Suppose U and V are subspace of R^n. Prove that orthogonal of ( U intersection V)=orthogonal of U+ orthogonal of V
for u,v∈U∩Vu,v\isin U\cap Vu,v∈U∩V :
if x∈(U∩V)⊥x\isin (U\cap V)^{\perp}x∈(U∩V)⊥ and U∩V≠0U\cap V \neq 0U∩V=0 then:
x⋅u=0x\cdot u=0x⋅u=0 or x⋅v=0x\cdot v=0x⋅v=0 , so
x∈U⊥+V⊥x\isin U^{\perp}+V^{\perp}x∈U⊥+V⊥
so,
(U∩V)⊥(U\cap V)^{\perp}(U∩V)⊥ is subset of U⊥+V⊥U^{\perp}+V^{\perp}U⊥+V⊥
if x∈U⊥+V⊥x\isin U^{\perp}+V^{\perp}x∈U⊥+V⊥ then:
x⋅u=0x\cdot u=0x⋅u=0 or x⋅v=0x\cdot v=0x⋅v=0 for u∈Uu\isin Uu∈U and v∈Vv\isin Vv∈V
then x∈(U∩V)⊥x\isin (U\cap V)^{\perp}x∈(U∩V)⊥
U⊥+V⊥U^{\perp}+V^{\perp}U⊥+V⊥ is subset of (U∩V)⊥(U\cap V)^{\perp}(U∩V)⊥
that is, (U∩V)⊥=U⊥+V⊥(U\cap V)^{\perp} =U^{\perp}+V^{\perp}(U∩V)⊥=U⊥+V⊥
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