Finding submodules of a module over a ring of matrices I'm studying for an algebra mid-term and was given this as one of my practice problems.
Let $E$ be an n-dimensional vector space over field k. And it has a basis $\{\beta_1, ..., \beta_n\}$. Let $M_n(k)$ be the ring of $n$ x $n$ matrices over $k$. We can say that $E$ is a $M_n(K)$ - module given the action $M_n(K)$ x $E \to E$ by $(A,v) \mapsto A\cdot v$. The question I have to answer this how many $M_n(K)$ submodules of $E$ exist. 
My answer: there are $n!$ submodules of $E$ that exist. Since for any linear transformation $A \in M_n(K)$, the image of the action is strictly determined on the action of $A$ on each $k_i\beta_i$ for $k_i \in k$. This is shown by $A\cdot v = A (k_1\beta_1 + ... + k_n \beta_n = Ak_1\beta_1 + ... + Ak_n\beta_n$. What I'm unsure of is each $A\cdot \beta_i$ a basis element of the image given by $A$? If it is, then can't I just say that the basis will be preserved by the transformation, so therefore any $m$-dimensional subspace of $E$ where $m\leq n$ is also a sub-module? 

Try to prove the following statement:

Let $v, w$ be any elements of $E$ with $v \neq 0$.  Then there exists an $A \in M_n(k)$ such that $Av = w$.

Now what can you say about the $M_n(k)$-submodules of $E$?
