I am studying the use of LMIs as an analysis tool for discrete-time dynamical systems. Consider the autonomous discrete-time system given by
$$
x_{_{k+1}} = A x_{_k} \tag{1} \label{sys}
$$
where $ x \in \mathbb{R}^n $ and $ A \in \mathbb{R}^{n \times n} $
In the textbook LMIs in Control Systems: Analysis, Design and Applications by Duan and Yu, the following condition is given for the stability of $\eqref{sys}$ using Lyapunov stability
$$
APA^{\text{T}} – P < 0 \tag{2} \label{lyap1}
$$
for some $ P = P^{\text{T}} < 0 $. In other words, if some permissible $P$ exists such that $\eqref{lyap1}$ is satisfied, then the system in $\eqref{sys}$ is stable.
I am having difficulty understanding where $\eqref{lyap1}$ comes from and here is why. First, we define the following Lyapunov function
$$
V_{_k} = x_{_k} P x_{_k} < 0
$$
Then, by Lyapunov stability theorem we need the following
$$
V_{_{k+1}} – V_{_{k}} < 0
\\
\implies x_{_{k+1}}^{\text{T}} P x_{_{k+1}} – x_{_{k}}^{\text{T}} P x_{_{k}} < 0
$$
Substituting $\eqref{sys}$ into the above inequality yields the following
$$
x_{_{k+1}}^{\text{T}} \left( A^{\text{T}} P A – P \right) x_{_{k}} < 0
$$
Which implies
$$
A^{\text{T}} P A – P < 0 \tag{3} \label{lyap2}
$$
The book refrenced previously uses $\eqref{lyap1}$ throughout the text. However, I have seen $\eqref{lyap2}$ in several other textbooks and I fully understand how it is derived. I suspect that $\eqref{lyap1}$ and $\eqref{lyap2}$ are equivalent; however I do not know how to show this.
My best guess is to note that if the matrix $ A $ is stable (i.e., eigenvalues reside within the unit circle), then so must the eigenvalues of $ A^{\text{T}} $ (since the eigenvalues of $A$ are the same as the eigenvalues of $A^{\text{T}}$). If we substitute $ A^{\text{T}} $ for $ A $ in $\eqref{lyap2}$ then we get $\eqref{lyap1}$; however, this reasoning seems weak. Is there a better way to show that $\eqref{lyap1}$ is equivalent to $\eqref{lyap2}$
Best Answer
Your argument is correct, but be aware that there is a potential pitfall:
If $A$ is Schur (all eigenvalues of $A$ are inside the open unit circle), then there exists $P>0$ and $Q>0$ such that
$$ A^TPA-P=-Q<0 \tag{1} $$
As you correctly stated, if $A$ is Schur, so is $A^T$, so there exist $\hat{P}>0$ and $\hat{Q}>0$ such that
$$ A\hat{P}A^T-\hat{P}=-\hat{Q}<0 \tag{2} $$
Note that by this argument, $(1)$ implies $(2)$, as you suspected, since both $-Q$ and $-\hat{Q}$ are negative definite. However, the equalities lead to different matrices in general. I.e. we have $$ \begin{align} P&\neq\hat{P}\\ Q&\neq\hat{Q} \end{align} $$ in general.
This pitfall is amplified by the fact that the Matlab function
dlyap, which is supposed to solve the discrete time Lyapunov equation, solves $(2)$ instead of $(1)$, so you have to pass $A^T$ as an argument to get the solution to $(1)$, which is usually what is desired.Edit: Btw, why do you think that "this reasoning seems weak"? I second the comment by KBS that your reasoning looks perfectly fine.