Consider two following equivalent statements about the stabilizability:
\begin{equation} \dot{x} = Ax + Bu \\ \end{equation} where $x \in R^n$.
$1.$ The above system is stabilizable if and only if no eigenvector of $A^T$ with eigenvalue $Re\{\lambda\} \geq 0$ belongs to null($B^T$).
$2.$The above system is stabilizable if and only if $rank[A-\lambda I \;\;\; B] = n$ for $\lambda$ satisfying $Re\{\lambda\} \geq 0$.
Now consider the following $A$:
A = $ \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $
$A^T$ has one eigenvalue which is $0$ with a multiplicity of $3$. The eigenvectors of $A^T$then are:
$v = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$
Now, if I choose $B^T$ as:
$B^T$ = $ \begin{bmatrix} 1 & 1 & 1 \end{bmatrix}$
then $v$ is not in the null space of $B^T$ since $B^Tv \neq 0$. However if you check
$rank[A-\lambda I \;\;\; B] = rank[A \;\;\; B] = 2$
which should be $3$ since the statement $1$ is satisfied with the chosen $B$.
It seems there is a contradiction between the statements. Can you tell me what's wrong in here?
If a matrix has repeated eigenvalues then any non-zero linear combination of their obtained eigenvectors would also satisfy as eigenvector for that eigenvalue. So in your case for example
$$ u = v_1 - v_2 = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} $$
also satisfies $A^\top u = \lambda\,u = 0$. However, $u$ does lie in the null space of $B^\top$.
So a more complete formulation for statement 1. would be:
A system is stabilizable if and only if for each eigenvalue $Re\{\lambda\}\geq0$ the intersect between the space spanned by the associated eigenvectors of $A^\top$ and null$(B^\top)$ is the origin.