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?
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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.
Rebellos
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it's easy to proof, that it's invertible, thank you, but why can we say that $AX + X = (A + 1)X$ and not $X(A + 1)$ ? – envy grunt Nov 24 '18 at 23:06
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1@envygrunt It really depends on how your matrices are defined, but generally no, the row of the multiplication operation regarding matrices is important. – Rebellos Nov 24 '18 at 23:08
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but in our problem it doesn't make sense, right? – envy grunt Nov 24 '18 at 23:11
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1Saying that $\mathbf{A\cdot B = B \cdot A}$ when talking about matrices isn't always true. For your given problem, it is : $$\mathbf{AX + X = A \cdot X + 1 \cdot X = (A+1)X}$$ Now, if more information can be yielded and you can show that the row of multiplication doesn't matter for your matrices, then it could be the other way as well, yes. – Rebellos Nov 24 '18 at 23:12
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thank you again, now it's clear – envy grunt Nov 24 '18 at 23:27
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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.
Matthew C
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