 Solution

ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2009/p/p4/a/solution

P = v v^{T} / (v^{T} v)

. Since

v \neq = 0

, it is well defined, symmetric and idempotent:

P^{T} = P

P^{2} = P

. Therefore

H_{v, α} = I - α P

is symmetric, and

H_{v, α}^{T} H_{v, α} = I + (α^{2} - 2 α) P .

Because

P \neq = 0

, this equals

I

exactly when

α (α - 2) = 0

. Hence

H_{v, α} orthogonal ⟺ α \in {0, 2}

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