Ytukyg

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?

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

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

which is the exact and unique solution to the problem.

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.