Lyapunov equations are the bread and butter of control theory, stability theory and engineering. There are two forms of the Lyapunov equation, continuous and discrete. The discrete Lyapunov equation looks like this:
\[\mathbf{A}^T\mathbf{M}\mathbf{A} - \mathbf{M} = -\mathbf{Q}\]and the continuous version looks like this:
\[\mathbf{A}^T\mathbf{M} + \mathbf{M}\mathbf{A} = -\mathbf{Q}\]where \(\mathbf{A},\mathbf{M},\mathbf{Q} \in \mathcal{R}^{n \times n}\) , and \(\mathbf{M}\) and \(\mathbf{Q}\) are restricted to be symmetric positive definite. I don’t think I’ve ever seen a proof directly going from one Lyapunov equation to the other. I’m sure many proofs exist, I just haven’t seen one. So here I’ll show how to go between the two by looking at the discretization of a continuous linear dynamical system, computing the discrete time Lyapunov equation, and then letting the time-step size approach zero and recovering the continuous time equations.
We start with the continuous-time linear dynamics:
\[\dot{\mathbf{x}} = \mathbf{A}\mathbf{x}\]And then discretize it as follows:
\[\dot{\mathbf{x}} \approx \frac{\mathbf{x}_{t+1}-\mathbf{x}_{t}}{\delta}\]Where \(\delta > 0\) indicates a small forward displacement in time. Substituting the bottom equation into the top and shuffling terms around, we get a discrete-time equation for \(\mathbf{x}_{t+1}\).
\[\mathbf{x}_{t+1} = \mathbf{x}_t + \delta \mathbf{A} \mathbf{x}_t = (\mathbf{I} + \delta\mathbf{A})\mathbf{x}_t = \mathbf{B}\mathbf{x}_t\]Where we’ve defined \(\mathbf{B} \equiv \mathbf{I} + \delta\mathbf{A}\). Now we can use the discrete time Lyapunov equation for \(\mathbf{B}\) :
\[\mathbf{B}^T\mathbf{M}\mathbf{B} - \mathbf{M} = -\delta\mathbf{Q}\]Plugging in our definition for \(\mathbf{B}\), we get:
\[(\mathbf{I} + \delta \mathbf{A})^T\mathbf{M}(\mathbf{I} + \delta \mathbf{A}) - \mathbf{M} = -\delta \mathbf{Q}\]Expanding this expression out yields:
\[(\mathbf{M} + \delta \mathbf{A}^T\mathbf{M}) (\mathbf{I} + \delta \mathbf{A}) = \delta(\mathbf{A}^T\mathbf{M} + \mathbf{M}\mathbf{A}) + \delta^2 \mathbf{A}^T\mathbf{M}\mathbf{A} = -\delta \mathbf{Q}\]Recall that \(\delta\) is a small displacement in time. Letting \(\delta\) go to zero brings us closer and closer to having continuous dynamics–and in the limit we achieve them. It stands to reason that we should also recover the continuous-time Lyapunov equations in the limit as well. Dividing through by \(\delta\) on both sides, and then letting \(\delta \rightarrow 0\) we find that:
\[\mathbf{A}^T\mathbf{M} + \mathbf{M}\mathbf{A} = -\mathbf{Q}\]which of course is the continuous-time Lyapunov equation.
I’ve also added this simple proof to the Wiki page for Lyapunov equations.