# Let, $A \in \mathbb{C}^{n}$. If $A$ is hermitian, then so is $e^{A}$.

Since $$A$$ is hermitian we have that $$A$$ is unitary diagonalizable; as in, there exists matrices $$S,D \in \mathbb{C}^{n \times n}$$ such that: $$A = SDS^{*},$$ where $$S$$ is unitary and $$D$$ is a diagonal matrix consisting of the eigenvalues of $$A$$. We have that, $$e^{A} = Se^{D}S^{*} = S\textrm{diag}(e^{d_1},\dots,e^{d_n})S^{*}.$$ Thus, $$(e^{A})^* = e^{A},$$ since the eigenvalues of a hermitian matrix are real, we have that: $$\overline{e^{d_i}} = e^{d_i}.$$ Hence, the conclusion follows. Any issues in my proof? Thank you!

• Your proof looks correct to me, but I think a simpler proof is to show that $(e^A)^\ast=e^{A^\ast}$ using the power series expansion of the exponential function. Jul 15, 2022 at 14:21
• An overkill would be to use (holomorphic) functional calculus, and the fact that $e^x\in\mathbb R$ whenever $x\in\mathbb R$. Jul 15, 2022 at 14:50

$$\left(e^A\right)^*=\left(S e^D S^*\right)^*=S\left(e^D\right)^*S^*$$ Because $$S$$ is supposed to be unitary.