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So I'm not seeing the reason for this potentially very simple fact. Why does this have to be true? Given a $2 \times 2$ matrix $A$ that is traceless, in one possible case it may have $0$ as both eigenvalue. From that how does it follow that $e^A$ has $1$ as eigenvalue?

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    If $A$ is $2\times 2$ and has $0$ as a double eigenvalue, what is $A$?2012-12-09

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