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I will explain what I know, and then I will ask my question. Let $V$ and $W$ be vector spaces such that at least one is finite dimensional. In class, we showed that if either $V$ or $W$ is finite dimensional, then $W \otimes V^* \cong \operatorname{Hom}(V,W)$. We set up $\hat{e} : W \times V^* \to \operatorname{Hom}(V,W)$ with $\hat{e}(w,f)(v) = f(v)w$. This induced the linear map $e : W \otimes V^* \to \operatorname{Hom}(V,W)$ where $\hat{e} = e \otimes$.

I understand why $e$ is injective, but I do not understand why it is surjective. I understand that any linear map $T: V \to W$ has finite rank (given that at least one of $V$ or $W$ has finite dimension), which gives me a finite basis of $im(T)$, but I do not know how to proceed. Any help would be great.

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Why should the map be injective? – user Mar 23 '13 at 15:23
@user: It is e that is injective; not e^. After moding out by the kernel, the map becomes injective. – user99680 Jun 4 '14 at 4:04
up vote 5 down vote accepted

Let $ T \in \text{Hom}(V,W) $. As $ \text{Range}(T) $ is a vector subspace of $ W $, we can find an ordered basis $ (\mathbf{w}_{i})_{i \in I} $ for $ \text{Range}(T) $.

As every element of a vector space has a unique expansion in terms of a given basis of the vector space, we see that for each $ i \in I $, there exists a unique linear functional $ T_{i} \in V^{*} $ such that $$ \forall \mathbf{v} \in V: \quad T(\mathbf{v}) = \sum_{i \in I} {T_{i}}(\mathbf{v}) \cdot \mathbf{w}_{i}. $$ Consider the sum $ \displaystyle \sum_{i \in I} \mathbf{w}_{i} \otimes T_{i} \in W \otimes V^{*} $ of pure tensors. This is a finite (hence well-defined) sum for the following reasons:

  • If $ W $ is finite-dimensional, then $ I $ is finite.

  • If $ V $ is finite-dimensional, then $ \text{Range}(T) $ is finite-dimensional, which makes $ I $ finite.

It is important to note that $ \displaystyle \sum_{i \in I} \mathbf{w}_{i} \otimes T_{i} $ can be viewed as a linear mapping from $ V $ to $ W $ in the following manner: $$ \forall \mathbf{v} \in V: \quad \left[ \sum_{i \in I} \mathbf{w}_{i} \otimes T_{i} \right](\mathbf{v}) ~ \stackrel{\text{def}}{=} ~ \sum_{i \in I} {T_{i}}(\mathbf{v}) \cdot \mathbf{w}_{i}. $$ This linear mapping is clearly $ T $ itself. Consequently, as $ T $ is arbitrary, we see that every $ T \in \text{Hom}(V,W) $ can be associated with an element of $ W \otimes V^{*} $, which gives us the surjectivity that we need.

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Thanks. This was helpful. – nigelvr Feb 24 '13 at 3:01
@nigelvr: You’re welcome! – Haskell Curry Feb 24 '13 at 3:42
Isn't e automatically injective when e^ is injective? – user99680 Jun 4 '14 at 4:06

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