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Thread: Bilinear Forms

  1. #1
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    Bilinear Forms

    f:VxV -> F
    we call f a withered form if there's a vector $\displaystyle v\ne 0$ for which f(v,u)=0 for all u in V.

    Could f be not-withered, while there still is a $\displaystyle v\ne 0$ for which f(u,v)=0 for all u in V?

    (BTW :$\displaystyle f(u,v)=[u]^t_{B}[f]_B[v]_B$ for any base B of V)
    Last edited by adam63; Jan 11th 2010 at 11:28 AM.
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    Quote Originally Posted by adam63 View Post
    f:VxV -> F
    we call f a withered form if there's a vector $\displaystyle v\ne 0$ for which f(v,u)=0 for all u in V.

    Could f be not-withered, while there still is a $\displaystyle v\ne 0$ for which f(u,v)=0 for all u in V?

    (BTW :$\displaystyle f(u,v)=[u]^t_{B}[f]_B[v]_B$ for any base B of V)

    Perhaps I'm missing something, but I'm afraid your question is nonsensical. You're given a definition (If A then B) and then you ask: could it be that not B but still A? Of course not! The definition tells you that if A (=exists a non zero vector s.t. f(u,v) = 0 ...etc.) then B(=the bil. form is called withered)...!!

    Tonio
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    Quote Originally Posted by tonio View Post
    Perhaps I'm missing something, but I'm afraid your question is nonsensical. You're given a definition (If A then B) and then you ask: could it be that not B but still A? Of course not! The definition tells you that if A (=exists a non zero vector s.t. f(u,v) = 0 ...etc.) then B(=the bil. form is called withered)...!!

    Tonio
    Please notice the difference, it's f(v,u) once, then f(u,v).
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    Quote Originally Posted by adam63 View Post
    f:VxV -> F
    we call f a withered form if there's a vector $\displaystyle v\ne 0$ for which f(v,u)=0 for all u in V.

    Could f be not-withered, while there still is a $\displaystyle v\ne 0$ for which f(u,v)=0 for all u in V?

    (BTW :$\displaystyle f(u,v)=[u]^t_{B}[f]_B[v]_B$ for any base B of V)
    the answer is no. to see this, suppose $\displaystyle \dim_F V = n$ and identify $\displaystyle V$ with $\displaystyle F^n.$ then $\displaystyle f(X,Y)=X^TAY,$ for some $\displaystyle A \in \mathbb{M}_n(F)$ and all $\displaystyle X,Y \in F^n.$ let $\displaystyle E_i$ be the $\displaystyle n \times 1$ vector with $\displaystyle 1$ in the $\displaystyle i$th row

    and $\displaystyle 0$ everywhere else. see that $\displaystyle E_i^T A$ is the $\displaystyle i$th row of $\displaystyle A$ and, for any $\displaystyle Y \in F^n,$ the $\displaystyle i$th row of $\displaystyle AY$ is $\displaystyle E_i^TAY.$ now suppose there exists $\displaystyle 0 \neq Y_0 \in F^n$ such that $\displaystyle X^TAY_0 = 0$ for all $\displaystyle X \in F^n.$ so

    $\displaystyle E_i^T A Y_0=0,$ for all $\displaystyle i.$ thus $\displaystyle AY_0 = 0$ and so $\displaystyle \det A = 0.$ but then $\displaystyle \det A^T =0$ and hence $\displaystyle A^T X_0 = 0,$ for some $\displaystyle 0 \neq X_0 \in F.$ thus $\displaystyle X_0^T A=0$ and so $\displaystyle X_0^TAY=0,$ for all $\displaystyle Y \in F^n,$ i.e. $\displaystyle f$ is withered.
    Last edited by NonCommAlg; Jan 12th 2010 at 12:06 AM.
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