Let U , W ⊆ V, vector subspaces then:1) U ⊆ W ⟹ W ⊥ ⊆ U ⊥ 2) ( U + W ) ⊥ = U ⊥ ∩ W ⊥

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Let   U  ,  W  ⊆  V, vector subspaces then:1)   U  ⊆  W    ⟹        W          ⊥        ⊆      U          ⊥      2)   (  U  +  W      )          ⊥        =      U          ⊥        ∩      W          ⊥

Julianne Mccoy · Accepted Answer

1)Let   v  ∈      W    ⊥  , which means that   (  ∀  w  ∈  W  )  :  ⟨  v  ,  w  ⟩  =  0. You want to prove that   v  ∈      U    ⊥  . Take   u  ∈  U. But then   u  ∈  W and therefore   ⟨  v  ,  u  ⟩  =  0. So,   v  ∈      U    ⊥  2)Since   U  ⊂  U  +  W and   V  ⊂  U  +  W,   (  U  +  W      )    ⊥    ⊂      U    ⊥   and   (  U  +  W      )    ⊥    ⊂      W    ⊥    .. Therefore   (  U  +  W      )    ⊥    ⊂      U    ⊥    ∩      W    ⊥  . Now, take   v  ∈  V  ∖  (  U  +  W      )    ⊥  . Then there is some   u  ∈  U and there is some   w  ∈  W such that   ⟨  v  ,  u  +  w  ⟩  ≠  0. Therefore, the numbers   ⟨  v  ,  u  ⟩ and   ⟨  v  ,  w  ⟩ cannoy be both equal to 0. In particular,   v  ∉      U    ⊥    ∩      W    ⊥

Let U,W sube V, vector subspaces then: 1) U sube W => W^(_|_) sube U^(_|_) 2) (U+W)^(_|_)=U^(_|_) nn W^(_|_)

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