Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)?

share|cite|improve this question
Bilinear forms are bilinear. – Mark Oct 4 '11 at 15:03
up vote 2 down vote accepted

The "pulling out of the sum" is a consequence of the properties of bilinear forms:

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.