bilinear form - proof

Question

I have to prove that the mapping $f(x,y)={\displaystyle \sum_{i=1}^{n}}{\displaystyle \sum_{j=1}^{n}}x_{i}y_{j}{f}(e_{i},e_{j})$

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!

Answer 1

Let's write $c_{ij}$ for $f(e_i,e_j)$ because it seems that the repeated appearance of $f$ is confusing. Then $$f(x+y,z)= \sum_{i=1}^{n}\sum_{j=1}^{n} c_{ij} (x_{i}+y_i)z_j \tag1$$ while $$f(x,z)+f(y,z)= \sum_{i=1}^{n}\sum_{j=1}^{n} c_{ij} x_{i} z_j + \sum_{i=1}^{n}\sum_{j=1}^{n} c_{ij} y_{i} z_j \tag2$$ Do you agree that (1) and (2) are equal?

Answer 2

Also, you should prove that $f(\alpha{\rm x},{\rm y}) = \alpha f({\rm x},{\rm y})$, which is almost obvious: $$ f(\alpha{\rm x},{\rm y}) = \sum_{i,j} \alpha x_i y_j f({\rm e}_i,{\rm e}_j)= \alpha \sum_{i,j} x_i y_j f({\rm e}_i,{\rm e}_j)= \alpha f({\rm x},{\rm y}). $$