Find the roots of the quadratic $2$ by $2$ Hermitian matrix equation

Question

I want to find the roots of the quadratic $2$ by $2$ hermitian matrix equation, which has the following form $$ X\Delta X + X\Lambda + \Lambda X - \Delta^{\dagger} = 0 $$ where $\Delta^{\dagger}$ is the Hermitian matrix of $\Delta$, and $\Lambda$ is a Hermitian matrix. Both $\Lambda$ and $\Delta$ are $2$ by $2$ matrices. So could I find the solution of the equation? What's the explicit expression of $X$?

Answer 1

$Z^*$ denotes the transconjugate of $Z$ and, by a change of notations, we consider the equation

$(*)$ $XAX+XS+SX-A^*$ where $S$ is hermitian.

$(*)$ is a particular Riccati equation. We consider the associated pseudo Hamiltonian matrix: $M=\begin{pmatrix}-S&-A\\-A^*&S\end{pmatrix}$; note that $M$ is hermitian and therefore is diagonalizable and has only real eigenvalues but the eigenvectors are not real (in general).

**We consider only the generic case when $M$ has $4$ distinct eigenvalues $(\lambda_i)_i$. Then $(*)$ has exactly $6$ distinct solutions.

In the above conditions, the $6$ $M$-invariant spaces of dimension $2$ are the $(span(e_i,e_j))_{1\leq i<j\leq 4}$ where $e_i$ is "the" eigenvector associated to $\lambda_i$.

At $span(e_i,e_j)$, we associate the matrix $[e_i,e_j]=\begin{pmatrix}U_{2,2}\\V_{2,2} \end{pmatrix}$ and finally, one of the $6$ required solutions $X_{i,j}=VU^{-1}$. $\square$

Remark 1. If $A,S$ are real, then all solutions in $X$ are real.

Remark 2. The principal part of the calculation is to find the eigenvalues and eigenvectors of $M$. That is, essentially, to solve a polynomial of degree $4$.