For a vector space $\mathcal{V}$, let $\mathcal{O}_\ell(\mathcal{V})$ denote the space of linear operators $\mathbb{O}:\mathcal{V}\rightarrow\mathcal{V}$. I'm wondering if $$\mathcal{O}_\ell(\mathcal{V})\simeq\mathcal{V}\otimes\mathcal{V}^*,$$ where $\mathcal{V}^*$ denotes the dual space of $\mathcal{V}$.
I'm thinking an isomorphism could be constructed by looking at the action of an operator $\mathbb{O}\in\mathcal{O}_\ell(\mathcal{V})$ on a basis $\{|e_\alpha\rangle\}_{\alpha<\lambda}$ for $\mathcal{V}$ and then choosing the member $\sum_{\alpha<\lambda}\mathbb{O}|e_\alpha\rangle\otimes\langle e_\alpha|\in\mathcal{V}\otimes\mathcal{V}^*$ to assign it to. In other words, we would define $\Phi:\mathcal{O}_\ell(\mathcal{V})\rightarrow\mathcal{V}\otimes\mathcal{V}^*$ by $$\Phi(\mathbb{O})=\sum_{\alpha<\lambda}\mathbb{O}|e_\alpha\rangle\otimes\langle e_\alpha|$$ for an arbitrary basis $\{|e_\alpha\rangle\}_{\alpha<\lambda}$ and check that it is an isomorphism in the desired sense, but I'm under the impression that some subtleties come into play if $\lambda\geq\omega$ so $\mathcal{V}$ is infinite dimensional. Does this isomorphism always work, and if not why does it fail for $\lambda\geq\omega$?
There is a natural imbedding $$V\otimes V^{\star} \to \mathcal{O}_l(V)$$ and the images is the space of operators with finite dimensional image.