Is $V$ a simple $\text{End}_kV$-module? Let $V$ be a finite-dimensional vector space over $k$ and $A = \text{End}_k V$. Is $V$ a simple $A$-module?
 A: A detailed answer was requested, so here it is.
The answer is yes (unless $V$ is the zero vectorspace). And we don't need finite-dimensionality of $V$, at least if we accept the axiom of choice. First, a convention: whenever $V$ is a $k$-vectorspace, write $\hat{V}$ for $V$ viewed as an $\text{End}_k V$-module. Let us now unpack the definitions.

Observation A. $\hat{V}$ is simple iff:
  
  
*
  
*$\hat{V}$ has a non-zero submodule, and
  
*every non-zero submodule $X$ of $\hat{V}$ has the property that $X=\hat{V}.$
  

That's just the definition. But with a bit of thought, we see that this can be written in terms of $V$ only.
In particular:

Observation B. $\hat{V}$ is simple iff:
  
  
*
  
*$V$ has a non-zero subspace, and
  
*every non-zero subspace $X$ of $V$ has the property that if for all endomorphisms $\varphi$ of $V$ we have $\varphi(X) \subseteq X$, then $X=V$.
  

To see that the two condition 1's are equivalent, notice firstly that since every submodule of $\hat{V}$ is a subspace of $V$, hence the forward direction is easy. The backward direction takes a tiny bit more work; start with a non-zero subspace $X$ of $V$, deduce that $V$ is non-zero, or in other words that $\hat{V}$ is non-zero, and hence that $\hat{V}$ is a non-zero submodule of itself.
The two condition 2's are seen to be equivalent simply by unpacking the definitions.
Clearly, unless $V$ has a non-zero subspace, it cannot be the case that $\hat{V}$ is simple.
So, let us conjecture:

Proposition. If $V$ has a non-zero subspace, then $\hat{V}$ is simple.

Proof.
We need to show (1) and (2) from Observation B. Condition (1) is immediate. So suppose we're given a non-zero subspace $X$ of $V$ such that for every endomorphism $\varphi$ of $V$, we have $\varphi(X)\subseteq X.$ The goal is to show that $X=V$. Since $X \subseteq V$ is obvious, we need to show $V \subseteq X$. So consider arbitrary $v \in V$. The goal is to show that $v \in X$.
Since $X$ is non-zero, we can find non-zero $x \in X$. I claim that, to achieve our goal, it suffices to find an endomorphism $\varphi$ of $X$ such that $\varphi(x)=v$. For suppose such an endomorphism exists. Then since $x \in X$, we have $\varphi(x) \in \varphi(X)$, so $\varphi(x) \in X$, so $v \in X$.
In summary, the set up is:

Assume
  
  
*
  
*$V$ is a vectorspace
  
*$x$ is a non-zero element of $V$
  
*$v \in V$
  
  
  Show 
  
  
*
  
*there exists an endomorphism $\varphi$ of $V$ such that $\varphi(x)=v$.
  

Now since $k$ is a field and $x$ is non-zero, hence $\{x\}$ is linearly independent. Hence by a fairly deep theorem of linear algebra, the set $\{x\} \subseteq V$ extends to a basis $B$ of $V$. This doesn't assume finite-dimensionality of $V$ (as long as we accept the axiom of choice; otherwise, we have to be more careful.)
Now define a function $f : B \rightarrow V$ as follows: $$f(b) = \begin{cases}x & b = x \\ 0 & b \neq x\end{cases}$$ Then there exists a unique linear transform $\varphi : V \rightarrow V$ that agrees with $f$ on $B$. We have: $$\varphi(x) = f(x) = \left(\begin{cases}x & x = x \\ 0 & x \neq x\end{cases}\right) = \left(\begin{cases}x & \mathrm{True} \\ 0 & \mathrm{False}\end{cases}\right) = x$$
Ergo, there exists an endomorphism $\varphi$ of $V$ with $\varphi(x)=v$. This completes the proof.
A: Hint: The answer is yes (as is often the case for finite dimensional vector spaces).  Note that for any $v \in V \setminus \{0\}$, we may select maps $T_1,\dots,T_n \in A$ so that $\{T_j(v)\}$ forms a basis (or a spanning set, if you prefer).
A: Simply look at the definition: Recall that a module $M$ over a ring $R$ is simple if the only non-zero submodule of $M$ is $M$ itself. Equivalently, for any $x\neq 0$ in $M$, the submodule $Rx=\left\{rx:r\in R\right\}$ generated by $x$ is $M$.
Now in your case: Take a nonzero $x\in V$ and any $y\in V$. Is it true that there exists $T\in A=\operatorname{End}(V)$ with $Tx=y$?
