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

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Shmuel (Seymour J.) Metz

Posts: 3,331
Registered: 12/4/04
Re: Orthogonal complement
Posted: Mar 4, 2013 11:07 AM
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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".


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