# Lyapunov Equations

This article is going to summarize the results for the following Lyapunov equation, \begin{align} A^*X+XA+Q=0 \end{align} with given real matrices A and Q.

The first observable constraint here is that this equation only has a unique solution if A has no complex conjugate pairs.

Lemma 3.18 Assume that A is stable, then the following statements hold:
$X$ = $\int_0^\infty e^{A^*t}Qe^{At}dt$
$X>0$ if $Q>0$ and $X \geq 0$ if $Q \geq 0$
if $Q \geq 0$, then ($Q,A$) is observable iff $X > 0$.

This last result states that if we have a stable $A$ matrix and $Q$ is positive-semidefinite then our solution $X$ must be positive-definite.

We can use this result to construct two different Lyapunov equations. The first looks at the pair ($C,A$) for observability and we can say our system is observable if the solution X to the following equation is positive-definite, which would imply $C^*C$ is at least positive-semidefinite. $A^*X + XA + C^*C = 0;$

The second is the pair ($A,B$), which is analogous for controllability. The solution X to the following equation is positive-definite, implies $BB^*$ is at least positive-semidefinite. $A^*X + XA + BB^* = 0;$

For each respective case we call the solution X the, observability Gramian and controllability Gramian, respectively.

A Gramian of a set of vectors ${\displaystyle v_{1},\dots ,v_{n}} v_{1},\dots ,v_{n}$ in an inner product space is the Hermitian matrix of inner products, whose entries are given by ${\displaystyle G_{ij}=\langle v_{i},v_{j}\rangle } G_{ij}=\langle v_{i},v_{j}\rangle$.
This means the Gramian is $G=G^*$ ie. Symmetric with real values on the diagonal.

Now looking at the case of the controllability gramian, lets take the X and partition it as $% $
Where $X_c$ is nonsingular and represents the controllable part of the system.