>>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 nondegeneracy:
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.
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