Show that the system $\dot{x}=Ax+Bu$ is controllable if the linear system of equations $XA-AX=0$, $XB=0$ admits only the trivial solution $X = 0$ 
Show that the system $\dot{x}=Ax+Bu$ is controllable if the linear system of equations $XA-AX=0$, $XB=0$ admits only the trivial solution $X = 0$ .

(The system is controllable iff the controllability matrix $[B, AB, A^{2}B, ..., A^{n-1}B]$ has full row rank)
 A: Note, this is not a complete answer: I don't know if this is related because I almost have no intuitive understanding of ranks but from the first equation we see that $X$ and $A$ commute. If we multiply the first equation by $B$ from the right we will obtain:
$$
X\,A\,B - A\,X\,B = 0 \implies X\,A\,B = 0.
$$
Then multiply from the left with $A$ and then use that $X$ and $A$ commute successively and write down the following equations:
$$
\begin{array}{c}
X\,B = 0 \\
X\,A\,B = 0 \\
X\,A^2B = 0 \\
\vdots \\
X\,A^{n-1}B = 0
\end{array}
$$
Or written as matrix
$$
X \begin{bmatrix}B & A\,B & A^2B & \cdots & A^{n-1}B \end{bmatrix}=0.
$$
Now knowing only little about the rank I would guess that the last equation implies what we are looking for, as it is similar to the case of an invertible matrix (which is square and has full rank). Hence, I would guess that if $X=0$ is the only solution to this equation then the system is controllable because the controllability matrix has to have full rank (similar to the invertible matrix).
Like I told this is just an attempt to solve this problem. We will have to wait for someone who is able to judge if this is a valid procedure.
