Let $X$ be a square, symmetric, positive definite matrix that can be decomposed into $p\times p$ block matrices:
$$X = \begin{bmatrix} X_{11} & X_{12} & \ldots & X_{1p}\\
X_{21} & X_{22} & \ldots & X_{2p}\\
\vdots & \vdots & \ddots & \vdots \\
X_{p1} & X_{p2} & \ldots & X_{pp}
\end{bmatrix}$$
Let $Y$ denote the inverse of $X$ that consists of $p \times p$ submatrices:
$$Y = X^{-1} = \begin{bmatrix} Y_{11} & Y_{12} & \ldots & Y_{1p}\\
Y_{21} & Y_{22} & \ldots & Y_{2p}\\
\vdots & \vdots & \ddots & \vdots \\
Y_{p1} & Y_{p2} & \ldots & Y_{pp}
\end{bmatrix}$$
I would like to write the diagonal submatrices $Y_{ii}$ for $1 \leq i \leq p$ as a function of the submatrices in $X$.
$\bf{p =2}$ case: Since $X_{11}, X_{22}$ are positive definite (and invertible), then
$$Y = \begin{bmatrix} (X_{11} – X_{12}X_{22}^{-1}X_{21})^{-1} & -(X_{11}-X_{12}X_{22}^{-1} X_{21})^{-1}X_{12}X_{22}^{-1} \\
-(X_{22} – X_{21} X_{11}^{-1} X_{12})^{-1} X_{21}X_{11}^{-1} & (X_{22}-X_{21} X_{11}^{-1} X_{12})^{-1} \end{bmatrix}$$
so
- $Y_{11} = (X_{11} – X_{12}X_{22}^{-1}X_{21})^{-1}$
- $Y_{22} = (X_{22}-X_{21} X_{11}^{-1} X_{12})^{-1}$
$\bf{p =3}$ case: Since $X_{11}, X_{22}, X_{33}$ are positive definite (and invertible), then (if my math is right) we have the following
- $Y_{11} = ((X_{11} – X_{13}X^{-1}_{11}X_{31}) – (X_{12}-X_{13}X^{-1}_{33}X_{32})(X_{22}-X_{23}X^{-1}_{33}X_{32})^{-1}(X_{21}-X_{23}X^{-1}_{33}X_{31}))^{-1}$
- $Y_{22} = ((X_{22} – X_{21}X^{-1}_{11}X_{12}) – (X_{23}-X_{21}X^{-1}_{11}X_{13})(X_{33}-X_{31}X^{-1}_{11}X_{13})^{-1}(X_{32} – X_{31}X^{-1}_{11}X_{12}))^{-1}$
- $Y_{33} = ((X_{33}-X_{31}X^{-1}_{11}X_{13}) – (X_{32}-X_{31}X^{-1}_{11}X_{12})(X_{22}-X_{21}X^{-1}_{11}X_{12})^{-1}(X_{23}-X_{21}X^{-1}_{11}X_{13}))^{-1}$
For the general $p \times p$ case, is it possible to write $$Y_{ii} = (Z_{ii})^{-1}$$
where $Z_{ii}$ can be written as a function of the submatrices in $X$? Would $Z_{ii}$ be a Schur complement of some sort?
Best Answer
Getting a direct expression isn't going to happen, because it is like asking for a direct expression for the components $L_{ij}$ and $U_{ij}$ of an LU decomposition given any general matrix $A$.
But you can get a symbolic result by using $p$ recursions, each time reducing the size of the matrix to be inverted by one order.
I will proceed with the concrete example of a $p=4$ matrix that will illustrate the process to make it into a $p=3$ matrix.
First split the given matrix ${\rm X}_{44}$ into 4 partitions
$${\rm X}_{44}=\left[\begin{array}{ccc|c} X_{11} & X_{12} & X_{13} & X_{14}\\ X_{21} & X_{22} & X_{23} & X_{24}\\ X_{31} & X_{32} & X_{33} & X_{34}\\ \hline X_{41} & X_{42} & X_{43} & X_{44} \end{array}\right]=\begin{bmatrix}{\rm X}_{33} & {\rm X}_{34}\\ {\rm X}_{43} & X_{44} \end{bmatrix}$$
A bit of nomenclature here, the italicized letter $X_{44}$ designates the sub-matrix as given, and the upright letters ${\rm X}_{33}$, ${\rm X}_{34}$, and ${\rm X}_{43}$ designate the block matrices that contain multiple sub-matrices. Each subscript $3$ on upright letters above actually represents a range $1 \ldots 3$ for the sub-matrices included.
More specifically, the above notation represents the following block matrices
$$\begin{aligned}{\rm X}_{33} & =\left[\begin{array}{ccc} X_{11} & X_{12} & X_{13}\\ X_{21} & X_{22} & X_{23}\\ X_{31} & X_{32} & X_{33} \end{array}\right]\\ {\rm X}_{34} & =\left[\begin{array}{c} X_{14}\\ X_{24}\\ X_{34} \end{array}\right]\\ {\rm X}_{43} & =\left[\begin{array}{ccc} X_{41} & X_{42} & X_{43}\end{array}\right] \end{aligned}$$
Do the same for the inverse ${\rm Y}_{44}$
$${\rm Y}_{44}=\left[\begin{array}{ccc|c} Y_{11} & Y_{12} & Y_{13} & Y_{14}\\ Y_{21} & Y_{22} & Y_{23} & Y_{24}\\ Y_{31} & Y_{32} & Y_{33} & Y_{34}\\ \hline Y_{41} & Y_{42} & Y_{43} & Y_{44} \end{array}\right]=\begin{bmatrix}{\rm Y}_{33} & {\rm Y}_{34}\\ {\rm Y}_{43} & Y_{44} \end{bmatrix}$$
Solve the matrix inversion problem for the components of ${\rm Y}_{44}$
$$\begin{bmatrix}{\rm Y}_{33} & {\rm Y}_{34}\\ {\rm Y}_{43} & Y_{44} \end{bmatrix}\begin{bmatrix}{\rm X}_{33} & {\rm X}_{34}\\ {\rm X}_{43} & X_{44} \end{bmatrix}=\begin{bmatrix}1_{\{3,3\}} & 0_{\{3,1\}}\\ 0_{\{1,3\}} & 1_{\{1,1\}} \end{bmatrix}$$
by splitting it up into four equations (one for each block in the partition) to get the result
$$\small \begin{aligned}{\rm Y}_{33}=\left[\begin{array}{ccc} Y_{11} & Y_{12} & Y_{13}\\ Y_{21} & Y_{22} & Y_{23}\\ Y_{31} & Y_{32} & Y_{33} \end{array}\right] & =\left(\left[\begin{array}{ccc} X_{11} & X_{12} & X_{13}\\ X_{21} & X_{22} & X_{23}\\ X_{31} & X_{32} & X_{33} \end{array}\right]-\left[\begin{array}{c} X_{14}\\ X_{24}\\ X_{34} \end{array}\right]X_{44}^{-1}\left[\begin{array}{ccc} X_{41} & X_{42} & X_{43}\end{array}\right]\right)^{-1}\\ {\rm Y}_{34}=\left[\begin{array}{c} Y_{14}\\ Y_{24}\\ Y_{34} \end{array}\right] & =-{\rm Y}_{33}\left[\begin{array}{c} X_{14}\\ X_{24}\\ X_{34} \end{array}\right]X_{44}^{-1}\\ Y_{44} & =\left(X_{44}-\left[\begin{array}{ccc} X_{41} & X_{42} & X_{43}\end{array}\right]\left[\begin{array}{ccc} X_{11} & X_{12} & X_{13}\\ X_{21} & X_{22} & X_{23}\\ X_{31} & X_{32} & X_{33} \end{array}\right]^{-1}\left[\begin{array}{c} X_{14}\\ X_{24}\\ X_{34} \end{array}\right]\right)^{-1}\\ {\rm Y}_{43}=\left[\begin{array}{ccc} Y_{41} & Y_{42} & Y_{43}\end{array}\right] & =-Y_{44}\left[\begin{array}{ccc} X_{41} & X_{42} & X_{43}\end{array}\right]\left[\begin{array}{ccc} X_{11} & X_{12} & X_{13}\\ X_{21} & X_{22} & X_{23}\\ X_{31} & X_{32} & X_{33} \end{array}\right]^{-1} \end{aligned}$$
Assemble ${\rm Y}_{44}$ from the four block matrices ${\rm Y}_{33}$, ${\rm Y}_{34}$, ${\rm Y}_{43}$ and $Y_{44}$ calculated above.
Note that to complete this step, you need to compute 3 block matrix inverses
So at this point, you would apply this method recursively to the 3 inversed above to get 3×3=9 inverses of $p=2$ order. And so on, until you get down to the component inverses $X_{ij}^{-1}$
You can summarize the number of inverses required for each size $p$ as follows
PS. Be mindful of the designation of a block matrix ${\rm X}_{ij}$ vs. a component sub-matrix $X_{ij}$ as they are of different sizes.