I have been looking at this specific case of a Sylvester equation for the square matrix $X$,
$$
AX-XA=-A,
$$
given a nilpotent square matrix $A$.
For a general Sylvester equation
$$
AX + XB = C,
$$
we should have a unique solution if and only if $A$ and $-B$ have distinct eigenvalue. So we definitely do not have uniqueness in our specific case.
But what can we say about existence?
What I have so far
1
I have tried writing our everything using the vectorization and
Kronecker product, but I'm not sure who to use this.Our equation becomes $$ \left( I_N \otimes (-A) + A^T \otimes I_N
\right)\mathbf{vec}(X)=-\mathbf{vec}(A), $$ where
\begin{equation}
\begin{aligned}
\left( I_N \otimes (-A) \right)_{n,m} &= -A_{n (mod N), m (mod N)} \\
\left( A^T \otimes I_N \right)_{n,m} &= A_{\lfloor m/N \rfloor, \lfloor n/N \rfloor}\, \delta_{n (mod N), m
(mod N)} \\ \mathbf{vec}(X)_{k} &= (X)_{k-\lfloor k/N \rfloor, \lfloor
k/N\rfloor+1} \end{aligned} \end{equation}2
I have one specific ad hoc solution for a given
$(A)_{i,j}= \delta_{i,j+1}$, $$ (X)_{i,j} = j \delta_{i,j} $$
Best Answer
If $A$ is not nilpotent, then no solutions (Jacobson). Let $A$ be $n\times n$ nilpotent. Then we may assume that $A=diag(J_{i_1},\cdots,J_{i_r})$ where $J_k$ is the nilpotent Jordan block of dimension $k$.
The general solution of our equation is $X=X_0+C(A)$ where $C(A)$ is the commutant of $A$ and $X_0$ is a particular solution. Then it remains to obtain some $X_0$; it suffices to obtain a particular solution of
$J_kX-XJ_k=-J_k$.
There is a diagonal solution $X_0$, which is
$diag(0,-1,-2,\cdots)$, and we are done.