My class version of the non-degeneracy definition states:
Let $V$ be a vector space over a field $\Bbb F$, equipped with a symmetric bilinear form $b : V \times V ā \Bbb F$ . Then $b$ is a non-degenerate bilinear form if $~ā\,uāV,\ b(v,u)=0 \implies v=0$.
I find myself confused when I try to apply it to proving all inner products are non-degenerate, and here is my confusion:
Given an inner product $b$. I let $u \in V$ and assume $b(v,u)=0$ and I am trying to show $v=0$.
I saw on a few texts that we are allowed to pick $v$ so that $v=u$ and thus we get $v=0$ from positivity. Why are we allowed to choose the value for $v$? I thought $v$ could just be any arbitrary vector and it doesn't have to happen to be $u$. To make it clearer, it doesn't make sense to me that $b(v,u)=0\Rightarrow v=u$ whatsoever. In my mind $v$ is fixed and does not vary with $u$.
Could someone kindly explain what is wrong with my reasoning?
Best Answer
You might find it easier to think of this: the radical of $V$ with respect to a (symmetric, if you may) bilinear form $b: V \times V \to \Bbb C$ is the subspace: $${\rm Rad}(V) = \{ u \in V \mid b(u,v) = 0,~\forall\,v \in V \}$$ We say that $b$ is non-degenerate on $V$ if ${\rm Rad}(V) = \{0\}$. Like a sort of kernel.
Now suppose that $b$ is an inner product, and take $u \in {\rm Rad}(V)$. So $b(u,v) = 0$ for all $v \in V$. Well, $u \in V$, so you can say that $b(u,u) = 0$. Now follows from the defintion of inner product (definite positivity) that $u = 0$. So ${\rm Rad}(V) = \{0\}$ and $b$ is non-degenerate on $V$.
Remark: to be precise, as I have written it, ${\rm Rad}(V)$ is the kernel of the linear map $V \ni u \mapsto b(u,\cdot) \in V^\ast$. You can ignore this, in any case.