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Topic: Orthogonal complement
Replies: 15   Last Post: Mar 5, 2013 6:48 PM

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Robin Chapman

Posts: 282
Registered: 5/29/08
Re: Orthogonal complement
Posted: Mar 5, 2013 6:16 AM
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On 04/03/2013 23:19, Kaba wrote:
> 5.3.2013 1:17, Kaba kirjoitti:
>> Claim
>> -----
> >
>> If S subset V is a non-degenerate subspace of V, then C(C(S)) = S.

>
> This should have read:
>
> If S subset V, where V is a non-degenerate symmetric bilinear space,
> then C(C(S)) = S.


This is true when
V is finite-dimensional
and
S is a vector subspace of V.

The bilinear form induces a linear map g: V -> S*
(S* being the dual space of S) which is surjective
(by non-degeneracy of S, g is injective on S, so
also surjective on S).
By definition Ker g is C(S), so by rank-nullity
dim C(S) = dim V - dim S* = dim V - dim S.

As g is surjective on S, for all u in V there is w in
S with g(u) = g(w), that is u - w is in C(S). Therefore
V = S + C(S) (dimension counting means the sum is direct).

Next we claim that C(S) is non-degenerate: any element of C(S)
orthogonal to C(S) is also by definition orthogonal to S
and so to V = S + C(S); thus it is zero.

Applying the above to C(S) gives dim C(C(S)) = dim(S).
As S is clearly a subset of C(C(S)) then S = C(C(S)).



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