Date: Mar 4, 2013 2:27 PM
Subject: Re: Orthogonal complement
3.3.2013 19:11, David C. Ullrich wrote:
> On Sun, 03 Mar 2013 17:43:57 +0200, Kaba <email@example.com> wrote:
>> 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 if and only if V = S + C(S).
>> Any ideas?
> One of us is missing something. Say f = 0. Then C(S) = V, hence
> V = S + C(S).
> Maybe the "+" was supposed to mean _direct_ sum? It seems likely
> that it's easy to show that f is non-degenerate if and only if
> S intersect C(S) = 0 for all S.
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).
The only change is the "on S".
> And it seems likely to me that if f is non-degenerate then it does
> follow that V = S + C(S). To give an inelegant proof, start
> with a basis for S, extend that to a basis for V, then
> apply Gram-Schmidt to convert to an orthonormal basis... ?
So it seems. I've spent the last two days trying different routes and
always get blocked by the need for theorems which seem equally hard.
I've also looked at some sources, but unfortunately those that I have
looked at seem incorrect.
Here's a relevant paper by Keith Conrad:
The Theorem 3.11 at the end of page 17. Unfortunately, the proof seems
to be lacking: it is relying on a generalization of the Riesz
representation theorem for Hilbert spaces. In particular, the proof
fails to show the surjectivity of the proposed map, which is the actual
hard part of it. The proof relies on Theorem 3.1 on page 16, which
should show that surjectivity. However, it did not seem so to me on the
last read. In particular the sentence "Since LB is a linear map between
vector spaces of the same dimension, injectivity, surjectivity, and
isomorphy are equivalent properties." sounds incorrect to me.
But perhaps I did not read carefully enough.