Solving for an unknown symmetric matrix using an answer found by a commutator.

Question

Suppose I have, for $A,X$ real square symmetric matrices, and $B$ skew-symmetric and real, $AX-XA=B$, with $B$ and $A$ known and $X$ unknown. What properties of $X$ need to be satisfied to find $X$ and how can I go about this? I am aware solutions are simple for $AX=B$ when $A$ is invertible, but I am unsure about $AX-XA=B$. I have only undergraduate knowledge of linear algebra.

Answer 1

Robert answered in the case when there is no hypothesis about $B$. When $B$ is skew ($B\in SK$), there are, in the generic case, always solutions in symmetric $X$ ($X\in S$).

Assume that the eigenvalues of $A$ are distinct and let $f:X\in S\rightarrow AX-XA\in SK$. Then $dim(\ker(f))=n$ (the set $P(A)$ of polynomials in $A$) and $rank(f)=(n^2+n)/2-n=(n^2-n)/2=dim(SK)$, that is, $f$ is onto. Finally, the set of solutions is in the form $X_0+P(A)$, where $X_0$ is a particular solution.

EDIT. $f$ is onto IFF the eigenvalues of $A$ are distinct.

Proposition. If $A$ has only $2$ distinct eigenvalues with multiplicities $p,q=n-p$, then $rank(f)=pq\leq n^2/4$.

Proof. We may assume that $A=diag(\lambda I_p,\mu I_q)$ with $\lambda\not=\mu$. It is easy to see that $dim(\ker(f))=(p^2+p)/2+(q^2+q)/2$ and to deduce that $rank(f)=(n^2+n)/2-dim(\ker(f))=pq$ that is $\leq n^2/4$.

Answer 2

This is a Sylvester equation. One necessary condition for a solution to exist is $\text{trace}(B) = 0$. The solution is never unique: you can always add a symmetric matrix that commutes with $A$ (e.g. any polynomial in $A$) to a solution and get another solution.

One way to find solutions when they exist is to think of the entries of $X$ as forming a vector of $(n^2+n)/2$ entries. The entries of $AX - XA - B$ give you $(n^2+n)/2$ linear equations in these entries. Solve this system.

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