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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!

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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 $l$.

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  • $\begingroup$ Yes, I essentially have to show that the rank of a map and the rank of the associated bilinear form are exactly the same. $\endgroup$ Aug 24, 2020 at 9:20
  • $\begingroup$ This also seems to be correct, but as we haven't seen the terms 'rank-non-increasing' and 'rank-preserving', I don't think we can use them... Thanks for your solution anyways! $\endgroup$ Aug 24, 2020 at 9:22
  • $\begingroup$ The wording of the question is confusing, since $\phi$ is also a linear map, and has a rank (equal to $n^2$, since it's an iso between two spaces $B(V)$ and $Hom(V, V^*)$ of dimension $n^2$). $\endgroup$
    – Max
    Aug 24, 2020 at 9:23
  • $\begingroup$ These are not "terms" you would expect to have seen, necessarily. These are the things you could define: "rank-non-increasing" meaning the rank of the image is less than or equal to the rank of the thing you start with; rank preserving means these ranks are equal. You can rewrite the solution without them. $\endgroup$
    – Max
    Aug 24, 2020 at 9:25
  • $\begingroup$ See the edited version. $\endgroup$
    – Max
    Aug 24, 2020 at 9:30

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