When does the equation $\dfrac{d}{dx}e^{A(x)}=A'(x)e^{A(x)}$ holds, where $A(x)$ is a matrix.

Question

When does the following equation holds true? $$\dfrac{d}{dx}e^{A(x)}=A'(x)e^{A(x)}.$$

In the latter:

$A(x)=\begin{pmatrix}a_1(x)&a_2(x)\\a_3(x)&a_4(x)\end{pmatrix}$,

$A'(x)=\begin{pmatrix}a_1'(x)&a_2'(x)\\a_3'(x)&a_4'(x)\end{pmatrix}$,

$e^{A(x)}=\displaystyle\sum\limits_{n=0}^\infty \dfrac{1}{n!}A^n(x)$.

I tried to solve this supossing that $A(x)=P(x)D(x)P^{-1}(x)$ where $D(x)$ is a diagonal matrix, thus $A^n(x)=P(x)D^n(x)P^{-1}(x)$, but I got stock in a part where I needed the product ${P^{-1}}'(x)P(x)$.

Answer 1

It almost never holds. The actual condition is that $[A(t), A(t')]=0$ where $t\neq{t'}$. The general expression is given by $$\frac{d}{dt}e^{A(t)}=e^{A(t)}\frac{1-e^{-ad_{A}}}{ad_{A}}\frac{dA(t)}{dt}=e^{A(t)}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(k+1)!}(ad_{A})^{k}\frac{dA(t)}{dt}$$ Where $ad_{A}F=[A, F]$

Answer 2

You don't need any diagonalization tricks here. Clearly it's true for the first term, but what about the second? We have $$ \frac{d}{dx}A^2 = \frac{d}{dx} \sum_{j=1}^2A_{ij}A_{jk} = \sum_{j=1}^2 A'_{ij}A_{jk} + A_{ij} A'_{jk}= A'A + AA' \ne 2A'A$$

Similarly, $$ \frac{d}{dx}A^3 = A^2A' + AA'A + A'A^2 \ne 3A'A^2$$

It should be clear from this that your formula doesn't work in general, but an easy sufficient condition for it to work is that $AA' =A'A.$