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?
why's the matrix exponential of an anti self-adjoint matrix unitary?
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linear-algebra
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1When in doubt, differentiate! – 2011-07-18
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0@Marek Following your pithy maxim, I started differentiating, but I got a bit lost. Can you elaborate? Thanks! – 2011-07-18
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1Differentiate $V(t) = \langle \exp{(tL)}v, \exp{(tL)}v \rangle$. You'll get $V' \equiv 0$ and since $V(0) = \|v\|^2$, you'll see that $\exp{(tL)}$ is unitary. – 2011-07-18
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0@Srivatsan: You are correct: [because you do not have 50 reputation points yet, you can only comment on your own questions and answers](http://meta.stackexchange.com/questions/19756/how-do-comments-work/19757#19757) (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. – 2011-07-18