Relationship between controllability and null space of A, B For the dynamic system $\dot x = Ax + Bu$ 
There's a saying that this system is controllable when $Ker(B) \in Ker(A)$, which means that $u$ have the control in every dimension of $x$.
I have no problem with this theorem when you have single input $u$, which means $u$ is just a number and $B$ is with $n\times1$ dimension.
But I cannot completely understand when $u$ is a vector. Look at this example:
$A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 
                     0 &0&0&0\\0 &0&0&0\\0 &0&0&0\\ \end{bmatrix}$ , 
$B = \begin{bmatrix} 0&0\\0&0\\0&0\\0&0\\\end{bmatrix}$
This is obvious uncontrollable. However, look at the null space:
$Ker(A)=span\{\begin{bmatrix} 0\\1\\0\\0\\ \end{bmatrix},
            \begin{bmatrix} 0\\0\\1\\0\\ \end{bmatrix},
            \begin{bmatrix} 0\\0\\0\\1\\ \end{bmatrix} 
            \}$
$Ker(B)=span\{\begin{bmatrix} 1\\0 \end{bmatrix},
              \begin{bmatrix} 0\\1 \end{bmatrix}
            \}
$
Thus we have $Ker(B) \notin Ker(A)$, but this system is uncontrollable.
Can anyone help me with this contradictory?
Thank you very much and I appreciate all of your replies!
 A: By system space if you mean "the space of reachable states", then it is the range space of the controllability matrix
$C=\begin{bmatrix}B & AB & A^2B & \dots & A^{n-1}B\end{bmatrix}$
It is relatively easy to see this for discrete time systems, but also not much difficult for continuous time systems. It follows from the Cayley-Hamilton Theorem for both cases.
If $C$ has full rank, this means we can reach any state in the space $\mathbb{R}^n$, hence the system is said to be "completely controllable" (or reachable, which is the same thing for continuous time case).
The left eigenvector condition follows from the fact that column space (or range) of a matrix is the orthogonal complement of its left null space, see this. It means that $\text{Im}C = \mathbb{R}^n$ is the same condition with $\text{Ker}C^T = 0$, which is equivalent to PBH theorem, see this for proof.
So, if $A$ has a left eigenvector $w$, which is orthogonal to all columns of $B$, then $w^T C=0$, which means we cannot reach the states in the direction of $w$, i.e. $w \notin \text{Im}C$. This also means that uncontrollable subspace is the space spanned by the left eigenvectors of $A$ that are orthogonal to the columns of $B$ and its orthogonal complement is the controllable subspace, which is the space spanned by the columns of $C$.
For these reasons, controllability condition is also equivalent to $\text{rank} \begin{bmatrix}\lambda I - A & B\end{bmatrix} = n$ for all $\lambda \in \mathbb{C}$. So, I would like to think that, whenever $\lambda I - A$ looses a rank for some $\lambda$ (which is an eigenvalue of $A$), it must be completed with one of the columns of $B$, so that we can drive the system to whichever state we want by using the input corresponding to that column. It cannot be done if the corresponding left eigenvector is orthogonal to the columns of $B$, because it means that we also cannot reach $w$ by using the columns of $B$, i.e. $w \notin \text{Im}B$ and $w \notin \text{Im}(\lambda I - A)$.
I hope this is the intuition you are looking for.
Note: These facts do not change if $A$ is not diagonalizable, as we are not interested in the vectors that are not left eigenvectors of $A$, because they are already in $\text{Im}(\lambda I - A)$ for all $\lambda \in \mathbb{C}$.
