Sum of diagonal blocks of inverse of block matrix

Question

I have an $(kn \times kn)$ dimensional square positive semi-definite symmetric matrix of the form:

$$A=\begin{bmatrix}A_{11} & A_{21} & ... & A_{n1} \\ A_{21}' & A_{22} & ... & A_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ A_{n1}' & A_{n2}' & ... & A_{nn} \\ \end{bmatrix}$$

Where the blocks, $\{A_{ij}\}_{i, j\in \{1,...n\}}$, are square and of dimension $(k \times k)$. Denote the inverse of $A$ as $B$, so that $B^{-1}=A$ and partition $B$ as:

$$B=\begin{bmatrix}B_{11} & B_{21} & ... & B_{n1} \\ B_{21}' & B_{22} & ... & B_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ B_{n1}' & B_{n2}' & ... & B_{nn} \\ \end{bmatrix}$$

Where the blocks, $\{B_{ij}\}_{i, j\in \{1,...n\}}$, are also square and of dimension $(k \times k)$. I want to find the sum of the diagonal blocks of $B$ in terms of the original blocks of $A$. That is, I want to find:

$$\sum_{i=1}^n B_{ii}$$

In terms of the $\{A_{ij}\}_{i,j \in \{1,...,n\}}$. Does anyone know of any nice formulae for this?

You forgot to mention that the diagonal blocks are symmetric. — Rodrigo de Azevedo, Aug 01 '20 at 21:40
There's a transpose on $A_{21}$. The diagonal blocks being symmetric also follows from me saying $A$ is symmetric no? — JDoe2, Aug 01 '20 at 22:13
The transposition should be on the blocks above the diagonal then. — Rodrigo de Azevedo, Aug 02 '20 at 10:33
Here's how "ugly" it looks for the $2 \times 2$ case. You could use some form of recursion to find the inverse in terms of the inverse of the northwest block. If you want a formula for, say, $n=10$, you might need a few pages to write it down. Things can get funny. — Rodrigo de Azevedo, Aug 03 '20 at 12:12

A learner · Answer 1 · 2020-08-01T19:25:03.133

0

For $k=1$ , As, $B=A^{-1}$, by the rule of trace of matrix, $\sum_{i=1}^n B_{ii} = \text{sum of all reciprocals of all eigenvalues of A} = (-1)^{n-1} \frac{\text{sum of all minors of A of order (n-1)}}{|A|} $

edited Aug 01 '20 at 19:25

answered Jul 31 '20 at 12:35

A learner

2,871

Have you tried anything? – A learner Aug 01 '20 at 18:10
Apologies I think maybe my question was not clear enough/misleading - the $B_{ij}$ are block partitions, not elements. I will rephrase to make this clear. – JDoe2 Aug 01 '20 at 19:02
@A learner, were you suggesting there is a more general result? – JDoe2 Jan 20 '22 at 01:28

Sum of diagonal blocks of inverse of block matrix

1 Answers1