# 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: $$\left[\begin{array}{rrrrrrrrrrrr} 1 & 0 & 0 & 0 & 0& 0& 0& 0& 0& 0& 0& 0\\ 0 & 1 &0.5 &0.5 & 0& 0& 0& 0& 0& 0& 0& 0\\ 0 & 0 &0.5 &0.5 & 1& 0.5& 0.5& 0& 0& 0& 0& 0\\ 0 & 0 & 0 & 0 & 0& 0.5& 0.5& 1& 0.5& 0.5& 0& 0\\ 0 & 0 & 0 & 0 & 0& 0& 0& 0& 0.5& 0.5& 1& 0\\ 0 & 0 & 0 & 0 & 0& 0& 0& 0& 0& 0& 0& 1 \end{array}\right]$$ 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?

• How big are the matrices you are working with? Oct 10 '15 at 14:06
• From tens of n by n matrices to hundreds of n by n matrices arranged in this pattern. Oct 10 '15 at 14:16
• If the matrices are on the order of 10 or 100, then they are not very large and using the svd to compute the Moore-Penrose pseudoinverse should be fast enough. Oct 10 '15 at 14:59

## 1 Answer

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.