```Date: Mar 4, 2013 6:17 PM
Author: Kaba
Subject: Re: Orthogonal complement

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-----If S subset V is a non-degenerate subspace of V, then C(C(S)) = S.Attempted proof sketch----------------------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-------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)) = S2) 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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