Matrix exponentials and their derivative

Question

Is the derivative of a exponential matrix $e^{At}$, with all elements as a function of $t$, be equal to the derivative of each element separately wrt $t$, assembled into the matrix?

Sorry I’m new to Linear algebra, please help me out.

Answer 1

The answer is yes. Here is an example that you might find helpful:

Suppose we are given the matrix $$ A = \pmatrix{-1&4\\1&-1}, $$ which has eigenvalues $\lambda = -1 \pm 2i$. We find that $$ e^{tA} = e^{-t}\left(\begin{array}{cc} \cos\left(2\,t\right) & 2\,\sin\left(2\,t\right)\\ -\frac{1}{2}\sin\left(2\,t\right) & \cos\left(2\,t\right) \end{array}\right). $$ Using the formula $\frac {d}{dt} e^{tA} = Ae^{tA}$, we should find that $$ \frac{d}{dt} e^{tA} = e^{-t}\pmatrix{-1&4\\1&-1}\left(\begin{array}{cc} \cos\left(2\,t\right) & 2\,\sin\left(2\,t\right)\\ -\frac{1}{2}\sin\left(2\,t\right) & \cos\left(2\,t\right) \end{array}\right) \\= e^{-t}\left(\begin{array}{cc} -\cos\left(2\,t\right) -2 \sin(2\,t)& 4\cos(2s)-2\sin\left(2\,t\right)\\ -\cos(2\,t)+\frac{1}{2}\sin\left(2\,t\right) & -\cos\left(2\,t\right) - 2 \sin(2\,t)\end{array}\right) . $$ We can check that the result for the derivative of the upper-left entry matches: by the product rule, $$ \frac d{dt}\left[e^{-t}\cos\left(2\,t\right)\right] = (-e^{-t}) \cos(2t) + e^{-t}(-2\sin(2t)) = e^{-t}(-\cos(2t) - 2\sin(2t)). $$