I am currently working on a problem for my course in linear algebra and now I am confronted with a problem I cannot solve myself. I hope you could give me some hints on how to solve this.
The Problem is as follows ($V$ is a vector space, $K$ is a field):
A $\zeta$-sesquilinear form $\sigma : V \times V \rightarrow K$ is 'not degenerate' if it fulfills these two properties:
- $\quad \forall x \in V \backslash \{0\} \, \, \exists b \in V : \sigma(x,b) \neq 0$
- $\quad \forall y \in V \backslash \{0\} \, \, \exists a \in V : \sigma(a,y) \neq 0$
Show that for $\text{dim}V < \infty$ each of those properties is equivalent to the bijectivity of the map $d_{\sigma}: V \rightarrow V^* : x \mapsto \sigma(x, \cdot )$
Due to some previous problem I additionally know the facts that:
- $\quad$ Property 1 $\quad \Longleftrightarrow \quad d_{\sigma}$ is injective
- $\quad$ Property 2 $\quad \Longleftrightarrow \quad \bigcap_{x \in V} \text{ker}\,d_{\sigma} = \{0\}$
Since I know that in finite dimension the injectivity of linear maps is equivalent to their bijectivity, I show one of the following cases:
$\quad \quad \quad$ Property 1 $\quad \Longleftrightarrow \quad$ Property 2
$\quad \text{or} \quad$ Property 1 $\quad \Longleftrightarrow \quad \bigcap_{x \in V} \text{ker}\,d_{\sigma} = \{0\}$
$\quad \text{or} \quad$ Property 2 $\quad \Longleftrightarrow \quad d_{\sigma}$ is injective
$\quad \text{or} \quad$ $d_{\sigma}$ is injective $\quad \Longleftrightarrow \quad \bigcap_{x \in V} \text{ker}\,d_{\sigma} = \{0\}$
Unfortunatley, I have a hard time trying to show one of these statements. Does anyone have any advice?
Thanks in advance!