Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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

 Messages: [ Previous | Next ]
 Shmuel (Seymour J.) Metz Posts: 3,473 Registered: 12/4/04
Re: Orthogonal complement
Posted: Mar 4, 2013 11:07 AM

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". 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".

--
Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel>

Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply to
domain Patriot dot net user shmuel+news to contact me. Do not