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!

  • 4
    $\begingroup$ 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. $\endgroup$
    – user1551
    Jul 15, 2022 at 14:21
  • $\begingroup$ An overkill would be to use (holomorphic) functional calculus, and the fact that $e^x\in\mathbb R$ whenever $x\in\mathbb R$. $\endgroup$ Jul 15, 2022 at 14:50

1 Answer 1


What you did looks good. I would just add the following element

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

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    $\begingroup$ I assumed that the word "Unitarian" was an auto-complete problem, changed it to "unitary". If you actually intended Unitarian, maybe as a joke(???), feel free to change it back... $\endgroup$ Jul 15, 2022 at 15:06
  • $\begingroup$ @DavidC.Ullrich... :) After all, the accepted phrase was/is "Weyl's unitarian trick"... :) $\endgroup$ Jul 15, 2022 at 16:40
  • $\begingroup$ @paulgarrett Fair enough - there's plenty of precedent for the word "unitarian" here. I didn't know that, thx. (But I still say "Unitarian" is wrong...) $\endgroup$ Jul 15, 2022 at 20:23
  • $\begingroup$ @DavidC.Ullrich... :) Yes, at least by this year, it doesn't make sense. :) $\endgroup$ Jul 15, 2022 at 20:51
  • $\begingroup$ @DavidC.Ullrich Yes it was auto-complete on iPad. Which by the way seems more religious than mathematician’s… $\endgroup$ Jul 16, 2022 at 7:04

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