why's the matrix exponential of an anti self-adjoint matrix unitary? Let L be a matrix, (.,.) a unitary scalar product, t a real number.
Apparently, if L is anti self-adjoint (i.e. (Lv,w) = -(v,Lw)) then exp(tL) is unitary.
Could anyone please tell me why? Is this always true?
 A: Anti self-adjoint for a matrix is the same as Skew-Hermitian, which means that $\overline{L^t}=-L$, and since $e^L e^{\overline{L^t}} = e^L e^{-L}=I$ you get the result you asked for. See http://en.wikipedia.org/wiki/Skew-hermitian  which also has an interpretation in terms of lie groups and algebras (Skew-Hermitian matrices form the lie algebra of U(n))
Edit : What I wrote before was confusing since $e^Ae^B = e^{A+B}$ is only true for matrices if $A$ and $B$ commute.
A: You can show, by continuity of $$\mathrm{End}(V)\rightarrow \mathrm{End}(V),~L\mapsto L^*$$ and the series expansion that defines $\mathrm{exp}$ for endomorphisms, that $\mathrm{exp}(L^*)=(\mathrm{exp}(L))^*$. When you apply this to $L$ anti self adjoint and $t\in \mathbb{R}$, then you  get $$(\mathrm{exp}(tL))^*=\mathrm{exp}(tL^*)=\mathrm{exp}(-tL)=\mathrm{exp}(tL)^{-1}$$ which exactly means that $\mathrm{exp}(tL)$ is unitary.
A: If $L$ is anti-self-adjoint then you can unitarily diagonalize it, so essentially everything reduces to what happens for diagonal matrices (which is really the same as what happens for complex numbers).  It's easy to check that all the eigenvalues $\lambda_i$ of $L$ must be purely imaginary.  The eigenvalues of $e^{tL}$ are of the form $e^{t \lambda_i}$; since $\lambda_i$ is imaginary, $e_{t \lambda_i}$ has magnitude 1, i.e. it lies on the unit circle.  This means that $e^{tL}$ is unitary.
This is more of a mnemonic than a proof, so you can fill in the details.  Since it relies on diagonalization it is less elementary than the other suggestions, but I think it is more memorable since you just have to remember how complex arithmetic works.
