# Bilinear Maps

• Sep 21st 2009, 04:05 PM
redsoxfan325
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?
• Sep 21st 2009, 06:07 PM
aliceinwonderland
Quote:

Originally Posted by redsoxfan325
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 $B(v_i, v_j)$.
• Sep 21st 2009, 06:32 PM
NonCommAlg
Quote:

Originally Posted by redsoxfan325
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: $B(u,v)=\sum_{i,j} b_ic_j B(e_i,e_j)=\sum_{i,j}b_ic_j[e_i]^T A [e_j]=(\sum_i b_i[e_i]^T)A (\sum_j c_j[e_j] )=[u]^TA[v].$