Structure of a inverse of a block matrix

Question

Let a block matrix $M$ like

$M=\left( \begin{matrix} A&B& 0 \\ B'&C &0 \\ 0 & 0 &D\end{matrix}\right)$

or any other block symmetric matrix (and invertible) $M$ such that has zeros blocks.

Taking the inverse of M, can I assure that I will always get an inverse with the same structure as M?

For clarify, $M^{-1}$ should be:

$M^{-1}=\left( \begin{matrix} \bar{A}&\bar{B}& 0 \\ \bar{B}'& \bar{C} &0 \\ 0 & 0 & \bar{D}\end{matrix}\right)$

More specifically, can I assure that I will always get an inverse with the structure as follows?

$M^{-1}=\left( \begin{matrix} \pm A^{-1}&B& 0 \\ B'& \pm C^{-1} &0 \\ 0 & 0 &\pm D^{-1}\end{matrix}\right)$

I just need an answer, not a proof. But an 'informal explanation' would be nice.

Answer 1

Let $\cal M$ denote the set of all block diagonal symmetric matrices that have the same size and block structure (with zero blocks) as your given matrix $M$. Note that $\cal M$ contains matrices (like the zero matrix) that are not invertible. This is not important.

Clearly any scalar multiple of a matrix in $\cal M$ is again in $\cal M$. And any finite sum of matrices in $\cal M$ is also in $\cal M$.

Suppose that positive powers of $M$ are also in $\cal M$. (This is true for the example you give). If $M$ is invertible, it follows that $M^{-1}$ is also in $\cal M$. This is because $M^{-1}$ is a sum of scalar multiples of positive powers of $M$, thanks to the Cayley Hamilton Theorem.

The additional structure for the inverse you wonder about --- this additional structure will generally not occur. As an example, take $$ M = \left[\begin{array}{err} 5 & 2 & 0 \\ 2 & 1 & 0 \\ 0& 0& 1 \end{array}\right]. $$ Then $$ M^{-1} = \left[\begin{array}{err} 1 & -2 & 0 \\ -2 & 5 & 0 \\ 0& 0& 1 \end{array}\right]. $$

Answer 2

Assume $A,D$ and $C-B^TA^{-1}B$ are invertible. Then by the proof below, we have $$ \pmatrix{A&B\\B^T&C}^{-1}=\pmatrix{A^{-1}+A^{-1}B(C-B^TA^{-1}B)^{-1}B^TA^{-1}&-A^{-1}B(C-B^TA^{-1}B)^{-1}\\-(C-B^TA^{-1}B)^{-1}B^TA^{-1}&(C-B^TA^{-1}B)^{-1}} $$ And $$ \pmatrix{A&B&0\\B^T&C&0\\0&0&D}^{-1}=\pmatrix{\pmatrix{A&B\\B^T&C}^{-1}&0\\0&D^{-1}} $$

The following is a proof of the above identity on the general submatrices inversion. There are several ways to get the solution. Here I give a simple way to prove it using the idea of row operations to get the inverse $$ (X,I_{})\longrightarrow(I_{}, X^{-1}) $$ First multiply a left matrix (similar to elementary row operation of adding 1st row by $-B^TA^{-1}$ to 2nd row) $$ \pmatrix{I&0\\-B^TA^{-1}&1}\pmatrix{A&B&I&0\\B^T&C&0&I}=\pmatrix{A&B&I&0\\0&C-B^TA^{-1}B&-B^TA^{-1}&I} $$ Then multiply a left matrix (similar to elementary row operation of adding 2nd row by $-B(C-B^TA^{-1}B)^{-1}$ to 2nd row) $$ \pmatrix{I&-B(C-B^TA^{-1}B)^{-1}\\0&I}\pmatrix{A&B&I&0\\0&C-B^TA^{-1}B&-B^TA^{-1}&I}=\pmatrix{A&0&I+B(C-B^TA^{-1}B)^{-1}B^TA^{-1}&-B(C-B^TA^{-1}B)^{-1}\\0&C-B^TA^{-1}B&-B^TA^{-1}&1} $$ Finally multiply a left matrix (similar to elementary row operation of time $A^{-1}$ to 1st row and $(C-B^TA^{-1}B)^{-1}$ to 2nd row) $$ \pmatrix{A^{-1}&0\\0&(C-B^TA^{-1}B)^{-1}}\pmatrix{A&0&I+B(C-B^TA^{-1}B)^{-1}B^TA^{-1}&-B(C-B^TA^{-1}B)^{-1}\\0&C-B^TA^{-1}B&-B^TA^{-1}&1}=\pmatrix{I&0&A^{-1}+A^{-1}B(C-B^TA^{-1}B)^{-1}B^TA^{-1}&-A^{-1}B(C-B^TA^{-1}B)^{-1}\\0&I&-(C-B^TA^{-1}B)^{-1}B^TA^{-1}&(C-B^TA^{-1}B)^{-1}} $$ So we have $$ \pmatrix{A&B\\B^T&C}^{-1}=\pmatrix{A^{-1}+A^{-1}B(C-B^TA^{-1}B)^{-1}B^TA^{-1}&-A^{-1}B(C-B^TA^{-1}B)^{-1}\\-(C-B^TA^{-1}B)^{-1}B^TA^{-1}&(C-B^TA^{-1}B)^{-1}} $$