Question

Prove Fact 5.1.8d: (V⊥)⊥ =V for any subspace V of R ⁿHint: Show that V⊆(V⊥) ⊥ by the definition of V ⊥

; then show that dim(V)=dim[(V ⊥)⊥
], by Fact 5.1.8c, i.e. that the dimensions of V and V ⊥sum to n.

Answer 1

To prove (V⊥)⊥ = V, one must first show that V is a subset of (V⊥)⊥, and then demonstrate that the dimensions of V and (V⊥)⊥ are equivalent. Both of these can be inferred from the definition of orthogonal complements and Fact 5.1.8c.

Step-by-step explanation:

Proof of Fact 5.1.8d

To prove that (V⊥)⊥ = V for any subspace V of ℝ¹, we need to show two things: (1) V is a subset of (V⊥)⊥ and (2) the dimensions of V and (V⊥)⊥ are equal, which we'll infer from Fact 5.1.8c.

Part 1: Showing V ⊆ (V⊥)⊥

By definition, V⊥ consists of all vectors in ℝ¹ that are orthogonal to every vector in V. Then (V⊥)⊥ consists of vectors that are orthogonal to the vectors in V⊥. Since every vector in V is orthogonal to every vector in V⊥, it must be included in (V⊥)⊥. Therefore, we can establish that V ⊆ (V⊥)⊥.

Part 2: Showing dim(V)=dim[(V⊥)⊥]

Based on Fact 5.1.8c, we know that the dimensions of V and V⊥ sum to n, the dimension of the entire space ℝ¹. This means that dim(V) + dim(V⊥) = n. Since (V⊥)⊥ is also a subspace of ℝ¹, we can similarly deduce that dim((V⊥)⊥) + dim(V⊥) = n. By transitivity, if dim(V) + dim(V⊥) = n and dim((V⊥)⊥) + dim(V⊥) = n, then dim(V) must be equal to dim((V⊥)⊥).