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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: 3,473
Registered: 12/4/04
Re: Orthogonal complement
Posted: Mar 4, 2013 11:19 AM
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In <5134c6c2$12$fuzhry+tra$>, on 03/04/2013
at 11:07 AM, Shmuel (Seymour J.) Metz
<> said:

>In <>, on 03/04/2013
> at 07:34 AM, David C. Ullrich <> said:

>>Hmm. According to the definition at

Whoops! I was looking at
<>, although the
definition in <> is

>>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."

A relevant quote from the article you cited would be "If V is
finite-dimensional then the rank of B1 is equal to the rank of B2. If
this number is equal to dim(V) then B1 and B2 are linear isomorphisms
from V to V*. In this case B is nondegenerate. By the rank-nullity
theorem, this is equivalent to the condition that the left and
equivalently right radicals be trivial. In fact, for finite
dimensional spaces, this is often taken as the definition of

Definition: B is nondegenerate if and only if B(v, w) = 0 for all
w implies v = 0."

>BTW, I only see "f"; where do you see "f_1".

Looking at the correct article, I see a B_1:V->V* but not an f_1.

Shmuel (Seymour J.) Metz, SysProg and JOAT <>

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