Natural Isomorphism between $T_1^1(V)$ and End$(V)$

Question

I'm a little stuck on showing that there is a natural isomorphism between the $\mathbb{R}$ vector space of $(1,1)$ tensors, and the $\mathbb{R}$ space of of linear maps $T:V\to V$. The hint is define the map $\phi:$End$(V)\to T_1^1(V)$ by $(\phi A)(\omega, X)=\omega(AX).$ I can show that this map is linear and injective, but I'm having trouble proving it's surjective without introducing a basis for $V$.

Answer 1

The space of tensors is $\newcommand\RR{\mathbb R}\newcommand\Bil{\operatorname{Bil}}\Bil(V^*,V;\RR)$, the space of bilinear maps $V^*\times V\to\RR$, and this is the same as $\hom(V^*\otimes V,\RR)$ which,by adjunction, is the same as $\hom(V^*,\hom(V,\RR))=\hom(V^*,V^*)$.

Your map is therefore a function $\phi:\hom(V,V)\to\hom(V^*,V^*)$, up to several natural identifications. In this avatar, it is given by $\phi(A)(\omega)(X)=\omega(AX)$. Computing, we find that this map is an antihomomorphism of $\RR$-algebras; this is no surprise, as this is just transposition.

Now the proper bilateral ideals of $\hom(V,V)$ are indexed by the set of infinite cardinals less than or equal the dimension of $V$ (to each such cardinal $\tau$ we attach the ideal of all maps $V\to V$ with image of dimension less than $\tau$); see this. Given that, the map $\phi$ cannot be bijective when $V$ is infinite dimensional because (I think!) the cardinal of the set of the ideals in $\hom(V,V)$ and in $\hom(V^*,V^*)$ are different.