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Thread: Bilinear Maps

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

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

    Does this statement require proof, and if so, how can I prove it?
    Last edited by redsoxfan325; Sep 21st 2009 at 06:40 PM. Reason: Changed to V x V
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    Quote Originally Posted by redsoxfan325 View Post
    I was told that any bilinear map $\displaystyle B: V\longrightarrow \mathbb{F}$ (where $\displaystyle V$ is an n-dimensional vector space and $\displaystyle \mathbb{F}$ is a field of scalars) can be expressed as $\displaystyle B(u,v)=u^TAv$, where $\displaystyle A$ is an $\displaystyle n\times n$ matrix.

    Does this statement require proof, and if so, how can I prove it?
    A bilinear form on V is a bilinear mapping $\displaystyle B: V \times V \longrightarrow \mathbb{F}$ 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, $\displaystyle B(u, v) = B(\sum_iv_ix_i, \sum_jv_jy_j)$.

    Using bilinearity, $\displaystyle B(\sum_iv_ix_i, \sum_jv_jy_j) = \sum_{i,j}x_iy_jB(v_i, v_j)=X^TAY$, where A is $\displaystyle B(v_i, v_j)$.
    Last edited by aliceinwonderland; Sep 21st 2009 at 06:20 PM. Reason: Added a description for a basis B
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    Quote Originally Posted by redsoxfan325 View Post
    I was told that any bilinear map $\displaystyle B: V\longrightarrow \mathbb{F}$ (where $\displaystyle V$ is an n-dimensional vector space and $\displaystyle \mathbb{F}$ is a field of scalars) can be expressed as $\displaystyle B(u,v)=u^TAv$, where $\displaystyle A$ is an $\displaystyle n\times n$ matrix.

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

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

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