Linear system of matrices

Question

I don't know Mathematica very well, I have an equation involving matrices of the following form: given two 5ドル\times5$ matrices $A,B$ where $B$ is nilpotent we want to find a matrix $X$ such that

k X + B.X-X.B + A = 0

for an integer $k$.

Is there a command solving such an equation? I tried with Solve and LinearSolve, but it does not work.

• azerbajdzan: What operation is [B,X]? What does not work?
• Henrik Schumacher: I guess [B,X] is supposed to be the commutator bracket. In Mathematica code that would be B.X - X.B.
• Vanja: yes exactly, it is the commutator. I have edited the question to clarify the notation

Answer 1

You can use LyapunovSolve which solves a.x + x.b = c:

SeedRandom[1] ;
A = RandomReal[{-1, 1}, {5, 5}];
B = RandomReal[{-1, 1}, {5, 5}];
k = RandomReal[{-1, 1}];
a = k * IdentityMatrix[5] + B ;
b = - B ;
c = - A ;
LyapunovSolve[a,b,c]

• Henrik Schumacher: Awesome. This should be the accepted answer.

Answer 2

This is a linear equation in X, so you can solve it, in principle by writing out the system matrix and the right-hand side like this:

A = RandomReal[{-1, 1}, {5, 5}];
B = RandomReal[{-1, 1}, {5, 5}];
k = RandomReal[{-1, 1}, {5, 5}];
X = Array[x, {5, 5}];
matrix = D[Flatten[k X + B . X - X . B], {Flatten[X], 1}];
rhs = Flatten[-A];
solution = ArrayReshape[LinearSolve[matrix, rhs], {5, 5}]

Of course, there might be way more clever solutions to this. E.g., we have not exploited, yet, that B is nilpotent. Probably you are supposed to multiply with B a couple of times from the left and from the right so that several terms cancel and the remainder can be used to eliminate some variables in the matrix X. Maybe the Jordan decomposition might be helpful.

• Vanja: nice thanks, I guess that the key is to use the flatten/arrayreshape.
• Henrik Schumacher: Yeah, Flatten reinterprets the 5ドル \times 5$-matrices as 25ドル$-vectors, D "linearizes" the system, and ArrayReshape undoes what Flatten did.

Answer 3

Without using any matrix theory just straightly writing an equation.

A = RandomInteger[{-10, 10}, {5, 5}];
B = RandomInteger[{-10, 10}, {5, 5}];
k = RandomInteger[{-10, 10}];
(* solution *)
# /. Solve[k # + B . # - # . B + A == 0] &@Array[x, Dimensions[A]]
(* verification *)
k # + B . # - # . B + A & /@ %

---

{{{0, 0, 0, 0, 0}, {0, 0, 0, 0, 0}, {0, 0, 0, 0, 0}, 
{0, 0, 0, 0, 0}, {0, 0, 0, 0, 0}}}