 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/6/c/solution

To avoid confusing the diagonalizing matrix with the exponential, write

A = W D W^{†}

from part (b). The convergent matrix exponential series gives

E = e^{A} = W e^{D} W^{†} = W diag (e^{λ_{1}}, \dots, e^{λ_{n}}) W^{†} .

Each

λ_{j}

is purely imaginary, so

∣ e^{λ_{j}} ∣ = 1

. The diagonal exponential is therefore a unitary matrix, and

E^{†} E = W diag (∣ e^{λ_{j}} ∣^{2}) W^{†} = I .

Hence the matrix called

U = exp (A)

in this part is unitary. The series identity

(e^{A})^{†} = e^{A^{†}} = e^{- A}

gives the same conclusion directly and also covers repeated eigenvalues.

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