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Suppose I have real square matrices $A$ and $B$, where $A$ is symmetric and $B$ is skew-symmetric. I would like to solve the following Sylvester equation in symmetric matrix $X$

$$ A X - X A = B $$

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.

Rodrigo de Azevedo
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A. Pfelzeit
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2 Answers2

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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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Robert Israel
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  • I had no idea this existed! So as long as $A$ does not have an eigenvalue $\lambda_i = -\lambda_j$ for some $i,j$ then this is solvable? Okay, excellent. I can certainly solve a system of equations like that, I think. – A. Pfelzeit May 10 '16 at 04:55
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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 $\leq n^2/4$.

  • Okay, neat, I hadn't even considered the mapping for it. That might be a problem for me since the eigenvalues of $A$ aren't distinct. So I guess that means $\textrm{dim(ker(}f))$ is the number of distinct eigenvalues of $A$? The current $A$ I'm working with has eigenvalues $\lbrace -1^{(n-1)},n-1 \rbrace$. Maybe I can alter the $A$ I am working with. Thanks! – A. Pfelzeit May 12 '16 at 18:20
  • As to the edit -- when you are right, you are right. There are many solutions to the equation for my situation (unfortunately! $I \otimes A - A \otimes I$ is singular) and the particular solution from matlab's pinv() function happens to result in a matrix that is not the solution I wanted to see (I already know what $X$ I want, but I want to get it without knowing). That last statement, that is from the discriminant of $px^2 + nx + q$? I've never seen that done before, if you have the chance, how did that arise? – A. Pfelzeit May 13 '16 at 16:41
  • Thank you!! That is very clear and I would not have expected that very algebraic answer. I appreciate your explanation!! – A. Pfelzeit May 14 '16 at 16:56