Block matrices

Two major techniques when it comes to matrices:

1. Decompositions: Decompose a complicated map into the composition of a chain of a chain of simple maps. (Can be seen as a decomposition at time dimension)
2. Block matrices: Decompose a matrix at space dimension.

Block Diagonal Matrix

Take this matrix as an example:

$$\left(\begin{array}{ccc}1& -1& 0\\ 1& 1& 0\\ 0& 0& 2\end{array}\right)$$

This matrix is not a diagonal matrix, but a block diagonal matrix.

If we take inverse of it:

$${\left(\begin{array}{ccc}1& -1& 0\\ 1& 1& 0\\ 0& 0& 2\end{array}\right)}^{-1}=\left(\begin{array}{ccc}0.5& 0.5& 0\\ -0.5& 0.5& 0\\ 0& 0& 0.5\end{array}\right)$$

This is basically taking the inverse of each diagonal block.

Intuitively, the upper left block only change the first two coordinates (the xy-plane), and the lower right block only change the last coordinate (the z-axis), so they behave independently!

Block Inverse

How to find the inverse of $\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$, provided $A$ is invertible? We can do a block LDU decomposition:

$$M=\left(\begin{array}{cc}I& O\\ C{A}^{-1}& I\end{array}\right)\left(\begin{array}{cc}A& O\\ O& (D-C{A}^{-1}B)\end{array}\right)\left(\begin{array}{cc}I& A^{-1}B\\ O& I\end{array}\right)$$

If $D-C{A}^{-1}B$ is also invertible, then

$$M^{-1}=\left(\begin{array}{cc}I& -A^{-1}B\\ O& I\end{array}\right)\left(\begin{array}{cc}{A}^{-1}& O\\ O& (D-C{A}^{-1}B{)}^{-1}\end{array}\right)\left(\begin{array}{cc}I& O\\ -C{A}^{-1}& I\end{array}\right)$$