 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2016-resit/10/b/solution

e^{B} = diag (e^{3}, e^{4}, 1) .

For the two-dimensional matrix

C

, direct multiplication gives

C^{2} = - C

, hence

C^{n} = (- 1)^{n - 1} C

for

n \geq 1

. Summing its defining power series gives

e^{C} = I + (1 - e^{- 1}) C = (1 1 - e^{- 1} 0 e^{- 1}) .

Finally,

e^{A} = V diag (e, e^{2}, e^{5}) V^{T}

, so each diagonal entry is a weighted sum of these exponential eigenvalues, with weights equal to squared eigenvector components:

(e^{A})_{11} = \frac{e ^{2}}{3} + \frac{2 e ^{5}}{3}, (e^{A})_{22} = (e^{A})_{33} = \frac{e}{2} + \frac{e ^{2}}{3} + \frac{e ^{5}}{6} .

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