
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 "nondegenerate" >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_ nondegenerate. 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 finitedimensional 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 nonsingular, and accordingly nondegenerate forms are also referred to as nonsingular 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>
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