Given I have:
$e^{tA} = e^t \begin{bmatrix} \cos(t)& -\sin(t) \\ \sin(t) & \cos(t) \end{bmatrix}$
How do I find A?
I believe to find A, I need to take the derivative of $e^{tA}$, but I don't see how that yields the original matrix A.
How to find A given e^{tA}?
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ordinary-differential-equations
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0What is the derivative of $e^{tA}$ with respect to $t$? Does this expose $A$ somehow? – 2017-02-28
1 Answers
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Use $ d(e^{tA}) / dt = A e^{tA}$. You can verify this by looking at the Taylor expansion for $e^{tA}$ (which is legitimate even for matrices!) or, even better, if $A$ is diagonalisable, go to an eigenvector basis where $A$ is diagonal.
Then $ A = d(e^{tA}) / dt |_{t = 0}$.
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0Ah, thanks! To take the derivative it's e^(tcos(t)), e^(-tsin(t))...etc? – 2017-03-02