Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 2 / 3 / b / ii

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- Solution ii

Solution

Now

B^{2} = ϵ^{2} I

, so splitting its matrix exponential into even and odd powers yields hyperbolic functions:

T = e^{B} = I cosh ϵ + \frac{sinh ϵ}{ϵ} B = (cosh ϵ sinh ϵ sinh ϵ cosh ϵ) .

The limiting coefficient at

ϵ = 0

is again one, and

T = I

. The eigenvectors

(1, 1)

and

(1, - 1)

have factors

e^{ϵ}

and

e^{- ϵ}

, respectively, showing how this hyperbolic transformation differs from a rotation.

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