Matrix Equation similar to Lyapunov

Question

I came across with the following type of matrix equation, \begin{equation} X^TA-AX=QX,\ \ \text{and} \ X^{-1}=X \hspace{15pt} (1) \end{equation} for $A,Q$ known symmetric matrices. This, reminds Lyapunov matrix equation, $$BX + XB^T + Q = 0.$$ Although, in Lyapunov the unknown matrix is different as in my equation. I am wondering if there is any method to solve (1).

Answer 1

Your equation has nothing to do with the Sylvester one. It is more complicated.

We assume that the matrices are real. Here $A,B=A+Q$ are $n\times n$ symmetric matrices and we consider the matrix system in the unknown $X$

$(*)$ $X^TA=BX,X^2=I_n$. In general, $(*)$ has no solutions.

EDIT

$\textbf{Proposition}$. Assume that $(*)$ has solutions.

i) Then $A,B$ have same signature.

ii) If moreover, $rank(A)=n$, we put $spectrum(A^{-1}B)=(\lambda_i)$. Then, there are $i,j$ s.t. $\lambda_i\lambda_j=1$.

iii) $\det(A)=\det(B)$.

$\textbf{Proof}$. i) is a consequence of $X^TAX=B$ with $X$ invertible. Note also that $X^TBX=A$.

ii) Let $U=A^{-1}B$. Then $X=UXU$ and $U^{-1}X-XU=0$.

The linear function $f:Y\mapsto U^{-1}Y-YU$ satisfies $spectrum(f)=(1/\lambda_i-\lambda_j)_{i,j}$; since $\ker(f)\not= 0$, we are done. $\square$

$\textbf{Remark}$. The $3$ above conditions do not suffice to obtain solutions. We give $2$ examples for $n=4$,

1. $A=diag(2,3,7,1/42),B=diag(2,\begin{pmatrix}5&3\\3&9\end{pmatrix},1/72)$. Although $A,B>0,\det(A)=\det(B)=1$ and $spectrum(U)=\{1,?,?,?\}$, $(*)$ has no solutions.

2. $A=diag(2,3,7,1/42),B=diag(2,\begin{pmatrix}5&2\\2&9\end{pmatrix},1/82)$. Here $A>0,B>0,\det(A)=\det(B)=1$ and $spectrum(U)=\{1,1,a,1/a\}$, where $a>0$; $(*)$ admits an infinity of slutions.