Kernel of a bilinear form

Question

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?

Answer 1

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?

Answer 2

This becomes consistent with the usual definition of the kernel — the preimage of zero — when instead of looking at the bilinear form $\alpha$ defined on $V\times V$ we look at the associated linear map $\alpha_1$ from $V$ to its dual $V^*$ defined by

$$ \alpha_1(x)(y) = \alpha(x, y).\tag2 $$

Then the definition of the kernel of $\alpha$ given in the question coincides with the kernel of $\alpha_1$

$$ \ker\alpha:=\ker\alpha_1 = \{x\in V\,|\,\alpha_1(x) = 0\}\tag2 $$

where $0\in V^*$ is the zero functional. Consequently, the answer to the second question is affirmative: the kernel of a bilinear form is the kernel of its matrix.