In http://math.stackexchange.com/questions/2735/solving-very-large-matrices-in-pieces there is a way shown to solve matrix inversion in pieces. Is it possible to turn it into a true divide and conquer algorithm?

I am refering to the answer in the referenced question that starts with:

$\textbf{A}=\begin{pmatrix}\textbf{E}&\textbf{F}\\\\ \textbf{G}&\textbf{H}\end{pmatrix}$

And computes the inverse via:

$\textbf{A}^{-1}=\begin{pmatrix}\textbf{E}^{-1}+\textbf{E}^{-1}\textbf{F}\textbf{S}^{-1}\textbf{G}\textbf{E}^{-1}&-\textbf{E}^{-1}\textbf{F}\textbf{S}^{-1}\\\\ -\textbf{S}^{-1}\textbf{G}\textbf{E}^{-1}&\textbf{S}^{-1}\end{pmatrix}$, where $\textbf{S} = \textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F}$

In this algorithm I will have some steps:

...

sequential

1. Compute inverse $\textbf{E}^{-1}$

2. Compute inverse $(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F})^{-1}$

...

Would be possible to device an algorithm that has steps:

...

parallel

1. Compute inverse $\textbf{X}^{-1}$

1. Compute inverse $\textbf{Y}^{-1}$

...

i.e. to identify two smaller matrices $\textbf{X}$ and $\textbf{Y}$ which are independent from their inverse?

Best Regards

Matrices(Springer GTM 216). I use it to show that the complexity of matrix inversion is the same as the complexity of matrix multiplication. I don't discuss the parallelization. I don't know the origin of this algorithm but it is not very new. – Denis Serre Oct 18 '11 at 5:26