TITLE: Derivative of matrix exponential of linear combination
QUESTION [6 upvotes]: Let's say $t$ is a real parameter and $\textbf{A}$ is $n \times n$ matrix. I know that
$$\frac{d}{dt} \exp(t\textbf{A}) = \textbf{A} \exp(t\textbf{A})$$
But what if there are multiple parameters $t_1, ..., t_n$ and multiple matrices $\textbf{A}_1, \dots , \textbf{A}_n$ that don't commute, does the same relation, namely
$$\frac{\partial}{\partial t_k}\exp \left(\sum_{i=1}^n t_i \textbf{A}_i \right) = \textbf{A}_k \exp \left(\sum_{i=1}^n t_i \textbf{A}_i\right)$$
still hold? If not, is there a closed form for the derivative?

REPLY [2 votes]: Let 
$$\eqalign{
t&=t_i,\quad A=A_i,\quad B=\sum_{k\ne i} t_kA_k \\
G(t) &= \frac{d}{dt}\exp(B+tA) \\
}$$
To evaluate $G(t)$ at $t=0$, use the block-triangular method.
$$\eqalign{
\exp\left(\left[\matrix{B&A\\0&B}\right]\right)
 &= \left[\matrix{\exp(B)&G(0)\\0&\exp(B)}\right] \\
}$$
To evaluate $G$ at $t=s,\,$ shift the definition of $B\to (B+sA)$
Define block-analogs of the standard basis vectors
$$e_1=\pmatrix{1\\0},\quad e_2=\pmatrix{0\\1}$$
by replacing the {${0,1}$} elements with the $n\times n\,$ {zero, identity} matrices
$$E_1=\pmatrix{I\\0},\quad E_2=\pmatrix{0\\I} \in{\mathbb R}^{2n\times n}$$
Then would this qualify as a closed-form solution?
$$\eqalign{
G(s) &= E_1^T\,\exp\left(\left[\matrix{B+sA&A\\0&B+sA}\right]\right)\,E_2 \\
}$$