 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2015/12/vi/solution

With only

α_{3}

nonzero,

T = diag (α_{3}, - α_{3})

and

U = diag (e^{i α_{3} /2}, e^{- i α_{3} /2}) .

For

α_{3} \in / 2 π Z

, the eigenvectors are the nonzero multiples of $(1, 0)^{T}$ and $(0, 1)^{T}$ . When

α_{3} = 2 πk

U = (- 1)^{k} I

and every nonzero vector is an eigenvector.

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