$\partial{y} = \text{Trace}(A)y$

Question

Let $Y$ be the fundamental solution matrix for an $nxn$-system $\partial{y} = Ay$. Also, with $\det(Y) \neq 0 $. Prove that $\det(Y)$ satisfies $\partial{y} = \text{Trace}(A)y$. Also show that the columns of $Y^{-1}$ satisfy the system of equations $\partial{y} = -A^ty$, where $A^t$ denotes the transpose of $A$.

I have minimal knowledge of differential equations, I am reading F. Beuker's notes in differential equations because I need them for my research in Number theory. Could someone please guide me how to solve the above problem? It is an exercise in Beuker's notes.

Thanks for your help in advance!

Answer 1

The first question asks you to prove Liouville's formula in the case of a constant matrix $A$. For the general case, there are plenty of proofs over the internet, e.g. on wikipedia.

For the constant case, a direct approach could be the following:

1. Show that $Y(t)=e^{(t-t_0)A}$ (check that $y(t)=e^{(t-t_0)A}y_0$ is the unique solution of the initial value problem when $y(t_0)=y_0$)
2. Show that for a square matrix $M$, $\det{(e^{M})}=e^{Tr(M)}$
3. Continue the calculation of $\frac{d}{dt}det (e^{(t-t_0)A}) = ...$ then conclude the proof.

The second question asks you about the "dual" of the system, or its "costate equation". To me, but I can be mistaken, there is a typo in Beukers' notes, and the exercise should be to prove that the rows of $Y^{-1}$ satisfy the given system. Knowing $Y(t)=e^{(t-t_0)A}$, you just have to leverage the fact that for a square matrix $M$, $(e^M)^{T}=e^{(M^{T})}$ and $(e^M)^{-1}=e^{-M}$.