The simple modules of upper triangular matrix algebras. Let k be a field. $A= \begin{pmatrix} k& k& k\\ 0&k&k\\ 0&0&k \end{pmatrix}$ is the upper triangular matrix algebra. Then the projective module $P(2)=Ae_2=\begin{pmatrix} k\\k\\0\end{pmatrix}$ has a submodule $P(1)= \begin{pmatrix} k\\0\\0\end{pmatrix}$. Also we know that the cokernel of the inclusion map $i: P(1) \rightarrow P(2)$ is $S(2)$(the simple module correspond to $e_2$), I thought $S(2)$ is of the form $\begin{pmatrix} 0\\k\\0\end{pmatrix}$, but find this is not an $A$-module, so what is $S(2)$, in other words, what are the simple modules of $A$ written as $3 \times 1$ matrix?
 A: It would be good to figure out first how to write them as quotients of maximal left ideals.  Since the simple left modules of $A$ are the same as those of $A/J(A)$, and $A/J(A)\cong k\times k\times k$, it's easy to find that the maximal left ideals look like this:
$\begin{bmatrix}0&k&k\\ 0&k&k\\ 0&0&k\end{bmatrix}$
$\begin{bmatrix}k&k&k\\ 0&0&k\\ 0&0&k\end{bmatrix}$
$\begin{bmatrix}k&k&k\\ 0&k&k\\ 0&0&0\end{bmatrix}$
The resulting quotients (let's call them $S_1, S_2, S_3$ respectively) "look" like 
$$\left[\begin{smallmatrix}k&0&0\\ 0&0&0\\ 0&0&0\end{smallmatrix}\right] \left[\begin{smallmatrix}0&0&0\\ 0&k&0\\ 0&0&0\end{smallmatrix}\right] \text{ and } \left[\begin{smallmatrix}0&0&0\\ 0&0&0\\ 0&0&k\end{smallmatrix}\right],$$ but these subsets of $A$ are not actually left ideals.
But still we can see each left ideal is only $1$-dimensional as a $k$ module, so we should be able to express them as $k$ with a funny action by $A$ on the left.
So in $S_1$, the action of $A$ on $S_1=k$ is given by $\left[\begin{smallmatrix}a&b&c\\ 0&d&e\\ 0&0&f\end{smallmatrix}\right]x:=ax$
in $S_1$, the action of $A$ on $S_2=k$ is given by $\left[\begin{smallmatrix}a&b&c\\ 0&d&e\\ 0&0&f\end{smallmatrix}\right]x:=dx$
in $S_1$, the action of $A$ on $S_1=k$ is given by $\left[\begin{smallmatrix}a&b&c\\ 0&d&e\\ 0&0&f\end{smallmatrix}\right]x:=fx$
I understand the idea to write them as a $3\times 1$ matrix because, for example, the simple left module of a simple Artinian ring embeds in the ring as a minimal left ideal, and so in that case you can exhibit a simple left module as columns within the matrix.
If that's what you had in mind, then the problem here is that upper triangular matrix rings (over fields, with $n>1$) do not contain copies of all their simple left modules as left ideals (that is, they aren't left Kasch.)
As you discovered, $S_1$ embeds as $\left[\begin{smallmatrix}k&0&0\\ 0&0&0\\ 0&0&0\end{smallmatrix}\right]$ and you could conceivably just use the first column to model $S_1$ with matrix multiplication on the left by $A$, but the analogous thing does not work for $S_2$ or $S_3$. It's not hard to see that the left socle of $A$ is $\left[\begin{smallmatrix}k&k&k\\ 0&0&0\\ 0&0&0\end{smallmatrix}\right]$ and is the direct sum of three copies of $S_1$, so there is no room for $S_2$ or $S_3$.
