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Math Help - Bilinear Forms

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    Bilinear Forms

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

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

    (BTW :  f(u,v)=[u]^t_{B}[f]_B[v]_B for any base B of V)
    Last edited by adam63; January 11th 2010 at 11:28 AM.
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  2. #2
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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 v\ne 0 for which f(v,u)=0 for all u in V.

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

    (BTW :  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 v\ne 0 for which f(v,u)=0 for all u in V.

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

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

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

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