 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2011/p/p4/b/solution

Let

r = ∥ u ∥_{2} > 0

, and choose

s = 1

u_{1} \geq 0

and

s = - 1

otherwise. Set

w = u + sr e_{1} .

This vector is nonzero, because

w_{1}

has magnitude

∣ u_{1} ∣ + r

. Direct computation gives

w^{T} u = r (r + s u_{1}), w^{T} w = 2 r (r + s u_{1}),

H (w) u = u - 2 w \frac{w ^{T} u}{w ^{T} w} = u - w = - s ∥ u ∥_{2} e_{1} .

The first entry is nonzero and every other entry is zero. The sign choice also avoids cancellation when forming the first entry of

w

, which is useful in numerical computation.

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