Moore-Penrose pseudoinverse of a matrix that is invertible in each column block 4I am not in major in math. But currently I am working with a couple of matrices in the form like this:
\begin{equation}
\left[\begin{array}{rrrrrrrrrrrr}
  1 &  0 &  0 &  0 &  0&   0&   0&   0&   0&   0&   0&  0\\
  0 &  1 &0.5 &0.5 &  0&   0&   0&   0&   0&   0&   0&  0\\
  0 &  0 &0.5 &0.5 &  1& 0.5& 0.5&   0&   0&   0&   0&  0\\
  0 &  0 &  0 &  0 &  0& 0.5& 0.5&   1& 0.5& 0.5&   0&  0\\
  0 &  0 &  0 &  0 &  0&   0&   0&   0& 0.5& 0.5&   1&  0\\
  0 &  0 &  0 &  0 &  0&   0&   0&   0&   0&   0&   0&  1
\end{array}\right] 
\end{equation}
If we consider only nonzero elements, 1-3 columns is invertible, 4-6 columns is invertible, and so on. Is there any efficient way to compute Moore-Penrose pseudoinverse for this type of matrices?
 A: Construction of a block pseudoinverse is an unanswered challenge in my research area. It's easy to build block inverses of Toeplitz matrices which are nonsingular.
Here is your data with symmetry axes:
$$
  \mathbf{A} =
\left(
\begin{array}{cccccc|cccccc}
 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 1 & \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & \frac{1}{2} & \frac{1}{2} & 1 & \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0
   & 0 \\\hline
 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 1 & \frac{1}{2} & \frac{1}{2} & 0
   & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 1 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{array}
\right), 
$$
$$
\mathbf{A}^{\dagger} =
\left(
\begin{array}{rrr|rrr}
 56 & 0 & 0 & 0 & 0 & 0 \\
 0 & 41 & -11 & 3 & -1 & 0 \\
 0 & 15 & 11 & -3 & 1 & 0 \\
 0 & 15 & 11 & -3 & 1 & 0 \\
 0 & -11 & 33 & -9 & 3 & 0 \\
 0 & -4 & 12 & 12 & -4 & 0 \\\hline
 0 & -4 & 12 & 12 & -4 & 0 \\
 0 & 3 & -9 & 33 & -11 & 0 \\
 0 & 1 & -3 & 11 & 15 & 0 \\
 0 & 1 & -3 & 11 & 15 & 0 \\
 0 & -1 & 3 & -11 & 41 & 0 \\
 0 & 0 & 0 & 0 & 0 & 56 \\
\end{array}
\right).
$$
Below are gray scale renderings; $\mathbf{A}$ on the left, $|\mathbf{A}^{\dagger}|$ on the right.


This is an open problem. 
