General solution of equation $AX + X + A = 0$ where A is nilpotent matrix.

Question

We need to find the general solution for the following equation: $$AX + X + A = 0$$ Here is the very beginning of my solution: $$AX + X = -A$$ but $A$ is nilpotent matrix, so $-A$ is also nilpotent, then $(AX + X)$ is nilpotent too, so $$\det(AX + X) = 0$$ what should i do next to solve this problem?

Answer 1

Hint :

"Re-phrase" your expression, into

$$\mathbf{AX + X + A = 0 \Leftrightarrow (A+1)X = -A}$$

where $\mathbf{1}$ denotes the identity matrix, such that $\mathbf{1 \cdot X = X}$. Now, it suffices to show that a solution exists if and only if $\mathbf{A+1}$ is invertible, thus :

$$\mathbf{X = -A(A+1)^{-1}}$$

The solution will also be unique.

Nerdy edit : An interesting fact about inversibility is the correlation with the geometric series. Specifically, if you can show that $\mathbf{(A+1)^{-1}}$ is invertible and $\|A\| < 1$, then

$$\mathbf{\frac{1}{A+1}} = \sum_{i=0}^\infty (-1)^n\mathbf{A}^n$$

which is the exact and unique solution to the problem.

Answer 2

Factor the right hand side:

$$ (A+I)X = -A $$

and then to find $X$ you can use the fact that the identity plus nilpotent is invertible.