Image of $f$ consists of all the maps of finite rank.

Question

I have the following map:

$f: W \otimes V^* \rightarrow \textbf{Hom}(V,W)$

Where:

$f(w\otimes g)(v)= g(v)w$.

Both $V,W$ are vector spaces. And I need to prove that $f$ consists of all the maps of finite rank. But I am not really sure how to approach this.

Answer 1

Each $f(w\otimes g)$ has rank $1$ because its image lie in the line spanned by $w$. An arbitrary element of $W\otimes V^*$ is a finite sum of pure tensors of the form $w\otimes g$ and so anything in the range of $f$ is a finite sum of things like $f(w\otimes g)$, hence has finite rank.

Answer 2

The key point here is that a map has finite rank if and only if it can be written as a finite sum of maps with rank $1$.

It's easy to see that a sum of rank-1 maps has finite rank. In particular, the image of $\sum_j g_j(\cdot )w_j$ must lie in the span of the vectors $w_j$. The reverse inclusion is tricky, but one approach is as follows. Note that a finite rank map $\phi:V \to W$ can be decomposed into a "rank-factorization" $\phi = \phi_2 \circ \phi_1$ where $r$ is the rank of $\phi$, $\phi_1:V \to \Bbb F^r$, and $\phi_2:\Bbb F^r \to W$. If $g_j,w_j$ are defined such that $$ \phi_1(x) = (g_1(x),\dots,g_r(x)), \quad \phi_2(c_1,\dots,c_r) = c_1 w_1 + \cdots + c_r w_r, $$ then it follows that $\phi = \sum_{j=1}^r g_j(\cdot )w_j$.