# small rank solution of a matrix equation

Consider the matrix equation $$AX-XA = R$$ where $A$ and $R$ are given square matrices such that $\operatorname{rank}(R)=r$. How to establish conditions (necessary, sufficient, or both) on $A$ and (or) on $R$ which ensure that there exists a solution $X$ of rank smaller or equal than $r$?

Some observations:

The equation involves the commutator $[A,X]=AX-XA$ between $A$ and $X$, thus the solution can not be unique. In fact if $X_0$ is a particular solution, than all the possible solutions are of the form $$X_0 + B$$ where $B$ is a matrix which commutes with $A$, i.e. $[A,B]=O$.

If we assume that $A$ is normal with simple eigenvalues, then a unitary $U$ and an invertible diagonal $D$ exist such that $A=UDU^*$. Therefore, we can vectorize the equation as $$(U\otimes U)(D\otimes I -I \otimes D)(U\otimes U)^* vec(X)=vec(R)$$ The $n^2 \times n^2$ diagonal matrix $D\otimes I -I \otimes D$ has exactly $n$ zero eigenvalues.

Does this "vectorized" equation ($n^2$ linear system) give us some more information? What can we ask about $D$ to ensure that a solution $vec(X)$, or equivalently $(U\otimes U)^*vec(X)$, has rank less or equal than $r$?

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There is no $B$ in your question. I assume you meant to write $AX-XB=R$. – Gil Feb 12 '13 at 17:00
Naive thougt: It's a linear transformation on X. Write the representing matrix in the standrat basis and use, e.g. Gauss elimination. – Lior B-S Feb 12 '13 at 17:59
@LiorB-S Not a terrible thought. Just a little naive since that is the idea on which the Kronecker product $\otimes$ is based. – adam W Mar 8 '13 at 21:27

The equation $AX-XB=R$ is known as the Sylvester equation.

In general, the solution to this equation is rather complicated and depends on the spectras of $A$ and $B$. Let $\sigma(A)$ and $\sigma(B)$ denote the spectras of $A$ and $B$. If their spectra is disjoint, then there is a unique solution.

If $\sigma(B)$ is inside a sphere whose centered at $0$ with radius $r$, and $\sigma(A)$ is outside the sphere, then the solution is given by:

$$X=\sum_{n=0}^{\infty} A^{-n-1}RB^n$$

Does it says something about the rank of $X$ ? I do not think so, because of the summation.

If $\sigma(A)$ is on the right half plane and $\sigma(B)$ is on the left half plane, the solution is given by:

$$X=\int_0^{\infty}e^{-tA}Re^{tB}dt$$

In general, there are many other cases and it's too long to write, but you can find it in "Matrix Analysis" by R. Bhatia, pages 203-210.

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Thank you for the reference. Note however that the equation I am considering involves the commutator of $A$ and $X$, i.e. is of the form $AX-XA$, thus the spectra $\sigma(A)$ and $\sigma(B)$ coincide and the solution can not be unique. – F. T. Feb 13 '13 at 9:49
That's true. When I read it at first I saw $B$ mentioned below. I probably just misread or confused it with $R$... – Gil Feb 14 '13 at 17:30