submodules of $V^n$ I'm stuck with Exercise 3 on page 25 of Kraft and Procesi's Primer on Invariant theory http://www.math.unibas.ch/~kraft/Papers/KP-Primer.pdf.  It goes as follows:
let $\rho:G\to GL(V)$ be an irreducible representation with $End_G(V)=K$, the ground field.  Show that every $G$-submodule of $V^n = V\otimes K^n$ is of the form $V\otimes U$ with a subspace $U\subset K^n$.
I'd be grateful for a hint...
 A: Since $V$ is irreducible, and by the uniqueness of direct sum decomposition, any $G$-submodule of $V^n$ is isomorphic to $V^m$ for some $m\leq n$.  Let $V^m\stackrel{j}{\to} V^n$ be the corresponding inclusion.  We want to prove that the image of $j$ has the form $V\otimes U$ for some subspace $U$ of $K^n$.
Since $End_G(V)=K$, we get that $Hom_G(V^m, V^n)$ is isomorphic to the space of $n\times m$ matrices with coefficients in $K$.  Thus $j$ is represented by a matrix $(\lambda_{ik})_{i=1,\ldots, m, k=1, \ldots, n}$, and the image of $j$ in $V^n$ is 
$$ Im(j) = \big\{ \big(\sum_{k=1}^m \lambda_{1k}v_k, \sum_{k=1}^m \lambda_{2k}v_k, \ldots, \sum_{k=1}^m \lambda_{nk}v_k \big) \ \big| \ (v_1, \ldots, v_m)\in V^m  \big\}. $$
Translating this as a submodule of $V\otimes K^n$, this gives
$$ \big\{ \sum_{k=1}^m \big( v_k \otimes (\sum_{\ell=1}^n \lambda_{\ell k}e_\ell) \big) \ \big| \ (v_1, \ldots, v_m) \in V^m   \big\}. $$
where $e_1, \ldots, e_n$ form the canonical basis of $K^n$.  Taking $U$ to be the subspace of $K$ generated by the $\sum_{\ell=1}^n \lambda_{\ell k}e_\ell$ for $k=1,\ldots, m$, the above set is equal to $V\otimes U$.
