Let $V$ be a vector space on the field $K$ and let $V^*$ be the dual space of $V$. For every bilinear form $\langle\cdot,\cdot\rangle$ on $V$ we define a linear map
\begin{equation}
L_{\langle\,\cdot,\cdot\,\rangle}: V \rightarrow V^* : v \mapsto \langle\cdot,v\rangle
\end{equation}
Let B(V) be the set of all bilinear forms on $V$ and consider the functions
\begin{equation}
\phi: B(V) \rightarrow \operatorname{Hom}_K(V,V^*) : \langle\cdot,\cdot\rangle \mapsto L_{\langle\cdot,\cdot\rangle}
\end{equation}
\begin{equation}
\psi: \operatorname{Hom}_K(V,V^*) \rightarrow B(V) : f \mapsto \langle \cdot,\cdot\rangle_f
\end{equation}
I have already shown that $\phi$ and $\psi$ are each other's inverse en are thus bijections.
I am stuck on the next question: "Show that the rank of $\phi$ and $\psi$ is equal to the rank of $\langle\cdot,\cdot\rangle$, resp. $f$, if we assume that $\dim V = n < \infty$."
I know that the rank of the bilinear form $\langle\cdot,\cdot\rangle$ is equal to the rank of a Grammian matrix of the bilinear form $\langle\cdot,\cdot\rangle$, but I couldn't get any further.
Thanks in advance!
Best Answer
Presumably this means that for any bilinear form $g\in B(V)$ the map $\phi(g)$ has the same rank (as a map) as the rank of $g$ (as a form); and that for any map $l\in Hom(V, V^*)$ the rank of $\psi(l)$ as a form is the same as rank of $l$.
This paragraph is motivation, ignore it if you dislike unfamiliar terms: Since you know $\phi$ and $\psi$ are inverses of each other, it is enough to show that each is rank-non-increasing (since then the only way the composition can be identity is if each one is rank-preserving).
Now suppose $g$ has null space $W$ (as a bilinear form). Check that then $\phi(g)$ restricted to $W$ is zero. So $rk(\phi(g))\leq rk(g)$.
Similarly, if $l$ has null space $U$ (as a map), then check that $\psi(l)$ has $U$ as part of null space (of the form). So $rk(\psi(l))\leq rk(l)$.
Combining the above two inequalities and the fact that $\psi\cdot \phi =Id$ we have for any $g$:
$$rk(g)=rk(\psi(\phi g))\leq rk(\phi(g))\leq rk (g).$$
Hence all the inequalities are equalities, and in particular $rk(\phi(g))= rk (g)$. Similar argument shows $rk (\psi(l))=rk (l)$ for all $l$.