Matrix inverse of block symmetric matrix

Question

Provided that $\mathbf{A}$ and $\mathbf{B}$ are $n\times n$ symmetric matrices, the matrix $\mathbf{C}$ is constructed by $\mathbf{A}$ and $\mathbf{B}$ with the form of \begin{align} \mathbf{C}= \begin{bmatrix} \mathbf{A} & \mathbf{B} &\cdots &\mathbf{B}\\ \mathbf{B} & \mathbf{A} &\cdots &\mathbf{B}\\ \vdots & \vdots &\ddots &\vdots\\\ \mathbf{B} & \mathbf{B} &\cdots &\mathbf{A}\\ \end{bmatrix} \end{align} Note that $\mathbf{C}$ is $mn\times mn$ matrix. Given the information above, do we get the expression of matrix inverse of $\mathbf{C}$. I guess that $\mathbf{C}^{-1}$ has the same structure as $\mathbf{C}$, but I can't get its expression.

Answer 1

Here is an approach using the fact that $B^{-1}$ exists. Note that $$ C = (I_m \otimes B)(I_m \otimes (B^{-1}A - I) + J_m \otimes I_n), $$ where $\otimes$ denotes the Kronecker product, $I_k$ denotes the identity matrix of size $k$, and $J_m$ is the size $m$ matrix whose entries are all equal to $1$. We note that $$ C^{-1} = (I_m \otimes (B^{-1}A - I) + J_m \otimes I_n)^{-1}(I_m \otimes B)^{-1} \\ = (I_m \otimes M + J_m \otimes I_n)^{-1}(I_m \otimes B^{-1}), $$ where $M = B^{-1}A - I$. Let $e \in \Bbb R^m$ denote the column-vector whose entries are all $1$; note that $J_m = ee^T$. If $M$ and $M + mI$ are invertible, then with the Woodbury matrix identity we can write $$ \begin{align} (I_m \otimes M + J_m \otimes I_n)^{-1} & = (I_m \otimes M + (e \otimes I_n)(e \otimes I_n)^T)^{-1} \\ & = (I_m \otimes M)^{-1} - (I_m \otimes M)^{-1}(e \otimes I_n)( \\ & \quad \qquad I_{n} + (e \otimes I_n)^T(I_m \otimes M)^{-1}(e \otimes I_n) \\ & \quad )^{-1}(e \otimes I_n)^T(I \otimes M)^{-1} \\ & = (I_m \otimes M^{-1}) - (I_m \otimes M^{-1})(e \otimes I_n) \\ & \quad \qquad \cdot(I_{n} + (e^Te \otimes M^{-1}))^{-1}(e \otimes I_n)^T(I \otimes M^{-1}) \\ & = (I_m \otimes M^{-1}) - (e \otimes M^{-1})(I_{n} + m \cdot M^{-1})^{-1}(e^T \otimes M^{-1}). \end{align} $$ We can simplify this a bit further: note that the second term can be reduced as follows. $$ (e \otimes M^{-1})(I_{n} + m \cdot M^{-1})^{-1}(e^T \otimes M^{-1}) = \\ (e \otimes M^{-1})M(M + m I_n)^{-1}(e^T \otimes M^{-1}) = \\ (e \otimes I_n)(M + mI_n)^{-1}(e^T \otimes M^{-1}) = \\ (e \otimes (M + mI_n)^{-1})(e^T \otimes M^{-1}) = \\ (ee^T) \otimes [(M + mI_n)^{-1}M^{-1}] = \\ J_m \otimes (M^2 + m\cdot M)^{-1}. $$ Putting this all together, we get $$ C^{-1} = (I_m \otimes M + J_m \otimes I_n)^{-1}(I_m \otimes B^{-1})\\ = (I_m \otimes M^{-1} - J_m \otimes (M^2 + m\cdot M)^{-1})(I_m \otimes B^{-1}) \\ = I_m \otimes (M^{-1}B^{-1}) - J_m \otimes ((M^2 + m\cdot M)^{-1}B^{-1}) \\ = I_m \otimes (A - B)^{-1} - J_m \otimes (B(B^{-1}A - I)^2 + m \cdot (A - B))^{-1}. $$ Because this expression is in the form $I_m \otimes P + J_m \otimes Q$, we can see that $C^{-1}$ will indeed have the same block-structure as $C$.

Answer 2

There is no way to construct the inverse of $C$. A simple counterexample is enough to demonstrate:

Let $$A=B=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$$ and $m=2$. Both $A$ and $B$ are invertible. Then $$C = \begin{bmatrix}A & B\\B & A \end{bmatrix} = \begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ \end{bmatrix} $$ which has pairs of identical rows and columns, and so has determinate $0$ (is degenerate) and cannot be inverted. Thus there is no general way to construct $C^{-1}$.