I have to prove that the mapping $f(x,y) = {\displaystyle \sum_{i=1} ^ {n} }{ \displaystyle \sum_{j=1}^{n} }x_iy_j{f}(e_i,e_j)$
is a bilinear form, that is, inter alia, the condition: $f(x+y,z)=f(x,z)+f(y,z)$
I have: $f(x+y,z)=f\left({\displaystyle \sum_{i=1}^{n}}x_{i}e_{i}+{\displaystyle \sum_{k=1}^{n}}y_{k}e_{k},{\displaystyle \sum_{j=1}^{n}}z_{j}e_{j}\right)=f\left({\displaystyle \sum_{i=1}^{n}}{\displaystyle \sum_{k=1}^{n}}\left(x_{i}e_{i}+y_{k}e_{k}\right),{\displaystyle \sum_{j=1}^{n}}z_{j}e_{j}\right)=\ldots$
and what's next?
Help!
First note, that $$ \sum_i x_i e_i + \sum_k y_k e_k \ne \sum_i \sum_k (x_i e_i + y_k e_k) $$ as the left hand side contains $n$ terms $x_1e_1$ (one for each $k$) and the right hand only one (analogously for the other terms). But, by distributivity, we have $$ \sum_i x_i e_i + \sum_k y_k e_k = \sum_i x_i e_i + \sum_i y_i e_i = \sum_i (x_i + y_i) e_i $$ Exploiting this, we have \begin{align*} f(x+y,z) &= f \left( \sum_i x_i e_i + \sum_i y_i e_i, \sum_j z_j e_j\right)\\ &= f\left(\sum_i (x_i + y_i)e_i, \sum_j z_j e_j\right)\\ &= \sum_i \sum_j (x_i + y_i)z_j f(e_i, e_j)\\ &= \sum_i \sum_j x_i z_j f(e_i, e_j) + \sum_i \sum_j y_i z_j f(e_i, e_j)\\ &= f(x,z) + f(y,z) \end{align*}