Kaba
Posts:
289
Registered:
5/23/11
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Re: Orthogonal complement
Posted:
Mar 4, 2013 6:17 PM
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4.3.2013 21:27, Kaba wrote:
> The plus is the direct sum. However, I also stated the claim incorrectly.
>
> Let V be a finite-dimensional vector space over K together with a
> reflexive bilinear form f : V^2 --> K, and let S subset V be a subspace
> of V. Let C(S) stand for the orthogonal complement of S. The bilinear
> form f is non-degenerate on S if and only if V = S + C(S).
Here is another related attempt at a proof I found:
Claim
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If S subset V is a non-degenerate subspace of V, then C(C(S)) = S.
Attempted proof sketch
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First it is shown that S subset C(C(S)), which is easy to see. Then,
without additional justifications, it is claimed that
dim(S) + dim(C(S)) = dim(V) = dim(C(S)) + dim(C(C(S))),
which implies dim(C(C(S)) = dim(S), and therefore C(C(S)) = S.
Related
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There is a similar argument in Lang's Algebra, equally mysterious. What
might be the source of such dimension arguments? I'm guessing the
rank-nullity theorem:
dim(V) = dim(f(V)) + dim(f^{-1}(0)),
for any linear f : V --> W.
The problem is that while it is possible to prove, for V non-degenerate,
that
S intersect C(S) = {0},
it seems hard to prove that S union C(S) = V. In general, these
questions seem related:
1) C(C(S)) = S
2) V = S + C(S) (direct sum)
3) dim(S) + dim(C(S)) = dim(V)
My intuition says that these properties hold only for finite-dimensional
spaces, and therefore any proof must necessary use a property which is
specific to finite dimensional spaces.
--
http://kaba.hilvi.org
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