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

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 David C. Ullrich Posts: 21,553 Registered: 12/6/04
Re: Orthogonal complement
Posted: Mar 4, 2013 2:47 PM

On Mon, 04 Mar 2013 11:07:30 -0500, Shmuel (Seymour J.) Metz
<spamtrap@library.lspace.org.invalid> wrote:

>In <6n89j89nt8lml3bpql31qkk5ggrb1q85ui@4ax.com>, on 03/04/2013
> at 07:34 AM, David C. Ullrich <ullrich@math.okstate.edu> said:
>

>>The two of us are evidently interpreting the word "non-degenerate"
>>differently. I was just sort of guessing what that meant...

>
>Were you thinking of "definite"? The Lorentz metric is not definite.
>

>>Hmm. According to the definition at
>> http://en.wikipedia.org/wiki/Bilinear_form

>
>>your f is _not_ non-degenerate. In the notation used there we have
>>f_1(s)(s) = 0, hence f_1 is not an isomorphism.

>
>There is a difference between "f_1(s)(s) = 0" and "f_1(s) = 0".

Oops. Sorry.

>Check
>the determinant, then note "If V is finite-dimensional then, relative
>to some basis for V, a bilinear form is degenerate if and only if the
>determinant of the associated matrix is zero ? if and only if the
>matrix is singular, and accordingly degenerate forms are also called
>singular forms. Likewise, a nondegenerate form is one for which the
>associated matrix is non-singular, and accordingly nondegenerate forms
>are also referred to as non-singular forms. These statements are
>independent of the chosen basis."
>
>BTW, I only see "f"; where do you see "f_1".

In the wikipedia article they say that if B is a bilinear
form then B_1 and B_2 are defined by etc. Here
the form is f, so the same notation leads to f_1
and f_2.