6

I am interested in stating existence of solution of a Sylvester equation

$$ AX - XB = C, $$

where $A$, $B$, $C$, and $X$ are $(n,n)$ matrices.

Existence of a unique solution $X$ is given, if $A$ and $B$ do not have an eigenvalue in common.

But what about the nonregular case. Are there results out there, that consider the case where $A$ and $B$ possibly share an eigenvalue.

I am only aware of the equivalence, that there exists a solution $X$, if and only if $$ \begin{bmatrix} A & C \\ 0 & B \end{bmatrix} \backsim \begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}, $$ where $\backsim$ is the equivalence relation. However, this didn't give me much insight into my particular setup.

Rodrigo de Azevedo
  • 23,223
Jan
  • 1,028
  • Did you look at the proof in the wikipedia article on the topic? It says there is no solution if $A$ and $B$ have any eigenvalue in common. – balddraz Jul 07 '15 at 12:53
  • Nonono. It says only that a possible solution can not be unique in this case. – Jan Jul 07 '15 at 13:39
  • Really? It literally says there is no solution, unique or non-unique, in the last sentence: "Then $AX+XB=C$ has no solution $X$, as is clear from the complex bilinear pairing $<(AX+XB)v,w>=<Cv,w>=<\overline {w},w>$; the right-hand side is positive whereas the left is zero." – balddraz Jul 07 '15 at 13:55
  • 1
    Ok. I think I am misunderstanding something here. Given $A$ and $B$ that have an eigenvalue in common, are you searching solutions for any $C$ of your choice? Or are you perhaps asking what form $C$ must have for $AX-XB=C$ to have a solution? If you are doing the former, then, as the last sentence of the article says, there is at least one $C$ for which you will get no solutions at all. If you are doing the latter, I retract my statements. – balddraz Jul 07 '15 at 14:14
  • Yes, I was just going through the proof. The difference is in 'for all $C$'. In fact, I have $C$ given and I look for conditions when it admits a solution. – Jan Jul 07 '15 at 14:18

0 Answers0