Semester 4 / Linear Algebra /Normal Space

Normal Space

2 min read Updated Fri Apr 24 2026 07:36:29 GMT+0000 (Coordinated Universal Time)
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Suppose WW is a subset of VV. W⊥W\perp is the set of all vectors that are orthogonal to all vectors of WW. Denoted by W⊥W^{\perp}

W⊥={ u∈V∣⟨u,w⟩=0  ∀w∈W }W\perp = \set { u \in V | \langle u,w \rangle = 0\; \forall w \in W }

If x∈W∩W⊥x \in W\cap W^\perp then x=0‾x = \underline{0}. W∩W⊥W\cap W^\perp can either be a null set or { 0‾ }\set{ \underline{0} }.

W⊥W^\perp is a subspace of VV for any WW.

(W⊥)⊥⊇W(W^\perp)^\perp \supseteq W. They are not equal though.

Theorem

Assume WW be a subspace of VV over FF having an orthonormal basis (wi)(w_i). Let:

  • u∈Vu \in V
  • P(u)=∑⟨wi,u⟩wi∈WP(u) = \sum \langle w_i, u \rangle w_i \in W

Then:

  • u−P(u)=Q(u)∈W⊥u - P(u) = Q(u) \in W^\perp
  • ∣∣u−P(u)∣∣≤∣∣u−w∣∣|| u - P(u)|| \le ||u - w|| for all w∈Ww \in W