System of equations wrt self-adjoint operators

Question

$X = \left( \begin{matrix} 2&s\\ 8&2 \end{matrix} \right)$ and $Y = \left( \begin{matrix} 2&-1\\ 2&2 \end{matrix} \right)$ are two operators wrt the same orthonormal basis $B$ in a 2D scalar product space. Furthermore, $s$ is a fixed, complex-valued parameter.

The associated system of equations is $P^2Q^{-1} = X$ and $PQ = Y$, where $P = P^*$ and $Q = Q^*.$

1) Assume that this system of equations is guaranteed to have a solution. Determine the value of $s$ and show that the solution for $P$ and $Q$ is unique.

2) Using the value of $s$ found in part (1), compute $P$ and $Q$ wrt $B$.

Answer 1

We have $XY^\ast=P^3$. Therefore $XY^\ast=\begin{pmatrix}4-s&2s+4\\14&20\end{pmatrix}$ is self-adjoint and hence $s=5$. Note that $P,Q$ are invertible because $X,Y$ are invertible. Also, since $XQ=P^2$ and $QY^{-1}=P^{-1}$ are self-adjoint, we have \begin{align} XQ&=QX^T,\tag{1}\\ QY^{-1}&=Y^{-T}Q,\tag{2}\\ XQ &= (YQ^{-1})^2.\tag{3} \end{align} Let $Q=\begin{pmatrix}a&b+yi\\b-yi&d\end{pmatrix}$. Using the above two equations, you will find that $Q$ and hence $P=YQ^{-1}$ are uniquely determined: $$ Q=\begin{pmatrix}0&1\\1&0\end{pmatrix}, \ P=\begin{pmatrix}-1&2\\2&2\end{pmatrix}. $$

Answer 2

From $PQ=Y$ you get $Y^*=(PQ)^*=Q^*P^*=QP$ and $XY^*=P^2Q^{-1}QP=P^3$. It follows that $XY^*$ is self-adjoint, which should be enough to get you the value of $s$. Then $XY^*=P^3$ should be enough to get you that value of $P$, and then from $PQ=Y$ you can get $Y$.

I haven't worked through the details myself. Check it out, and see whether it works.