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Given $A$ is a skew-hermitian, ( i.e $A^H = -A$ ) then how do you prove the matrix exponential $e^A$ is unitary. To prove the unitary property of the matrix, I need to show $(e^A)^{*}(e^A) = (e^A)(e^A)^{*}= I$. Can any one help me how to proceed and prove the result?

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up vote 1 down vote accepted

Since $A$ is skew-Hermitian, $A$ is diagonalizable and all eigenvalues of $A$ are pure imaginary.

For matrices $A,U,J$ such that $A=U^{-1}JU$ $$e^A = U^{-1}e^{J} U \quad , \quad e^{A^H}= U^{-1}e^{J^{H}} U $$ where $J$ is diagonal and $e^J$ can be computed element by element.

Then, $$e^Ae^{A^{H}} = U^{-1}e^{J} U U^{-1}e^{-J} U = U^{-1}e^{J} e^{J^{H}} = U^{-1}\pmatrix{e^{ia}e^{-ia} \\&e^{ib}e^{-ib}\\&&\ddots}U = U^{-1}U=I$$

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Look at this: matrix exponential. $$(e^A)^He^A=e^{A^H}e^A=e^{-A}e^A=e^0=I$$

A longer answer is you expand the matrix exponential. And for a real skew-symmetric matrix $A$, $e^A$ is not only an orthogonal matrix but also a rotation matrix because $\det e^A=1$. In addition, for every vector $a$, there is an associating skew-symmetric matrix $A$. Denote $A=[a]_{\times}$. Then $e^A$ can be expressed as

$$e^A=I+\frac{[a]_{\times}}{\|a\|}\sin \|a\|+\left(\frac{[a]_{\times}}{\|a\|}\right)^2(1-\cos\|a\|)$$ which actually is the Rodrigues' rotation formula.

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