# Is the space of linear operators on a vector space $\mathcal{V}$ isomorphic to $\mathcal{V}\otimes\mathcal{V}^*$?

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 $\lambda\geq\omega$?

• Be aware that which kind of basis (e.g. Hamel vs. Schauder), and which kind of dual space (e.g. the regular dual space vs. the continuous dual space), and which kind of tensor product are likely to affect the answer. – user14972 Sep 23 '17 at 9:37
• @Hurkyl I think that these are the subtleties that I'm unaware of -- I am not familiar with any delineation you mentioned aside from the regular vs. continuous dual. An answer which highlighted the possible consequences of different choices would be greatly appreciated! – Alec Rhea Sep 23 '17 at 9:42

## 1 Answer

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.

• So all members of $\mathcal{V}\otimes\mathcal{V}^*$ correspond to linear operators on $\mathcal{V}$ with finite dimensional images, but members of $\mathcal{O}_\ell(\mathcal{V})$ with infinite dimensional images will not correspond to members of $\mathcal{V}\otimes\mathcal{V}^*$? – Alec Rhea Sep 23 '17 at 9:35
• @Alec Rhea: correct. Like the identity operator if $V$ is infinite dimensional – orangeskid Sep 23 '17 at 9:36
• Thanks; is $\mathcal{V}^∗$ the space of bounded linear operators here or just the space of all (possibly unbounded) linear operators? – Alec Rhea Sep 23 '17 at 9:37
• $V^{\star}$ is here the space of linear functionals ( so no topology on $V$). If you want to talk about a topological vector space ( say Hausdorff to avoid problems), you can take to $V^{\star}$ to be the continuous linear functionals. Then in this case the image will be the space of linear continuous operators of finite rank – orangeskid Sep 23 '17 at 9:41
• Much appreciated! – Alec Rhea Sep 23 '17 at 9:43