Stability of discrete-time dynamical systems using Lyapunov stability

Question

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 $\eqref{lyap2}$

Answer 1

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.

Answer 2

Just an addition to the previous good answer and an extension of my comment. We have that $A$ is Schur stable if and only if there exists a symmetric positive definite matrix $P$ such that $$A^TPA-P\prec0$$ where the inequality symbol means "negative definite".

If we apply once the Schur complement, we get that those conditions are equivalent to

$$\begin{bmatrix} -P & A^T\\A & -P^{-1} \end{bmatrix}\prec0$$

and applying it once more yields

$$P^{-1}\succ0,\ AP^{-1}A^T-P^{-1}\prec0.$$

Since $P$ is positive definite, then so is $P^{-1}$ and this proves the equivalence between the LMI conditions.

We can also state the following result:

Let $P\succ0$ be given, then the following statements are equivalent:

• $V(x)=x^TPx$ is a Lyapunov function for the system $x_{k+1}=Ax_k$
• $W(y)=y^TP^{-1}y$ is a Lyapunov function for the system $y_{k+1}=A^Ty_k$.

Note that this is only true when the system is LTI, the story is a bit different in the LTV case.