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Thread: basic bilinear forms

  1. #1
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    basic bilinear forms

    Let f:V VF be a bilinear form on a vector space V, where F is the field of scalars. 3 conditions must be fulfilled:
    1. f(u1+u2,v)=f(u1,v)+f(u2,v)
    2. f(u,v1+v2)=f(u,v1)+f(u,v2)
    3.f(u,tv)= f(tu,v)=tf(u,v)

    If F=Q (as the set of rational numbers) then if f fulfills 1. and 2. then it must fulfill 3.
    I was wondering how to prove that?
    Another question is to find an example of a function/map that is not a bilinear form (and precise on which field it is) because it fulfills 1. and 2. but not 3.

    Thanks a lot for any help or suggestions!

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  2. #2
    Junior Member nimon's Avatar
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    Edinburgh, UK
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    Hey there,

    I had an exam question on this the other day! To show that 1.,2. imply 3. on $\displaystyle \mathbb{Q}$ it's easiest to show it for $\displaystyle \mathbb{N}$ first, then $\displaystyle \mathbb{Z}$, and then $\displaystyle \mathbb{Q}$. I'll give you some hints!

    (n) Try induction to show that $\displaystyle f(nu,v) = nf(u,v)$. This is obvious for $\displaystyle n=1$, and so assume it is true for $\displaystyle n\in\mathbb{N}$. Then $\displaystyle f((n+1)u,v) = f(nu + u,v) = \ldots$ From here you can use 1. and the inductive hypothesis. The symmetric case, i.e. for $\displaystyle f(u,nv)$ follows from 2.

    (z) Then for $\displaystyle \mathbb{Z}$, it suffices to show that $\displaystyle f(-u,v) = -f(u,v)$ (why?). Consider $\displaystyle f(u-u,v)$ and use 1. Again use 2. for the symmetric case.

    (q) Now for $\displaystyle \mathbb{Q}$, it suffices to show $\displaystyle f(\frac{1}{n}u,v) = \frac{1}{n}f(u,v)$ (why?). Now consider $\displaystyle f(u,v) = f(n\frac{1}{n}u,v) = \ldots$ and use (n).

    Hope this is helpful, write back if you get stuck

    (I don't know why I just used a smiley I think they're stupid but I was endeared by the yellowness.)
    Last edited by nimon; May 4th 2010 at 02:48 AM. Reason: 'sufficues' is not a word
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