$$A=\begin{bmatrix} 0 & 0 & 1\\ 1 & -1 & 0\\ -1 & 1 & 1 \end{bmatrix}$$.
My attempt: The characteristic polynomial of this matrix is $x^{3}=0$. Now, at $x=0$, the $N(A-xI)=1$, this implies that at least one block of order 1.
$N(A-xI)^{2}=2$, this implies that at least one block of order 2.
the $N(A-xI)^{3}=3$, this implies that at least one block of order 1.
So, the resultant matrix is $$A=\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix}$$.
Is this solution is correct?