Exercise 6.5.31

Answers

1.

Check that

H_{u} (x + cy) = x + cy - 2 ⟨ x + cy, u ⟩ u

= (x - 2 ⟨ x, u ⟩ u) + c (y - 2 ⟨ y, u ⟩ u)

= H_{u} (x) + c H_{u} (y) .

2.

Compute

H_{u} (x) - x = - 2 ⟨ x, u ⟩ u .

The value would be zero if and only if $x$ is orthogonal to $u$ since $u$ is not zero.

3.

Compute

H_{u} (u) = u - 2 ⟨ u, u ⟩ u = u - 2 u = - u .

4.

We check

H_{u}^{∗} = H_{u}

by computing

⟨ x, H_{u}^{∗} (y) ⟩ = ⟨ H_{u} (x), y ⟩ = ⟨ x - 2 ⟨ x, u ⟩ u ⟩

= ⟨ x, y ⟩ - 2 ⟨ x, u ⟩ \cdot ⟨ y, u ⟩

and

⟨ x, H_{u} (y) ⟩ = ⟨ x, y - 2 ⟨ y, u ⟩ u ⟩

= ⟨ x, y ⟩ - 2 ⟨ x, u ⟩ \cdot ⟨ y, u ⟩ .

Also, compute

H_{u}^{2} (x) = H_{u} (x - 2 ⟨ x, u ⟩ u)

H_{u} (x) - 2 ⟨ x, u ⟩ H_{u} (u)

(x - 2 ⟨ x, u ⟩ u) + 2 ⟨ x, u ⟩ u = x .

Combining $H_{u} = H_{u}^{∗}$ and $H_{u}^{2} = I$ , we have $H_{u} H_{u}^{∗} = H_{u}^{∗} H_{u} = I$ and so $H_{u}$ is unitary.

jephianlin

2011-06-27 00:00