Differentiation involving matrix exponentials

Question

I am having trouble understanding differentiation involving matrix exponentials. Suppose we want to compute the derivative of $e^{-At}\mathbf{f}(t)$, where $A$ is a $2\times 2$ real matrix and $$ \mathbf{f}(t)=\begin{pmatrix} f_1(t) \\ f_2(t) \end{pmatrix}. $$ Using the product rule, I (naively) thought that $$ \frac{d}{dt}\left(e^{-At}\mathbf{f}(t)\right)=e^{-At}\frac{d\mathbf{f}(t)}{dt}-\mathbf{f}(t)Ae^{-At}, $$ but I noticed that $\mathbf{f}(t)Ae^{-At}$ does not make much sense as $\mathbf{f}(t)\in\mathbb{R}^{2\times 1}$ while $A\in\mathbb{R}^{2\times 2}$ and so multiplication is not well-defined. What am I missing?

Answer 1

You have put $\mathbf{f}(t)$ on the wrong side, that's all. It is $$\frac{d}{dt}\left(e^{-At}\mathbf{f}(t)\right) =e^{-At}\frac{d\mathbf{f}(t)}{dt} - Ae^{-At}\mathbf{f}(t) $$

Answer 2

Let $X$ and $Y$ be Banach spaces and $A:\mathbb R\to\mathcal{L}(X;Y)$ and $B:\mathbb R\to X$ be continuously differentiable functions. Here, $\mathcal L(X;Y)$ is the Banach space of bounded linear functions from $X$ to $Y$. So let's define $$F(t):=A(t)B(t)=A(t)\circ B(t)=A(t,B(t)).$$ Now, we can use the chain rule $$F'(t)=\partial_t A(t,B(t))+\partial_BA(t,B(t))B'(t).$$ Note: We have $\partial_tA(t,B(t))=A'(t)B(t)$ and since $A(t)$ is a bounded linear map in $B$ for every $t$, its derivative with respect to $B$ is just the linear map itself: $$\partial_BA(t,B(t))=A(t).$$ Put together we get $$F'(t)=A'(t)B(t)+A(t)B'(t).$$

So, this does really look like the product rule, although it is not. Note also, how you cannot switch the order of $A$ and $B$, so this is consistent with Reinhard Meier's answer.