Linear System $X'=AX-XA$

Question

Let $A \in M_n(\mathbb{R})$ and consider the matrix differential equation $$X'=AX-XA$$ with initial condition $X(0)=X_0 \in M_n(\mathbb{R})$.

Show that the initial value problem above has a single solution defined in $ \mathbb{R} $.

I think I need to write the above equation in the form $Y'=BY$, and then use the theorem of existence and oneness. But I can not do it. Any idea?

Thank you.

Answer 1

Define a basis $E_{11}$, $E_{12}$, ..., $E_{nn}$. Then, $$ X = \sum x_{ij} E_{ij} $$ Gather these $\{x_{ij}\}$ into a single, $n^2$-dimensional vector $\vec{x}$.

Now, you just need to prove that $AX-XA$ is linear in $X$. (Hint: consider $X \to aX + bY$, easy enough to show.) Thus, there exists a $n^2\times n^2$ matrix $B$ such that $AX-XA$, when expressed in our matrix basis, is equal to $B\vec{x}$ and subsequently $\vec{x}'=B\vec{x}$.

Answer 2

A fundamental result on the theory of ODE states that the solution of

$$ X’=f(X) $$

for a given initial condition, is unique provided $f$ is Lipshitz. All you have to do is check this requirement for $f(X)=[A,X]$.