Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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

share|improve this answer

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.