# tensor product with dual space

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

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