Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $r$?

share|cite|improve this question
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:

\begin{equation} X=\sum_{n=0}^{\infty} A^{-n-1}RB^n \end{equation}

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:

\begin{equation} X=\int_0^{\infty}e^{-tA}Re^{tB}dt \end{equation}

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.