A blinear form $H$ is non-degenerate if and only if $H(v,w) = 0 \ \forall v \in V \Leftrightarrow w = 0$

Question

Question: Given a bilinear form $H: V \times V \rightarrow F$ on a finite dimensional V, show that $H$ is nondegenrate if and only if the only $w \in V$ s.t. $H(v,w) = 0 \quad \forall v \in V$ is $w = 0$

(i.e. H is nondegenrate if and only if $H(v,w) = 0 \ \forall v \in V \Leftrightarrow w = 0$)

I know that $H$ is non-degenerate if and only if the following hold:

1. $H(v,w) = 0 \ \forall v \in V \Leftrightarrow w = 0$
2. $H(v,w) = 0 \ \forall w \in V \Leftrightarrow v = 0$

What this questions seems to be asking me to prove is that one of the above conditions implies the other (specifically, the first condition implies the second).

Since it suffices to prove that 1) implies 2):

Assume $H(v,w) = 0 \ \forall v \in V \Leftrightarrow w = 0$. Prove that $H(v,w) = 0 \ \forall w \in V \Leftrightarrow v = 0$

I am able to prove one direction for the second if and only if, but unable to get the other direction:

The direction I can do:

If $v = 0$, then $H(v,w) = H(0,w) = H(a,w) - H(a,w) = 0 \quad$ (for some $a \in V$)

$\Rightarrow$ if $v=0$ then $H(v,w) = 0$

The direction I can't do:

If $H(v,w) = 0 \quad \forall w \in W$:

1. $H(0,w) = H(v,w) - H(v,w) = 0 - 0 = 0$

2. $H(v,v) = 0$

I feel like one of these above statements should somehow imply that $v = 0$, but I'm not sure how to get there...

Any help would be appreciated, thank you!

EDIT: To add, I would like to know if there is a way to prove this without using a dual space/dual basis argument. Intuitively, I feel like there should be a way to prove it without the dual. (i.e. just proving by going forward then backward direction)

Answer 1

Let's denote by $V^\vee$ to the dual space of $V$, and define linear maps $H_l$ and $H_r$ from $V$ to $V^\vee$ by $H_l(v) = H(v,-)$ and $H_r(w) = H(-,w)$ for all $v,w \in V$.

Note then that the condition $$H(v,w)=0 \textrm{ for all } v \in V \!\iff\! w=0$$ can be written as $H(-,w)=0 \!\iff\! w=0$, which is to say that $H_r$ is injective, and then (because $V$ is finite-dimensional) it is an isomorphism.

Thus, the induced map $H_r^\vee : V^{\vee\vee} \to V^\vee$ of $H_r$ is also an isomorphism, and if we denote by $\operatorname{ev} : V \to V^{\vee\vee}$ to the canonical isomorphism, from $H_r^\vee \circ \operatorname{ev} = H_l$ we also have that $H_l$ is an isomorphism.

In particular, $H_l$ is injective, which is to say that $H(v,-)=0 \!\iff\! v=0$, in other words, $$H(v,w)=0 \textrm{ for all } w \in V \!\iff\! v=0.$$