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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?

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When in doubt, differentiate! – Marek Jul 18 '11 at 17:08
@Marek Following your pithy maxim, I started differentiating, but I got a bit lost. Can you elaborate? Thanks! – Srivatsan Jul 18 '11 at 19:07
Differentiate $V(t) = \langle \exp{(tL)}v, \exp{(tL)}v \rangle$. You'll get $V' \equiv 0$ and since $V(0) = \|v\|^2$, you'll see that $\exp{(tL)}$ is unitary. – t.b. Jul 18 '11 at 19:13
@Srivatsan: You are correct: because you do not have 50 reputation points yet, you can only comment on your own questions and answers (once you gain 50 points, the "add comment" button will appear for you). I've converted your answer, and Theo's comment on that answer, to comments on the main question. – Zev Chonoles Jul 18 '11 at 19:22

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.

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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 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.

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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.

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