Exponential and Q matrice question For any matrix $Q$, show that $$\det(e^{Q})=e^{\operatorname{tr}Q}$$ where tr represents the trace and det is the determinant

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.