Kernel of bilinear form It is written in book, I read: kernel of bilinear form is space consisting of vectors $y$, such:
$$Ker(\alpha)=\{y\in V:\alpha(x,y)=0,\ \forall x\in V\}$$
Nice I get it, but then it is said, that definition on top is equal to:
$$Ker(\alpha)=\{y\in V:\alpha(e_i,y)=0,\ i=1,...,n\},$$
where $e_1,..e_n$ are basis vectors. That is the part I don't understand. I think, that we can transform definition using linear properties of $\alpha$:
$$x=x_1e_1 + ... + x_ne_n\\
Ker(\alpha)=\{y\in V:x_1\alpha(e_1,y)+..+x_n\alpha(e_n,y)=0,\ i=1,...,n, \ \forall x\in V\}$$
So please, how are these definitions equivalent?
Thank you!

Let $\ker_1(\alpha)$ be the first definition and $\ker_2(\alpha)$ be the second. We want to prove that $\ker_1(\alpha) = \ker_2(\alpha)$. It is clear that $\ker_1(\alpha)\subseteq \ker_2(\alpha)$. Now let $y \in \ker_2(\alpha)$. Let $x \in V$ be an arbitrary vector. Write $x = \sum_i x_ie_i$. Then: $$\alpha(x,y) = \alpha\left(\sum_ix_ie_i,y\right) = \sum_ix_i\alpha(e_i,y) = \sum_ix_i\cdot 0 = 0,$$ hence $y \in \ker_1(\alpha)$. So $\ker_1(\alpha) = \ker_2(\alpha)$ and the definitions are equivalent.
