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I have a matrix $\boldsymbol{Q} \in \mathbb{R}^{M \times M}$ in Hankel form and I would like to decompose it into Observability matrix $\boldsymbol{\Gamma} \in \mathbb{R}^{M \times M}$ times the controllability matrix $\boldsymbol{\Delta} \in \mathbb{R}^{M \times M}$. Here, $\boldsymbol{\Gamma}$ is anti-upper triangular matrix and $\boldsymbol{\Delta}$ is a lower triangular matrix.
$\boldsymbol{Q} = \begin{bmatrix} \boldsymbol{c}^T\boldsymbol{b} & \boldsymbol{c}^T\boldsymbol{Ab} & \dots & \boldsymbol{c}^T\boldsymbol{A}^{M-1}\boldsymbol{b} \\ \boldsymbol{c}^T\boldsymbol{Ab} & & \boldsymbol{c}^T\boldsymbol{A}^{M-1}\boldsymbol{b} & 0\\ \vdots & \iddots & \iddots & \vdots\\ \boldsymbol{c}^T\boldsymbol{A}^{M-1}\boldsymbol{b} & 0 & \dots & 0 \end{bmatrix} $
$ \boldsymbol{Q} = \begin{bmatrix}\boldsymbol{c}^T \\ \boldsymbol{c}^T\boldsymbol{A} \\ \vdots \\ \boldsymbol{c}^T\boldsymbol{A}^{M-1} \end{bmatrix}\begin{bmatrix}\boldsymbol{b} & \boldsymbol{Ab} & \dots & \boldsymbol{A}^{M-1}\boldsymbol{b} \end{bmatrix} = \boldsymbol{\Gamma}\boldsymbol{\Delta}. $
To the best of my knowledge, there exists no such decomposition. However, the structure of $\boldsymbol{\Gamma}$ and $\boldsymbol{\Delta}$ is based on the prior knowledge about my Hankel matrix. Is there any possibility, to use this knowledge in the decomposition?