(Bilinear Forms) Prove that the following statements are equivalent Let $V$ be a vector space of finite dimension $n$, $F$ a field.
Let $f\in Bil(V, V; F)$.
Prove that the following statements are equivalent:


*

*$f$ is non degenerate

*$rank(f) = n$

*$Ker_r(f) = \{0\}$

*$Ker_l(f) = \{0\}$

*$\phi^{-1}:V\to V^{*}$ which is defined by $\phi^{-1}(v)(u) = f(v, u)$ is an isomorphism.


I have successfully shown the following:
$1 \iff 2$, $2 \implies 3$, $2 \implies 4$.
But I'm stuck with the rest.
Can you give me a hand please?
Thanks!
 A: Let $V$ be an $n$-dimensional vector space over a field $F$ and let $f : V \times V \to F$ be a bilinear form. You're presumably defining $f$ to be non-degenerate if for every non-zero $v \in V$ there exists some $w_L, w_R \in V$ such that $f(v,w_L) \neq 0$ and $f(w_R,v) \neq 0$. For convenience, let's set the following terminology and notation:


*

*the rank of $f$ is the rank of the matrix $[f]_\beta := \left(f(e_i,e_j)\right)_{i,j=1}^n$ for any basis $\beta = \{e_1,\dotsc,e_n\}$ of $V$;

*the left kernel or radical of $f$ is $\ker_L(f) := \{v \in V \mid \forall w \in V, \; f(v,w) = 0\}$;

*the right kernel or radical of $f$ is $\ker_R(f) := \{v \in V \mid \forall w \in V, \; f(w,v) = 0\}$;

*the linear transformation $f_L : V \to V^\ast$ is defined by $f_L(v)(w) := f(v,w)$;

*the linear transformation $f_R : V \to V^\ast$ is defined by $f_R(v)(w) := f(w,v)$;


and let's make the following observations:


*

*for every $v \in V$, $v \notin \ker_L(f)$ if and only if there exists some $w_L \in V$ such that $f(v,w_L) \neq 0$;

*for every $v \in V$, $v \notin \ker_R(f)$ if and only if there exists some $w_R \in V$ such that $f(w_R,v) \neq 0$;

*$\ker_L(f) = \ker(f_L)$;

*$\ker_R(f) = \ker(f_R)$;

*$f_R = f_L^\ast$ (i.e., $f_R$ is the transpose of $f_L$).


What you want to show, then, are that the following are equivalent:


*

*$f$ is non-degenerate;

*$\operatorname{rank}(f) = n$;

*$\ker_R(f) = \{0\}$;

*$\ker_L(f) = \{0\}$;

*$f_L : V \to V^\ast$ is an isomorphism.


Now, you've shown that 1 $\iff$ 2 (as well as that 2 $\implies$ 3, and that 2 $\implies$ 4). Notice that by our first two observations above, $f$ is non-degenerate if and only if for every non-zero $v \in V$, $v \notin \ker_L(f)$ and $v \notin \ker_R(F)$, if and only if $\ker_L(f) = \{0\}$ and $\ker_R(f) = \{0\}$, i.e., 1 $\iff$ (3 AND 4). Hence, it's enough to show that 3, 4, and 5 are all equivalent.
Now, by the observation that $f_R = f_L^\ast$, it follows that $f_L$ is an isomorphism if and only if $f_R$ is, i.e., 5 $\iff$ $f_R$ is an isomorphism. Given that $\ker_L(f) = \ker(f_L)$ and $\ker_R(f) = \ker(f_R)$, you can now use the rank-nullity theorem to show that 3 $\iff$ $f_R$ is an isomorphism and 4 $\iff$ 5.
