn't provide a bound in terms of the information about $\tilde A$. I think you are right about perturbation theory; off the top of my head, Kato's text on perturbation theory and Bhatia's Matrix Analysis might have usable results. $\endgroup$ – Ben Grossmann Dec 26 '16 at 17:23
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Ben Grossmann Dec 26 '16 at 17:34
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A bit aside, but regarding the if and only if part it is not true. The Sylvester equation (see e.g. wiki) $AX + XB =C$ has a unique solution iff $A$ and $-B$ has no eigenvalues in common. In your case this reduces to the condition that for any two eigenvalues $\lambda_i,\lambda_j$ of $A$ we must require $\lambda_i+\lambda_j\neq 0$. If you have access to numerically inverting matrices it is in fact very simple to solve the Sylvester (or Lyapunov) equation. Just construct $X \mapsto f(X) = AX + XB$ as a linear map on an $n^2$ dimensional space and invert it.

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