Exercise 6.2.13

Answers

1.

x \in S^{⊥}

then we have that

x

is orthogonal to all elements of

S

, so are all elements of

S_{0}

. Hence we have

x \in S_{0}^{⊥}

2.

x \in S

, we have that

x

is orthogonal to all elements of

S^{⊥}

. This means

x

is also an element in

{(S^{⊥})}^{⊥}

. And

span (S) \subset {(S^{⊥})}^{⊥}

is because $span (S)$ is the smallest subspace containing $S$ and every orthogonal complement is a subspace.

3.

By the previous argument, we already have that

W \subset {(W^{⊥})}^{⊥}

. For the converse, if

x \notin W

, we may find

y \in W^{⊥}

and

⟨ x, y ⟩ \neq 0

. This means that

W \supset {(W^{⊥})}^{⊥}

4.

By Theorem 6.6, we know that

W = W + W^{⊥}

. And if

x \in W \cap W^{⊥}

, we have

⟨ x, x ⟩ = ∥ x ∥^{2} = 0 .

Combine these two and get the desired conclusion.

jephianlin

2011-06-27 00:00