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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 $A$? –  Brian M. Scott Dec 9 '12 at 21:39

3 Answers 3

Hint: $e^A$ is defined in terms of a power series, so what happens when we apply the power series to an eigenvector of $A$ with eigenvalue $0$?

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We have then that $A(v) = 0(v) = 0$ so $e^Av = Iv$ so $e^A$ has 1 as an eigenvalue. Thanks –  Poppy Dec 9 '12 at 21:47
    
Exactly. No problem! –  Tom Oldfield Dec 9 '12 at 21:50

This works for square matrices of any size.

If you write $A=SJS^{-1}$, with $S$ invertible and $J$ a Jordan matrix, then the eigenvalues of $A$ appear in the diagonal of $J$. Note that $e^{A}=Se^JS^{-1}$, so $e^J$ has the same eigenvalues as $A$.

As $J$ is triangular, its diagonal will be exactly the exponentials of the eigenvalues of $A$. In other words, if the eigenvalues of $A$ are $\lambda_1,\ldots,\lambda_k$, then the eigenvalues of $e^A$ are $e^{\lambda_1},\ldots,e^{\lambda_k}$.

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Hint: Take a look at the JNF of $A$. If both eigenvalues are equal to $0$, then $A$ is similar to $\left(\begin{array}&0&0\\0&0\end{array}\right)$ or $\left(\begin{array}&0&1\\0&0\end{array}\right)$

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