Endomorphism Ring for a Module (Specifically, a Vector Space) I am currently trying to work through and understand an example provided regarding endomorphism rings for modules . I understand both concretely and intuitively that $End_{F}(F^n)=M_n(F)$. Here, $F$ is a field, and $n\in\mathbb{N}$.
To summarize my understanding of this, the way I see it is that we have module homomorphisms $\phi:F^n\rightarrow F^n$, and each of these maps can be represented by an $n\times n$ matrix with entries from the field $F$, and matrix multiplication gives us a linear map in the standard manner, thus giving us the aforementioned equality.
My confusion now comes in a following example provided to me, which states that $End_{M_n(F)}(F^n)=F$. I understand that $F^n$ can be seen as a module over the matrix ring $M_n(F)$. However, I do not believe I understand this example correctly. Applying a similar thought process as with the previous example, this seems to tell me that considering $F^n$ as a module over $M_n(F)$ only gives us maps that adhere only to scalar multiplication by an element of $F$, which does not seem true to me.
I am hoping that someone is able to provide intuition as to where my thinking is incorrect, along with perhaps a proper way to think about these examples, with perhaps some direction to a computation on this latter equality. I have seen a couple other posts, including this one, but still am a bit lost on the matter.
Thank you in advance for any light that can be shed on this matter!
 A: The ring $End_{M_n(F)}(F^n)=F$ consists of maps $\alpha\colon F^n\to F^n$ which are linear with respect to ${M_n(F)}$.  This means that:

*

*$\alpha(v_1+v_2)=\alpha(v_1)+\alpha(v_2)$ for all $v_1,v_2\in F^n$.


*$\alpha(Bv)=B\alpha(v)$ for all $B\in {M_n(F)}$ and $v\in F^n$.
If we just consider point 1. and point 2. in the special case where $B$ is a scalar matrix $B=\lambda1_n$ with $\lambda\in F$, then we see that:
$$End_{M_n(F)}(F^n)\subseteq End_{F}(F^n)=M_n(F)$$
That is, for every $\alpha\in End_{M_n(F)}(F^n)$, we have $A\in M_n(F)$ with $\alpha(v)=Av$ for all $v\in F^n$.
Now let us consider point 2. in full generality.  What does it say about $A$:$$ABv=BAv$$ for all $v\in F^n$.
That is, $A$ cannot be just any matrix, but rather it must be a matrix which commutes with every $B\in M_n(F)$.
Suppose $B$ has a single non-zero entry, and it is a $1$ on the diagonal.  Then $AB$ has a single column of $A$, with all other entries $0$.  Conversely $BA$ has a single row of $A$, with all other entries $0$.  So for $A$ to commute with this $B$, all its off-diagonal entries must be $0$.
That is $A=\left(\begin{array}{ccccc} \lambda_1\\&\lambda_2\\&&\lambda_3\\&&&\lambda_4\\&&&&\ddots
\end{array}\right)$
Now let $B$ be a matrix with a single non-zero entry, $B_{ij}=1$, for some $i\neq j$.  Then $AB=\lambda_i B$, whilst $BA=\lambda_j B$.
We conclude $\lambda_i=\lambda_j$ for all $i\neq j$, so $A=\lambda1_n$ with $\lambda\in F$.  Thus $End_{M_n(F)}(F^n)=F$.
