Computing quotient representations and Hom set fort wo representations Consider the representation $M$ defined by

We want to find all subrepresentations quotient representations of $M$, and $\mathrm{Hom}(M,N)$, where $N$ is a representation with  $N \cong M$. 
I put B in Jordan normal form $\begin{pmatrix}0&1\\0&0\end{pmatrix}$, I assumed $v=\begin{pmatrix}x\\y \end{pmatrix} \in K^2$, and I computed $\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}x\\y \end{pmatrix}=\begin{pmatrix}y\\0 \end{pmatrix}$ which forces us to choose $y=1$ reducing $A$ to $\begin{pmatrix}0\\1 \end{pmatrix}$ .  So we get $M \cong M'$, where $M'$ is the representation given by

Since $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ is indecomposable, $M'$ is indecomposable. Moreover, we have three choices for the nonzero subrepresentations of $M$:

I could not figure out how to find neither quotient representations of $M$ nor $\mathrm{Hom}(M,M')$. Could you guys help me to understand this problem?
thanks,
 A: Your working is wrong.  Your task is to find subspaces of these vector space, and to find a mapping for $A,B$ compatible with the given one.  I will show you working in the following.
Also, (this is not important, but...) you ask for $Hom(M,N)$ with $N\cong M$...so why don't you say you want $Hom(M,M)$?
Now, first of all, what field $K$ are you working on?  
If char $K=2$, then the matrices $A$ and $B$ are all zero.  This means $M$ is just a semi-simple representation.  With 1 copy of simple corresponding to the first vertex, and two copies of the simple corresponding to the second vertex.  Any direct summand of this semi-simple rep is obviously the (non-trivial) subreps of $M$.  Quotients are now easy to see.
Since $M$ is semi-simple, say $M=M_1\oplus M_a\oplus M_b$ where $\dim M_1=(1,0)$ and $\dim M_a=\dim M_b=(0,1)$, the Hom-space $Hom(M,M)=Hom(M_1,M_1)\oplus Hom(M_2\oplus M_3,M_2\oplus M_3)$, which is a 5-dimensional $K$-space.
If char $K\neq 2$, then choose new basis for $K^2 := Kv_1\oplus Kv_2$ by $K(v_1+v_2)\oplus K(v_1-v_2)$.  Now with respect to this new basis, $B$ is rewritten $\begin{pmatrix} 0 & 0 \\ 4 & 0\end{pmatrix}$ and $A$ needs to be rewritten to $\begin{pmatrix}4\\0\end{pmatrix}$.
Note: I prefer such a choice of basis as it makes the action of the quiver on the representation much more apparent.  You can still do the following calculation using the original basis and matrices $A$ and $B$.
Start by picking $M_1=K(v_1-v_2)$ then it is closed under $A$ and $B$'s action, so is a subrep of $M$.  You may wonder if you can pick $K(v_1+v_2)$ as a subrep of $M$, but applying $B$ to this space gets you to $K(v_1-v_2)$, so the next subrep you can get is $M_2=K(v_1+v_2)\oplus K(v_1-v_2)$.  EDIT: In general, you pick a vector $\lambda(v_1+v_2)+\mu(v_1-v_2)$ and consider the $A$ and $B$'s action on it to see that if $\lambda,\mu$ both non-zero, you will just get back $M_2$.
Summarising, subrep's of $M$ are: $0\subset M_1\subset M_2\subset M$
Quotienting $M$ by these subrep's we have: $M\twoheadrightarrow M/M_1\twoheadrightarrow M/M_2 \twoheadrightarrow 0$.  Here $M/M_2$ is the simple rep with dimension vector $(1,0)$; $M/M_1$ is the representation with dimension vector $(1,1)$ and the map replacing $A$ is just the non-zero entry of $A$ and map replacing $B$ is zero.
From $M$ to itself, the obvious map is the identity mapping.  You can try to find other maps: so let's put a non-zero linear map, mapping the 1-dim $K$-space at the first vertex, so this can only be a scalar multiple of identity, now commutation of diagram forces you to take the map between the $K$-spaces of the second vertex to be also a scalar multiple of the identity $id_M$.
Next thing you try will be to put $0$ as the map between the first vertex, commutation forces you to get zero map between the second vertex.  So we can conclude $End(M)=K\cdot id$.
EDIT: Correction for calculation in char $K\neq 2$ case. Thanks Julian for pointing out the error.
