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?
If a $2 \times 2$ matrix $A$ has $0$ as an eigenvalue why does $e^A$ have $1$ as an eigenvalue?
2
$\begingroup$
matrices
-
3If $A$ is $2\times 2$ and has $0$ as a double eigenvalue, what is $A$? – 2012-12-09