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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?

Good Morning Captain
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  • You can't say that $A\cdot {\beta_i}$ is also a basis, for instance what if $A: \beta_i \mapsto \beta_i, \ i \neq j$ and $A : \beta_j \to 0$, which defines a linear transformation. – Daniel Donnelly Mar 07 '16 at 03:33

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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 $E$?

D_S
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  • Without finishing the proof, I wanted to think of the implication of the statement first. Does it mean that a $M_n(k)$-submodule must be closed under all transformations? In other words, the only submodules are the zero and whole module? – Good Morning Captain Mar 07 '16 at 06:31
  • That's correct. – D_S Mar 07 '16 at 13:23