exactly one bilinear form

Question

Let $E = \{e_1, ..., e_n\}$ be a basis of a $K$-vector space $V$. Let $A: E \times E \to K$ be any map. Then, there is exactly one bilinear form $B: V \times V \to K$ such that $B (e_i, e_j) = A (e_i, e_j)$ for each $e_i, e_j \in E$.

How do you prove it?

Answer 1

Given two vectors in $V$, expand each as a (unique) linear combination of the basis vectors, then use the bilinearity of $B$.

Answer 2

Can I write it so?:

Let $A(e_{i},e_{j})=a_{ij}$ - any transformation. Since $B(e_{i},e_{j}) = A(e_{i},e_{j})$, so $A(x,y)=\sum_{i, j = 1}^{n}a_{ij}x_{i}y_{j}=B(x,y)$

which proves the uniqueness bilinear form $B:V \times V \rightarrow K$.