1
$\begingroup$

Suppose

$$ R = \begin{bmatrix} A & B\\ C & D\end{bmatrix} $$

is a $2 \times 2$ block matrix of real numbers, where $A$ and $D$ are squared diagonal matrices.

Is it possible that the following four conditions hold simultaneously?

  1. $R$ is invertible

  2. $D$ is nonsingular

  3. the Schur complement of $D$, $A-BD^{-1}C$ is singular.

  4. $A$ is singular.

If so, could you please provide a way to find the inverse $R$ in terms of the partitions of $R$?

$\endgroup$
0

2 Answers 2

0
$\begingroup$

It's true in general since Volker Strassen found the following formula :

$$\begin{pmatrix}A&B\\C&D\end{pmatrix}^{-1}=\begin{pmatrix}\left(A-B D^{-1}C\right)^{-1}&-\left(A-B D^{-1}C\right)^{-1} BD^{-1}\\-D^{-1}C\left(A-B D^{-1}C\right)^{-1}&D^{-1}+D^{-1}C\left(A-BD^{-1}C \right)^{-1}BD^{-1}\end{pmatrix}$$

$\endgroup$
1
  • $\begingroup$ The formula you mention only holds if $D$ and $A-BD^{-1}C$ are invertible. However, I'm curious what will happen if $D$ is invertible but $A-BD^{-1}C$ and $A$ are not invertible? $\endgroup$
    – VivianX
    Commented Sep 13, 2022 at 4:33
0
$\begingroup$

Suppose $A$ is a $1\times 1$ equal to zero. Let $D$ be any invertible matrix and $C$ any column vector. Choose $B$ to be a row vector which is orthogonal to $D^{-1}C.\;$ These choices will meet all of the stated conditions.

Assume that the initial elements of $\{B,C,D\}$ are $\{1,2,3\},$ respectively. Then repartitioning produces $A=\pmatrix{0&1\\2&3}$ or some other invertible matrix depending on those initial elements.

The important point is that the invertibility of new $A$ means that $R^{-1}$ can be calculated using its Schur complement.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .