# Kernel of a bilinear form

bilinear-formlinear algebra

Let $$\alpha$$ be a bilinear form on $$V$$,then we define $$\ker(\alpha)=\{y\in V:\alpha(x,y)=0,\forall x\in V\}$$. Why this is said to be the kernel of $$\alpha$$? Is there any relation of this kernel with the null space of the matrix of the bilinear form?

I want to explain this with an example:

Take the bilinear form $$f(x,y)=2x_1y_1+x_2y_2+9x_3y_3+3x_1y_2+3x_2y_1+5x_1y_3+5x_3y_1+4x_2y_3+4x_3y_2$$

Now,$$Ker(f)=\{x\in V: f(x,y)=0 ,\forall y\in V\}$$

So,we need to find those $$(x_1,x_2,x_3)\in \mathbb R^3$$ such that $$f((x_1,x_2,x_3),(y_1,y_2,y_3))=0$$ for all $$y_1,y_2,y_3\in \mathbb R$$.

i.e. we need to find $$(x_1,x_2,x_3)$$ such that $$(2x_1+3x_2+5x_3)y_1+(3x_1+x_2+4x_3)y_2+(5x_1+4x_2+9x_3)y_3=0$$ for all $$y_1,y_2,y_3\in \mathbb R$$.

i.e. the above is an identity in $$y_1,y_2,y_3$$

So,we have to find $$(x_1,x_2,x_3)$$ such that,

$$2x_1+3x_2+5x_3=0$$

$$3x_1+x_2+4x_3=0$$

$$5x_1+4x_2+9x_3=0$$

Which is equivalent to finding the kernel of the matrix corresponding to $$f$$.Now does this make sense?