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The question is probably elementary, but i don't really know how to do it. I'm reading this paper on bilinear forms and in trying to show that there's a unique matrix representing each bilinear form given a fixed basis, one of the proofs has the following step:

Given $v,w \in V$, $B$ is a bilinear form on $V$, and $\{ b_1, \ldots, b_n \}$ is some basis for $V$,

$$B(v,w) = B(\sum_i v_i b_i, \sum_j w_j b_j) = \sum_i v_i B( b_i, b_j)\sum_j w_j=\ldots$$

How do I demonstrate equality between the last two expressions (pull the sum out)?

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    $\begingroup$ Bilinear forms are bilinear. $\endgroup$ – Mark Oct 4 '11 at 15:03
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The "pulling out of the sum" is a consequence of the properties of bilinear forms:

http://en.wikipedia.org/wiki/Bilinear_form

If $B$ is a bilinear form, then certainly $B(v_1\mathbf{b}_1 + v_2\mathbf{b}_2, \mathbf{w}) = B(v_1\mathbf{b}_1, \mathbf{w}) + B(v_2\mathbf{b}_2, \mathbf{w}) = v_1B(\mathbf{b}_1, \mathbf{w}) + v_2B(\mathbf{b}_2, \mathbf{w})$. By induction then,

$$ B(v_1\mathbf{b}_1 + v_2\mathbf{b}_2 + \cdots + v_n\mathbf{b}_n, \mathbf{w}) = v_1B(\mathbf{b}_1, \mathbf{w}) + v_2B(\mathbf{b}_2, \mathbf{w}) + \cdots v_nB(\mathbf{b}_n, \mathbf{w}).$$

In other words,

$$ B\left( \sum_i v_i\mathbf{b}_i, \mathbf{w} \right) = \sum_i v_i B(\mathbf{b}_i, \mathbf{w}).$$

In a similar way,

$$ B\left( \mathbf{v}, \sum_{j} w_i\mathbf{b}_j \right) = \sum_j B(\mathbf{v}, \mathbf{b}_j)w_j.$$

So, if $\{ \mathbf{b}_1, \ldots, \mathbf{b}_n\}$ is a basis, we can write uniquely,

$$ \mathbf{v} = \sum_i v_i\mathbf{b}_i $$

and (using a different index variable to avoid confusion later):

$$ \mathbf{w} = \sum_j w_j\mathbf{b}_j $$

The result you quote above follows from combining it all into one expression.

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