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Math Help - Bilinear Maps

  1. #1
    Super Member redsoxfan325's Avatar
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    Bilinear Maps

    I was told that any bilinear map B: V\times V\longrightarrow \mathbb{F} (where V is an n-dimensional vector space and \mathbb{F} is a field of scalars) can be expressed as B(u,v)=u^TAv, where A is an n\times n matrix.

    Does this statement require proof, and if so, how can I prove it?
    Last edited by redsoxfan325; September 21st 2009 at 07:40 PM. Reason: Changed to V x V
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  2. #2
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    Quote Originally Posted by redsoxfan325 View Post
    I was told that any bilinear map B: V\longrightarrow \mathbb{F} (where V is an n-dimensional vector space and \mathbb{F} is a field of scalars) can be expressed as B(u,v)=u^TAv, where A is an n\times n matrix.

    Does this statement require proof, and if so, how can I prove it?
    A bilinear form on V is a bilinear mapping B: V \times V \longrightarrow \mathbb{F} such that

    B(u+u', v) = B(u,v) + B(u',v),
    B(u, v+v') = B(u,v) + B(u,v'),
    B(tu, v) = B(u, tv) = tB(u,v), where t is a scalar in the field F.

    Let u= BX, v=BY, where B = (v_1, v_2, ... , v_n) is a basis of V and X,Y are coordinate vectors.

    Then, B(u, v) = B(\sum_iv_ix_i, \sum_jv_jy_j).

    Using bilinearity, B(\sum_iv_ix_i, \sum_jv_jy_j) = \sum_{i,j}x_iy_jB(v_i, v_j)=X^TAY, where A is B(v_i, v_j).
    Last edited by aliceinwonderland; September 21st 2009 at 07:20 PM. Reason: Added a description for a basis B
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  3. #3
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    Quote Originally Posted by redsoxfan325 View Post
    I was told that any bilinear map B: V\longrightarrow \mathbb{F} (where V is an n-dimensional vector space and \mathbb{F} is a field of scalars) can be expressed as B(u,v)=u^TAv, where A is an n\times n matrix.

    Does this statement require proof, and if so, how can I prove it?
    that B is actually from V \times V to \mathbb{F}. also u^TAv has no meaning unless you fix a basis \mathcal{B}=\{e_1, \cdots , e_n \} for V and then by u,v we'll mean [u], \ [v], the coordinate vectors of u,v with respect to

    the basis \mathcal{B}. anyway, the proof is quite easy: define the matrix A=[a_{ij}] by a_{ij}=B(e_i,e_j). now [e_k], \ 1 \leq k \leq n, is an n \times 1 vector with 1 in its k-th row and 0 is in other rows. it's easy to see

    that [e_i]^T A [e_j]=a_{ij}=B(e_i,e_j). finally if u=\sum_i b_ie_i, \ v=\sum_i c_ie_i, then using bilinearity of B we get: B(u,v)=\sum_{i,j} b_ic_j B(e_i,e_j)=\sum_{i,j}b_ic_j[e_i]^T A [e_j]=(\sum_i b_i[e_i]^T)A (\sum_j c_j[e_j] )=[u]^TA[v].
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