Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 61 / 2 / a

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- Solution a

Solution

From

A^{2} = I

, induction gives

A^{2 n} = I

and

A^{2 n + 1} = A

. The matrix exponential converges absolutely in finite dimension, so its even and odd terms may be separated:

e^{- i θ A} = [\sum_{n \geq 0} \frac{( - 1 ) ^{n} θ ^{2 n}}{( 2 n )!}] I - i [\sum_{n \geq 0} \frac{( - 1 ) ^{n} θ ^{2 n + 1}}{( 2 n + 1 )!}] A = cos θ I - i sin θ A .

This involution exponential formula requires no Hermiticity assumption on

A

. Hermiticity is needed if the exponential is additionally asserted to be unitary.

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