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.

  • $\begingroup$ Why should the map be injective? $\endgroup$ – user Mar 23 '13 at 15:23
  • $\begingroup$ @user: It is e that is injective; not e^. After moding out by the kernel, the map becomes injective. $\endgroup$ – 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.

  • $\begingroup$ @nigelvr: You’re welcome! $\endgroup$ – Haskell Curry Feb 24 '13 at 3:42
  • $\begingroup$ Isn't e automatically injective when e^ is injective? $\endgroup$ – user99680 Jun 4 '14 at 4:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.