Phase portrait when the matrix is not diagonalizable I have the following matrix 
$$\begin{bmatrix} 0 & 0\\ -1& 0\end{bmatrix}$$
which has one eigenvalue $0$ with eigenvector $\begin{bmatrix} 0\\ 1\end{bmatrix}$.
I need to draw the phase diagram, but I don't know how to do this with only one eigenvalue equal to $0$.
 A: The linear system corresponding to the given matrix, which I shall denote by $A$, is
$\begin{pmatrix} \dot x \\ \dot y \end{pmatrix} = A\begin{pmatrix} x \\ y \end{pmatrix} =  \begin{bmatrix} 0 & 0 \\ -1  & 0 \end{bmatrix}\begin{pmatrix} x \\ y \end{pmatrix} , \tag 1$
that is,
$\dot x = 0, \tag 2$
$\dot y = -x; \tag 3$
from (2) it follows that $x(t)$ is constant:
$x(t) = x(t_0) = x_0, \tag 4$
and then (3) becomes
$\dot y = -x_0, \tag 5$
leading to
$y(t) = -x_0 t + y_0; \tag 6$
alternatively, one can rely on the well-known exponential formula
$\begin{pmatrix} x(t) \\  y(t) \end{pmatrix} = e^{At} \begin{pmatrix} x_0 \\ y_0 \end{pmatrix} = \exp \left (t \begin{bmatrix} 0 & 0 \\ -1  & 0 \end{bmatrix} \right )\begin{pmatrix} x_0 \\ y_0 \end{pmatrix}; \tag 7$
we evaluate the matrix exponential in (7) via the observation that
$A^2 = 0, \tag 8$
readily verified by direct calculation; thus the exponential power series
$e^{At} = \exp \left (t \begin{bmatrix} 0 & 0 \\ -1  & 0 \end{bmatrix} \right ) = \displaystyle \sum_0^\infty \dfrac{(At)^n}{n!} \tag 9$
is truncated after the first-order term, and so
$e^{At} = \exp \left (t \begin{bmatrix} 0 & 0 \\ -1  & 0 \end{bmatrix} \right ) = I + At = \begin{bmatrix} 1 & 0 \\ -t & 1 \end{bmatrix}; \tag{10}$
thus (7) becomes
$\begin{pmatrix} x(t) \\  y(t) \end{pmatrix} = \begin{bmatrix} 1 & 0 \\ -t & 1 \end{bmatrix}\begin{pmatrix} x_0 \\ y_0 \end{pmatrix} = \begin{pmatrix} x_0 \\ -x_0t  + y_0 \end{pmatrix} \tag{11}$
in accord with (4), (6).
Having developed an exact solution to (1), constructing/drawing the phase portrait is a relatively straightforward matter.  As pointed out by Moo in his/her comment to the question itself, $x$ is constant along each flow line (integral curve), so they are all lines parallel to the $y$-axis; since the velocity along each line is a constant $-x_0$, the lines run in the direction of decreasing $y$ in the right half-plane, and in the direction of increasing $y$ in the left.  Of course, the $y$-axis consists enirely of equilibria, since there $x_0 = 0$.
