How to obtain the state matrix of this trajectory? Continuous-time LTI case.
I have a problem getting the state matrix of this trajectory.

One element of the state matrix is known.
$$ A = \begin{pmatrix} a & 4  \\c & d  \end{pmatrix}
$$
I know that one Eigenvalue is 0, $s_1 = 0$, and one Eigenvector is $p_1=\begin{pmatrix} 1\\1  \end{pmatrix}$.
With the equation $s_1 - a*p_{11} + s - 4*p_{12} = 0$, i get $a = -4$.
I also found out that $c=-d$.
But I am stuck now. I don't know how to get the other elements.
Is there more information to get from the image?
Help appreciated...
German version:
Für ein freies System  2. Ordnung der Form $\frac{dx}{dt} = A*x$ ist das dazugehörige Trajektorienbild gegeben.
Außerdem ist die zugehörige Systemmatrix  A (teilweise) bekannt:
$$ A = \begin{pmatrix} a & 4  \\c & d  \end{pmatrix}
$$
Bestimmen Sie die fehlenden Elemente der Systemmatrix  A . 
(HINWEIS: Benützen Sie das angegebene Trajektorienbild!) 
Translation:
The trajectory image associated with a free system of second order of the form $\frac{dx}{dt} = A*x$ is given.
Further, the associated system matrix $A$ is (partially) known:
$$ A = \begin{pmatrix} a & 4  \\c & d  \end{pmatrix}
$$
Determine the missing elements of the system matrix $A$.
(HINT: Use the given trajectory image!)
 A: If ruhezone is the set of the equilibrium points, then $\dot x=0$ when $x_1(t)=x_2(t)$, and if those are arrows pointing towards the $x=y$ line, the elements of the eq. point set are stable. The question, as far as I understand, wants you to make a qualitative assesment of how this line is the attractor of the trajectories in its neighborhood. 
From the first hint, $A_{12} = 4$ we obtain 
$$
\pmatrix{\dot x_1\\\dot x_2} = \pmatrix{-4 &4\\ c&d}\pmatrix{x_1\\  x_2}
$$
Just test for yourself, whenever $x_1>x_2$, $\dot x_1<0$ and $x_1<x_2$, $\dot x_1>0$. 
Now, this is where I might mistaken: if those arrows are for vector field annotation and horizontal arrow means the vector field has no $x_2$ contribution, that means $\dot x_2$ is zero everywhere no matter what. Hence $c=d=0$.
$$
\pmatrix{\dot x_1\\\dot x_2} = \pmatrix{-4 &4\\ 0&0}\pmatrix{x_1\\  x_2}
$$
Indeed, 0 is an eigenvalue and $\begin{pmatrix} 1\\1  \end{pmatrix}$ is the eigenvector together with -4 and $\begin{pmatrix} 1\\0  \end{pmatrix}$.
