= Solution
The <finite-dimensional spectral theorem> gives a <spectral decomposition> of the <Hermitian matrix>:
$$
A=QDQ^\dagger,\qquad D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n),\quad\lambda_j\in\mathbb R.
$$
The <matrix exponential respects similarity>, so
$$
e^A=Q\operatorname{diag}(e^{\lambda_1},\ldots,e^{\lambda_n})Q^\dagger.
$$
The diagonal entries are real, so the <conjugate transpose> of this expression equals itself. Its <eigenvalues> are $e^{\lambda_j}>0$, proving
$$
\boxed{(e^A)^\dagger=e^A,\qquad \operatorname{spec}(e^A)=\{e^{\lambda_j}\}_j\subset(0,\infty)}.
$$
In particular, for any nonzero $v$, $v^\dagger e^Av=\sum_j e^{\lambda_j}|(Q^\dagger v)_j|^2>0$. Thus the <exponential of a Hermitian matrix> is a <Hermitian positive-definite matrix>, even when $A$ itself has negative <eigenvalues>.
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