Inverse of a special block matrix

Question

I have a special $NM \times NM$ matrix of the form

\begin{align*} S = \left[ \begin{array}{cccc} V + \lambda I & V & \cdots & V \\ V & V + \lambda I & \cdots & V \\ \cdots & \cdots & \cdots & \cdots \\ V & V & \cdots & V + \lambda I \end{array} \right] \end{align*} where $V$ is a symmetric $N\times N$ matrix and $I$ is an identity matrix of size $N$.

I want to know if there is any way to express $S^{-1}$ in a simpler form involving $V^{-1}$ and $(V+\lambda I)^{-1}$.

Answer 1

Using the results from your other question and the idea of Jean Marie's answer: $S = (I_M\otimes\lambda I_N) + ({\bf 1}{\bf 1}^T \otimes V)$

We can rewrite @Accumulation's answer in the stringent form:

\begin{align} S^{-1} &= \tfrac{1}{\lambda}\big(I_{mn} - (-\tfrac{1}{\lambda}{\bf 1}_m{\bf 1}_m^T \otimes V)\big)^{-1} \\&=\tfrac{1}{\lambda}\sum_{k=0}^{\infty} (-\tfrac{1}{\lambda}{\bf 1}_m{\bf 1}_m^T \otimes V)^k \\ &=\tfrac{1}{\lambda}(I_m\otimes I_n) + \tfrac{1}{m\lambda}\sum_{k=1}^{\infty}(-\tfrac{m}{\lambda})^k ({\bf 1}_m{\bf 1}_m^T \otimes V^k) \\ &=\tfrac{1}{\lambda}(I_m\otimes I_n) + \tfrac{1}{\lambda}\big(\tfrac{1}{m}{\bf 1}_m{\bf 1}_m^T \otimes \sum_{k=1}^{\infty}(-\tfrac{m}{\lambda}V)^k\big) \\ &=\tfrac{1}{\lambda}(I_m\otimes I_n) +\tfrac{1}{\lambda}\big(\tfrac{1}{m}{\bf 1}_m{\bf 1}_m^T \otimes (I_n+\tfrac{m}{\lambda}V)^{-1}-I_n\big)\\ &= \tfrac{1}{\lambda}\Big((C_m\otimes I_n)+ \big(\tfrac{1}{m}{\bf 1}_m{\bf 1}_m^T \otimes (I_n+\tfrac{m}{\lambda}V)^{-1}\big)\Big)\end{align}

Note that the matrix $C_m = I_m - \tfrac{1}{m}{\bf 1}_m {\bf 1}_m^T $ is known as the (geometric) centering matrix

Here the Neumann series converge if $\| \tfrac{1}{\lambda}{\bf 1}_M{\bf 1}_M^T \otimes V\|<1 $ and $\|\tfrac{M}{\lambda} V \|<1$. If we take the Frobenius norm, both are equivalent to $\|V\|_{F}<\tfrac{|\lambda|}{M}$

EDIT: fixed mistake with handling $k=0$ term

Answer 2

I think the following analysis might work:

Expressing $S$ as $\lambda(I_{nm \times nm}-\frac {-1} {\lambda} W)$ where $W$ is your matrix, except without the $\lambda I$ (that is, $W$ is a block matrix where each block is $V$), $S^{-1}$ is equal to the geometric series $\frac 1 {\lambda} \sum_{k=0}^{\infty}(\frac{-1}{\lambda}W)^k$.

$W^k$ is a block matrix where each block is $m^{k-1}V^k$ (I prefer lowercase variable names for integers, to distinguish them from matrices). So then each block of the inverse is $\frac 1 {\lambda}\sum_{k=0}^{\infty}(\frac{-1}{\lambda})^km^{k-1}V^k=\frac 1 {m\lambda}\sum_{k=0}^{\infty}(\frac{-1}{\lambda})^km^kV^k=\frac 1 {m\lambda}\sum_{k=0}^{\infty}(\frac{-m}{\lambda}V)^k$, which is yet another geometric series, and it converges to $\frac 1 {m\lambda}(I+\frac{m}{\lambda}V)^{-1}=(m\lambda+m^2V)^{-1}$.