# Specific Sylvester equation. Existence without uniqueness

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{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}

2

I have one specific ad hoc solution for a given $$(A)_{i,j}= \delta_{i,j+1}$$, $$(X)_{i,j} = j \delta_{i,j}$$

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.