Transposing in quadratic forms (example: Lyapunov equation) I'm wondering why the Lyapunov equation is derived as follows:
Consider in the System
$$\dot{x}=Ax$$
The following Lyapunov Function is chosen
$$V(x)=x^TPx$$
where
$$\dot{V}(x)=\dot{x}^TPx+x^TP\dot{x}$$
follows. Using the first equation delivers
$$\dot{V}(x)=x^T(A^TP+PA)x=-x^TQx$$
Because I can transpose a quadratic form, I could also write the following
$$\dot{V}(x)=2\dot{x}^TPx=2x^TA^TPx$$
Which would mean that $2A^TP = -Q$ but I suppose that's nonsense. But what am I doing wrong?
 A: First of it can be noted that $2\,A^\top P = -Q$ is only correct if $P$ is symmetric. For non-symmetric matrices it would become $A^\top (P + P^\top) = -Q$.
Your approach is valid, however it is more convenient to work with symmetric matrices. Namely:


*

*matrices are positive definite if their symmetric parts has only positive eigenvalues (this requires less steps if you start with a symmetric matrix). 

*For large dimensions symmetric matrices require about only half the memory to store them compared to a non-symmetric matrix. 

*For a given $A$ and $Q$ the Laypunov equation $A^\top P + P\,A=-Q$ guarantees to have a positive definite solution for $P$ if $A$ is Hurwitz and $Q$ is positive definite and there a lot of implementations which can solve this. I am not sure if this would also hold for your equation as well.


But if you have found a positive definite $P$ and $Q$ such that $A^\top (P + P^\top) = -Q$ then that would be sufficient to show exponential stability of $\dot{x}=A\,x$.
