 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2017/8/b/ii/solution

The normalized vectors

u_{1} = (1, 1)^{T} / 2

and

u_{2} = (- 1, 1)^{T} / 2

I_{0}

and

3 I_{0}

. Therefore

U = \frac{1}{2} (11 - 1 1), D = (I_{0} 0 0 3 I_{0}) .

The columns are orthonormal, so

U^{- 1} = U^{T} = U^{*}

and direct multiplication verifies

U^{- 1} M U = D

. Signs of individual eigenvectors are free; the displayed signs also make

det U = + 1

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