 Solution

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

T v = t v

with

v \neq = 0

, induction gives

T^{k} v = t^{k} v

for every nonnegative integer

k

. Substituting into the matrix exponential series gives

Uv = \sum_{k = 0}^{\infty} \frac{( i /2 ) ^{k} T ^{k} v}{k !} = e^{i t /2} v .

Thus each eigenvector of $T$ is an eigenvector of $U$ , with eigenvalue

e^{i t /2}

. Different eigenvalues of

T

can produce the same exponential eigenvalue.

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