# Show that the rank of $\phi$ and $\psi$ is equal to the rank of $\langle\cdot,\cdot\rangle$, resp. $f$

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 $$$$L_{\langle\,\cdot,\cdot\,\rangle}: V \rightarrow V^* : v \mapsto \langle\cdot,v\rangle$$$$

Let B(V) be the set of all bilinear forms on $$V$$ and consider the functions $$$$\phi: B(V) \rightarrow \operatorname{Hom}_K(V,V^*) : \langle\cdot,\cdot\rangle \mapsto L_{\langle\cdot,\cdot\rangle}$$$$

$$$$\psi: \operatorname{Hom}_K(V,V^*) \rightarrow B(V) : f \mapsto \langle \cdot,\cdot\rangle_f$$$$

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.

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 $$l$$.

• Yes, I essentially have to show that the rank of a map and the rank of the associated bilinear form are exactly the same. Aug 24, 2020 at 9:20
• 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! Aug 24, 2020 at 9:22
• 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 $n^2$).
– Max
Aug 24, 2020 at 9:23
• 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.
– Max
Aug 24, 2020 at 9:25
• See the edited version.
– Max
Aug 24, 2020 at 9:30