Bilinear form (question about a step in a proof )

Question

I'm having trouble understanding a step in a proof about bilinear forms

Let $ \mathbb{F}:\,\mathbb{R}^{n}\times\mathbb{R}^{n}\to \mathbb{R}$ be a bilinear functional.

$x,y\in\mathbb{R}^{n}$

$x=\sum\limits^{n}_{i=0}\,x_{i}e_{i}$

$y=\sum\limits^{n}_{j=0}\;y_{j}e_{j}$

$F(x,y)=F(\sum\limits^{n}_{i=1}\,x_{i}e_{i},\sum\limits^{n}_{j=1}\;y_{j}e_{j})=\sum\limits^{n}_{1=i,j}F(x_{i}e_{i},y_{j}e_{j})=\sum\limits^{n}_{1=i,j}x_{i}y_{j}F(e_{i},e_{j})=\sum\limits^{n}_{1=i,j}x_{i}y_{j}a_{ji}$ $=\sum\limits^{n}_{ji}x_{i}y_{j}a_{ji}$

Could someone explain to me how we got that

$F(e_{i},e_{j})=a_{ji}$

thank you

Edit:

wrote my whole question down .

$A=[a_{ij}]_{i,j=1,...,n}$

$<Ax,y>=<A(\sum\limits^{n}_{i=1}x_{i}e_{i}),\sum\limits^{n}_{j=1}y_{j}e_{j}>=\sum\limits^{n}_{i=1}x_{i}Ae_{i},\sum\limits^{n}_{j=1}\;y_{j}e_{j}>=\sum\limits^{n}_{i,j=1}x_{i}y_{j}<Ae_{i},e_{j}>$

$=\sum\limits^{n}_{ji}x_{i}y_{j}a_{ji}$

I understand why $\sum\limits^{n}_{ji}x_{i}y_{j}a_{ji}=<Ax,y>$, however I dont understand why $<Ax,y>=F(x,y)$

In other words I don't understand why there $F(x,y)=x^{t}Ay$

Answer 1

I believe that is just definition, i.e. we set $F(e_i, e_j) := a_{ji}$. This we do, because for a bilinear functional $F$, the entire values of $F$ are determined by the values at $(e_i, e_j)$. This is clear from the identity you wrote on $F(x,y)$. Any $(x,y)$ is a linear combination of $(e_i,e_j)$'s and $F$ is bilinear so once you know what $a_{ji}$'s are, you know what $F(x,y)$ are.

Answer 2

you could say $F(x,y) = x^TAy$

with A as a symmetric matrix. A has an orthonormal basis.

If we express $x,y$ in this basis then

$F(x,y) = x^T\Lambda y$ With diagonal $\Lambda$