For any matrix $Q$, show that $$\det(e^{Q})=e^{\operatorname{tr}Q}$$
where tr represents the trace and det is the determinant
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2Reduce $Q$ to Jordan form. – PAD Oct 07 '12 at 12:44
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3As a startinmg point: If $Q$ is a diagonal matrix then an eigenvector of $Q$ with eigenvalue $\lambda$ is eigenvektor of $e^Q$ with eigenvalue $e^\lambda$. Therefore the product of eigenvalues of $e^Q$ is the exponential of the sum of eigenvalues of $Q$. – Hagen von Eitzen Oct 07 '12 at 12:44
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WLOG Assume that $Q$ is upper triangular with eigenvalues $\lambda_i$'s.Then $Q^k$ is also upper triangular with eigenvalues $\lambda_i^k$. Thus,$e^Q=\sum \frac{Q^k}{k!}$ is also upper triangular with diagonal entries as $e^{\lambda_i}$.
Now just take determinants and we are done.
TheJoker
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