What is the rank of this linear map defined on big and abstract spaces.

Question

Let $V$ be a finite-dimensional space, and let ${\cal L}(V)$ denote the space of all endomorphisms of $V$.

For any $\phi \in {\cal L}({\cal L}(V))$, there is a unique bilinear map ${\cal L}(V) \otimes {\cal L}(V) \to {\cal L}(V)$ (which we denote by ${\Phi}(\phi)$), satisfying

$$ \Phi(\phi) (a \otimes b)=\phi(ab) $$

for any $a,b\in {\cal L}(V)$. The $\Phi$ thus defined is obviously linear in $\phi$. What is its rank ?

Answer 1

Write $n=dim_k V$. Then $dim \mathcal{L}(\mathcal{L}(V))=n^4$ (and $dim Hom(\mathcal{L}(V)\otimes \mathcal{L}(V), \mathcal{L}(V)=n^8$).

We have $rank(\Phi)=dim Im \Phi=\dim V-\dim ker\Phi$. So we compute the kernel of $\Phi$:

\begin{align} ker(\Phi)&=\{\phi|\Phi(\phi)=0\}\\ &=\{\phi|\Phi(\phi)(a\otimes b)=0\forall a,b\}\\ &=\{\phi|\phi(a\circ b)=0\forall a,b\}\\ &=\{0\} \end{align}

Hence $\Phi$ is injective and the dimension of the image is $n^4$.