I know that
$$\frac{d}{dt}e^{At} = Ae^{At}$$
However, in one lecture, I found the following
$$\frac{d}{dt}e^{A^Tt} = e^{A^Tt}A^T$$
The lecture is as follows
How to show the second case? Why not $A^Te^{A^Tt}$?
I know that
$$\frac{d}{dt}e^{At} = Ae^{At}$$
However, in one lecture, I found the following
$$\frac{d}{dt}e^{A^Tt} = e^{A^Tt}A^T$$
The lecture is as follows
How to show the second case? Why not $A^Te^{A^Tt}$?
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Note, that for any matrix $C$, $C$ and $e^C$ commute. One can see this as follows: $$ Ce^C = C \cdot \sum_{k=0}^\infty \frac 1{k!} C^k = \sum_{k=0}^\infty \frac 1{k!} C^{k+1} = \sum_{k=0}^\infty \frac 1{k!} C^k \cdot C = e^C C $$ Hence, $$ A^T e^{A^T} = e^{A^T} A^T. $$