General solution to linear system with general form I am trying to find the general solution of the system 
$$X' = \begin{bmatrix}
    a & b  \\
    c & d
  \end{bmatrix}X$$
where $a+d \not= 0 $ and $ad-bc=0$
I find that the eigenvalues are 0 and $a+d$ with corresponding eigenvectors both $[0,0]$ which implies that the general solution is simply $X(t)=0$ but this solution does not seem right to me. Is there something I have done wrong? 
 A: We are given
$$X' = \begin{bmatrix}
    a & b  \\
    c & d
  \end{bmatrix}X\\\text{where}~ ~~a+d \ne 0 , ~~ad-bc=0$$
We can find the eigenvalues using the characteristic polynomial by solving $|A - \lambda I| = 0$, yielding
$$\lambda_{1,2} = \frac{1}{2} \left(-\sqrt{a^2-2 a d+4 b c+d^2}+a+d\right),\frac{1}{2} \left(\sqrt{a^2-2 a d+4 b c+d^2}+a+d\right)$$
We can then find the associated eigenvectors by solving $[A-\lambda_i]v_i = 0$, yielding
$$v_1 = \begin{pmatrix}
    -\dfrac{\sqrt{a^2-2 a d+4 b c+d^2}-a+d}{2 c}\\1
  \end{pmatrix}, v_2 = \begin{pmatrix} -\dfrac{-\sqrt{a^2-2 a d+4 b c+d^2}-a+d}{2 c}\\1\end{pmatrix}$$
Now, we can use the conditions we are given, namely $a+d \not= 0 ,ad-bc=0 $, by substituting $bc = ad$ in each eigenvalue/eigenvector pair yielding
$$\lambda_1 = 0, v_1 = \begin{pmatrix} -\dfrac{d}{c}  \\ 1 \end{pmatrix} \\ \lambda_2 = a + d , v_2 = \begin{pmatrix} \dfrac{a}{c} \\ 1 \end{pmatrix}$$
We can now write
$$X(t) = c_1~ e^{\lambda_1 t} ~v_1 + c_2 ~e^{\lambda_2 t}~ v_2 = c_1 \begin{pmatrix} -\dfrac{d}{c}  \\ 1 \end{pmatrix} + c_2~ e^{(a+d)t} \begin{pmatrix} \dfrac{a}{c} \\ 1 \end{pmatrix}$$
