Derivative of matrix exponential with two matrices

Question

Suppose $A$ and $B$ are non-commuting matrices and $t$ is a real parameter. Is the $t$-derivative of the matrix exponential $e^{A + t B}$ equal to the matrix $B \,e^{A+tB}$? Or is this not true due to Baker-Campbell-Hausdorff? Relatedly, is it true that the $t$-derivative at $t=0$ is given by $$ \frac{d^n}{dt^n} e^{A+tB} \mid_{t=0}\,\stackrel{?}{=} B^n e^A $$

Answer 1

The answer is no.

Here is a counter-example. Let $A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $B=\begin{pmatrix}0&1\\0&0\end{pmatrix}$. Then you can explicitly calculate $$ \left.\frac{d}{dt}e^{A+tB}\right|_{t=0}=\begin{pmatrix}0&e-1\\0&0\end{pmatrix}, $$ while $$ B\,e^A=\begin{pmatrix}0&1\\0&0\end{pmatrix}. $$