>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."
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 reply to firstname.lastname@example.org