TITLE: $A$ is skew hermitian, prove $e^A$ is unitary
QUESTION [6 upvotes]: 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?

REPLY [4 votes]: 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, 
$$e^Ae^{A^{H}} = U^{-1}e^{J} U U^{-1}e^{-J} U = U^{-1}e^{J} e^{J^{H}}  = U^{-1}\pmatrix{e^{ia}e^{-ia} \\&e^{ib}e^{-ib}\\&&\ddots}U = U^{-1}U=I$$