 Solution

The finite-dimensional spectral theorem gives a spectral decomposition of the Hermitian matrix:The matrix exponential respects similarity, soThe diagonal entries are real, so the conjugate transpose of this expression equals itself. Its eigenvalues are $e^{{\lambda }_j}>0$, provingIn particular, for any nonzero $v$, $v^{†}{e}^{A}v={\sum }_j{e}^{{\lambda }_j}|(Q^{†}v{)}_j{|}^2>0$. Thus the exponential of a Hermitian matrix is a Hermitian positive-definite matrix, even when $A$ itself has negative eigenvalues.