= Exponential of a Hermitian matrix
{title2=$A=A^\dagger\implies e^A=(e^A)^\dagger>0$}
A <Hermitian matrix> has a unitary <spectral decomposition> $A=Q\operatorname{diag}(\lambda_j)Q^\dagger$ with real <eigenvalues> $\lambda_j$. The <matrix exponential respects similarity>, giving $e^A=Q\operatorname{diag}(e^{\lambda_j})Q^\dagger$. Its <eigenvalues> are positive and its <conjugate transpose> equals itself. In particular $v^\dagger e^Av=\sum_j e^{\lambda_j}|(Q^\dagger v)_j|^2>0$ for $v\ne0$, so it is a <Hermitian positive-definite matrix>.
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