$X'=AX$ with $\det A=0$ - Plane solutions Let $A$ be a $3\times3$ real matrix with $\det A=0$, we consider the system of differential equations defined by $X'=AX$. How can I show that each solution is contained in a plane ?

What I have so far :
$\det A=0\Rightarrow$ there are constant solutions to $AX=0,X\neq0$. Let $\lambda_0$ be the eigenvalue for $0$ and $E_0$ the set of associated eigenvectors.
If $A=0$, it's clear.
If $A\neq0$:


*

*if $\dim E_0=1:E_0=span(X_0)$. If we were in $\Bbb{C}$, we could say that there are two more eigenvalues $\lambda_1,\lambda_2$ and thus the solutions of $X'=AX$ would be $c_0E_0+c_1E_{\lambda_1}e^{\lambda_1t}+c_2E_{\lambda_2}e^{\lambda_2t}$, which defines a plane. However, I doubt that by going back to $\Bbb{R}$ by taking the real part, that is still the equation of a plane.

*if $\dim E_0=2$ :  ?

 A: Let $\chi_A\in\mathbb R[X]$ be the characteristic polynomial of $A$. Since $\det(A)=0$, $$\chi_A(X)=X^kR(X)$$ where $1\leq k \leq 3$ is the multiplicity of $0$ in $\chi_A$ and $R\in \mathbb R[X]$.
Since $X^k$ and $R$ are coprime by definition, we get a decomposition of $\mathbb R^3$ by stable subspaces w.r.t. $A$ : $$\mathbb R^3= \ker(R(A))\oplus\ker(A^k).$$


*

*If $k=1$, then $R$ is a real polynomial of degree $2$ and the previous decomposition gives us $$A\sim \begin{pmatrix}\star &\star&0\\ \star&\star&0 \\ 0 & 0& 0\end{pmatrix}.$$

*If $k=2$, the $R$ is of degree $1$ and thus $$A\sim \begin{pmatrix} \alpha &0&0\\ 0&0&1 \\ 0&0&0\end{pmatrix} \text{ or } A\sim \begin{pmatrix}\alpha&0&0\\0&0&0\\0&0&0\end{pmatrix}$$
where $\alpha$ the unique root of $R$.

*If $k=3$ then $A$ is nilpotent and $$A = 0 \text{, } \quad A\sim \begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix} \quad \text{ or } \quad A\sim \begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}$$


So, up to a change of basis, the last coordinate of $X$ is constant and thus the solution leaves in a plane.
A: There is a left kernel vector $u$ with $u^TA=0$. The differential equation along this direction reads then as 
$$
(u^TX)'=u^TX'=u^TAX=0
$$
thus the component $u^TX$ is a constant and thus a first integral, the solution remains in this affine plane.
