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)) = 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.

--
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