Finding the Jordan Canonical Form of a Classical Adjoint of a Jordan Block Let $A$ be a size $n$ Jordan matrix with $0$ on its diagonal, that is 
$$A = J_n(0) = [a_{ij}] = \begin{cases}
1, &j=i+1\\
0, &\text{elsewhere}
\end{cases}
$$
What is the Jordan Canonical Form of the classical adjoint of A, $\text{adj} A$?
Can we start with the fact that $A$ is singular and $A (\text{adj} A) = 0_n?$

If you just start computing the classical adjoint for $n=2,3,4...$ you should notice a pattern as to what they look like.
$$adj\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & - 1\\ 0 & 0\end{pmatrix}$$
$$adj\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$
$$ adj \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & -1\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}$$
Once you prove that this pattern holds, the Jordan Form is straightforward to compute.
