Nicoletta Bof - Luca Fabietti (Edited by AF)
Proposition. Let A ∈ Rn×n and B ∈ Rm×m . If at least one of the two
matrices A and B is nilpotent, then the Stein equation P − AP B = Q admits
a unique solution P ∈ Rn×m for all Q ∈ Rn×m . This solution can explicitly
written as
P = Aν−1 QB ν−1 + Aν−2 QB ν−2 + . . . + AQB + Q.
(1)
where ν is the minimum between the nilpotency indices of A and B if both A
and B are nilpotent, it is nilpotency index of A if only A is nilpotent, and it is
nilpotency index of B if only B is nilpotent.
Proof: Let P be given by (1). We can compute P − AP B and, by taking
into account that at least one of Aν and B ν is zero (in view of the nilpotency
assumption), we easily see that P − AP B = Q, i.e. P given by (1) is a solution
of the Stein equation.
Uniqueness is clearly equivalent to the fact that equation P − AP B = 0
admits only the solution P = 0. This equation can be written as P = AP B
which immediately gives AP B = A2 P B 2 so that P = AP B = A2 P B 2 and by
iterating this argument, P = Aν P B ν = 0, where we have used the nilpotency
assumption.
Remark In the following we detail an explicit computation that leads to the
form (1) of the solution (which, in the statement of the proposition, was pulled
out of a cylinder). In doing this we suppose A nilpotent and that ν is its
nilpotency index: Aν = 0. By multiplying on the left by Aν−1 we get:
Aν−1 (P − AP B) = Aν−1 Q
⇒
Aν−1 P = Aν−1 Q.
Now by multiplying on the left by Aν−2 , and by using the previous equivalence we get:
Aν−2 (P −AP B) = Aν−2 Q ⇒ Aν−2 P = Aν−1 P B+Aν−2 Q = Aν−1 QB+Aν−2 Q = Aν−2 (AQB+Q).
Repeating a similar procedure:
Aν−3 (P −AP B) = Aν−3 Q ⇒ Aν−3 P = Aν−2 P B+Aν−3 Q = Aν−2 (AQB+Q)B+Aν−3 Q
Aν−3 P = Aν−3 (A2 QB 2 + AQB + Q).
Finally we get:
Aν−(ν−1) P = AP = A(Aν−2 QB ν−2 + Aν−3 QB ν−3 + . . . + AQB + Q)
so
P − AP B = P − A(Aν−2 QB ν−2 + Aν−3 QB ν−3 + . . . + AQB + Q)B = Q
then
P = Aν−1 QB ν−1 + Aν−2 QB ν−2 + . . . + AQB + Q.
is a solution of (1).
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