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The following is a theorem:

If $A$ is a self-adjoint matrix (i.e. $A^\dagger = A$), then $U = e^{iA}$ is a unitary matrix.

This is easy to prove: $(e^{iA})^\dagger = e^{-iA^\dagger} = e^{-iA}$, with the last step a consequence of self-adjointness of $A$. Since powers of matrices commute with themselves, $e^{iA}$ commutes with $e^{-iA}$, and so we can write $e^{iA}e^{-iA}=e^{iA-iA}=e^{0}=I$, where $I$ is the identity matrix. This shows that $(e^{iA})^{-1}=(e^{iA})^\dagger$, and hence that $e^{iA}$ is unitary.

This is all well and good. However, the following (the converse) is not a theorem:

If $U$ is a unitary matrix, then the (a?) matrix $A$ defined by $e^{iA} = U$ is self-adjoint.

I have been led to believe that the best way to disprove this is by counterexample, but I have no idea how to construct a non-self-adjoint matrix $A$ for which $e^{iA}$ is still unitary.

An explicit counterexample would be great, but an approach to figuring out how to find a counterexample would be even beter.

senshin
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  • @DanielFischer In the 1-by-1 case (scalars), $e^{i(2\pi i)}=e^{-2\pi}$ isn't unitary (inverse is $e^{2\pi}$; conj. transpose is $e^{-2\pi}$, and those aren't equal). Am I misunderstanding you? – senshin Sep 26 '13 at 19:25
  • Oops, thought of $e^A$, sorry. – Daniel Fischer Sep 26 '13 at 19:28
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    $\large U^{\dagger}U = 1 \quad\Longrightarrow\quad {\rm e}^{-{\rm i}A^{\dagger}}{\rm e}^{{\rm i}A} = 1\quad$. In general, $\large{\rm e}^{-{\rm i}A^{\dagger}}{\rm e}^{{\rm i}A} \not= {\rm e}^{{\rm i}\left(-A^{\dagger} + A\right)}$ unless $\large\left[A, A^{\dagger}\right] = 0$. – Felix Marin Sep 26 '13 at 22:37
  • See Baker Campbell Hausdorff formula in wikipedia. – Felix Marin Sep 26 '13 at 22:45

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$U$ is unitary, hence diagonalisable. So, $A$ must be diagonalisable too, because matrix exponentials of nontrivial Jordan blocks are not diagonalisable. Let $A=PDP^{-1}$ be an eigendecomposition. It follows from $U=e^{iA}$ that diagonal entries of $D$ must be real. Hence $e^{iD}$ is a unit diagonal matrix.

But then $e^{iA}=Pe^{iD}P^{-1}$ is in general not unitary. To make $e^{iA}$ unitary without using a unitary $P$, one good way is to make $e^{iD}=I$, so that $U=e^{iA}=Pe^{iD}P^{-1}=I$ for all choices of $P$. Yet $D$ cannot be a scalar multiple of $I$, otherwise $A$ would become a scalar multiple of $I$, which is Hermitian. Therefore, $D$ should be chosen so that its diagonal entries are different integer multiples of $2\pi$. For such choices, practically almost all randomly chosen $P$ would make $A=PDP^{-1}$ become non-Hermitian. For instance, $$ A=\pmatrix{1&1\\ 0&1}\pmatrix{2\pi&0\\ 0&0}\pmatrix{1&1\\ 0&1}^{-1} =\pmatrix{2\pi&-2\pi\\ 0&0}\ \Rightarrow\ e^{iA}=I. $$

user1551
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