Finding the Jordan Canonical Form of a Classical Adjoint of a Jordan Block

Question

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 $A (\text{adj} A) = 0_n?$

Answer 1

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.

Answer 2

I did a calculation for the $2\times 2$ block $A$ acting on the $2\times 2$ matrices. We have $\dim \operatorname{ker} \operatorname{adj}(A) = 2$, $\dim \operatorname{ker} \operatorname{adj}(A)^2 = 3$,$\dim \operatorname{ker} \operatorname{adj}(A)^3 = 4$ ( if $\operatorname{char} k = 0$ then we get $\operatorname{adj}(A)^2= 0$ already). So in the $2\times 2$ case ($0$ characteristic) we see a Young diagram with columns of size $1$ and $3$.

I think that for characteristic $> n$ or $0$, we have a Young diagram with columns of sizes $1$, $3$, $\ldots$, $2n-1$, and thus Jordan cells of these sizes. I need to get back with more details.

$\bf{Added:}$ The result above is correct. It could be understood as follows. $A= X$ is part of an $sl2$-triple, and $X$ acts on $S_{n-1}$ of dimension $n$. Now the action on the $n\times n$ matrices is just $S_{n-1} \otimes S_{n-1}$ which decomposes according to Clebsch-Gordan.