It's relatively easy to show that any eigenvector of an arbitrary $n \times n$ matrix $A$ is also an eigenvector of it's matrix exponential, $B = e^A$. But how does one show the reverse is true: that any eigenvector of $B = e^A$ is also an eigenvector of $A$?
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1This is not entirely true. For example you may have a matrix which has complex eigenvalues and whose exponential has real eigenvalues, and thus there is no way to relate the eigenvalues of the exponential with those of original matrix, even if you try to say, use a logarithm. – rubikscube09 Apr 03 '19 at 22:16
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A counterexample is $$ A=\left[\begin{array}{cc}0&-2\pi\\2\pi&0\end{array}\right],\hspace{8mm}e^A=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]. $$ Then for example $[1,0]^T$ is an eigenvector of $e^A$ but not of $A$.
Julian Rosen
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