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A Hermitian matrix has a unitary spectral decomposition $A=Q\mathrm{diag}({\lambda }_j)Q^{†}$ with real eigenvalues ${\lambda }_j$. The matrix exponential respects similarity, giving $e^A=Q\mathrm{diag}(e^{{\lambda }_j})Q^{†}$. Its eigenvalues are positive and its conjugate transpose equals itself. In particular $v^{†}{e}^{A}v={\sum }_j{e}^{{\lambda }_j}|(Q^{†}v{)}_j{|}^2>0$ for $v\ne 0$, so it is a Hermitian positive-definite matrix.

If $A$ is Skew-Hermitian, then $-iA$ is a Hermitian matrix. The finite-dimensional spectral theorem gives $A=Q\mathrm{diag}(i{\theta }_j)Q^{†}$, with real ${\theta }_j$ and a unitary matrix $Q$. Since the matrix exponential respects similarity, $e^A=Q\mathrm{diag}(e^{i{\theta }_j})Q^{†}$. Each diagonal entry has modulus one, so $e^A$ is a unitary matrix. Thus a skew-Hermitian matrix generates norm-preserving linear evolution.

Induction shows $(PA{P}^{-1}{)}^k=P{A}^k{P}^{-1}$ for every nonnegative integer $k$: in a product, adjacent factors $P^{-1}P$ cancel. At $k=0$, this is $I=PI{P}^{-1}$. Hence each partial sum of the matrix exponential satisfiesThe series is absolutely convergent, since a submultiplicative matrix norm bounds it by ${\sum }_k\Vert A{\Vert }^k/k!=e^{\Vert A\Vert }$. Passing to the limit therefore proves that the matrix exponential respects similarity:No diagonalizability assumption is required.

If $A^{†}=-A$, then $H=-iA$ is a Hermitian matrix, because $H^{†}=i{A}^{†}=-iA=H$. By the finite-dimensional spectral theorem, there is a unitary matrix $Q$ withThus $A=Q\mathrm{diag}(i{\theta }_j)Q^{†}$. Since the matrix exponential respects similarity,Each diagonal entry has modulus one, so $C^{†}C=I$. Using $Q^{†}Q=Q{Q}^{†}=I$,Hence the exponential of a skew-Hermitian matrix is unitary.

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.