Defining a degenerate bilinear form

Question

In this Wikipedia entry, a degenerate bilinear form $f$ on a (arbitrary) vector space $V$ is defined as one for which the map $$ v \mapsto (x \mapsto f(x,v)) $$ is not an isomorphism. But we may as well define degeneracy in terms of the map $x \mapsto f(v,x)$ instead of $x \mapsto f(x,v)$. This leads to the question if the following result always holds:

The map $v \mapsto (x \mapsto f(x,v))$ is an isomorphism if and only if the map $v \mapsto (x \mapsto f(v,x))$ is an isomorphism.

Is this true? If yes, how can we prove it?

Answer 1

Let $f_*:w\in V\mapsto f(_-,w)\in V^*$ and ${_*}f:v\mapsto f(v,_-)\in V^*$.

Let $\chi:v\in V \mapsto (\varphi\in V^*\mapsto \varphi(v))\in V^{**}$.

Let $r:\rho\in V^{**}\mapsto \rho\circ f_*\in V^*$.

For all $v,w\in V$, we have

$(r(\chi(v)))(w)=(\chi(v)\circ f_*)(w)=\chi(v)(f_*(w))=f_*(w)(v)=f(v,w)={_*f}(v)(w).$

Hence, $r\circ\chi={_*}f$.

Now we can prove the following thm.

Thm. Assume that $V$ is finite dimensional. Then for all $f:V\times V\to K$, ${_*}f$ is bijective $\iff$ $f_*$ is bijective.

Proof. Since $V$ is finite dimensional, $\chi$ is bijective. Then ${_*}f$ is bijective $\iff$ ${_*}f\circ \chi^{-1}=r$ is bijective $\iff f_*$ is bijective.

I'm pretty sure that the converse is true, but I only have a counterexample with two different vector spaces : if $V$ is not finite dimensional, and if $f: (v,\varphi)\in V\times V^*\mapsto \varphi(v)\in K$, then $f_*=Id_{V^*}$ which is bijective, and ${_*}f=\chi$ which is not bijective if $V$ is not finite dimensional.

[Edit: typo fixed, following the comment of EpsilonDelta.]

Answer 2

The following reason applies in finite dimensions because I use that a linear map is either both, injective and surjective, or neither. Then your are right, $v\mapsto(x\mapsto f(x,v))$ is an isomorphism $V\to V^*$ if and only if $v\mapsto(x\mapsto f(v,x))$ is one.

Proof.

If $v\mapsto(x\mapsto f(x,v))$ is not an isomorphism, then there is some non-zero $v_0\in V$ with $v_0\mapsto 0\in V^*$, or in other world, $f(x,v_0)=0$ for all $x$. But then $v\mapsto(x\mapsto f(v,x))$ cannot be an isomorphism, since any linear form $\mu\in V^*$ with $\mu(v_0)=1$ cannot be attained.

The same reasoning works the other way around, achieving the equivalence. $\square$

---

Whether and how this works in infinite dimensions, I have currently no idea.