Inverse of a special type of block matrix

Question

Let

$$M = I_{n}\otimes A + J_{n}\otimes B$$

where $A = aI_{m} + bJ_{m}$ and $B = cI_{m}$. Here, $I_s$ is the identity matrix of order $s$ and $J_s$ is the matrix of all ones of order $s$. Is it possible to get an analytical solution of $M^{-1}$?

Answer 1

We note that both the matrices $I$ and $J$ are circulant. So, if we take $F$ to denote the DFT matrix, we find that $$ (F_n \otimes F_m)^*M(F_n \otimes F_m) = I_n \otimes (a I_m + bm \,K_m) + cn\, K_n \otimes I_m \\ = a I_m \otimes I_n + bm\, I_n \otimes K_m + cn\, K_n \otimes I_m $$ Where $K_s$ denotes the size $s$ matrix which has a $1$ as its 1,1 entry and zeros everywhere else. Let $P = (F_n \otimes F_m)^*M(F_n \otimes F_m)$, the matrix expanded above.

The matrix $P$ is diagonal, so it's easy to come up with an expression for its inverse. In particular, we find that $$ P = \pmatrix{aI_n + bm \,K_n + cn\, I_n \\ & aI_n + bm \,K_n \\ & &\ddots \\ &&& a I_n + bm\, K_n} \\ = \pmatrix{(a + bm + cn)\,K_m + (a + cn)(I_m - K_m) \\ & (a + bm)\,K_m + a\,(I_m - K_m) \\ & &\ddots \\ &&& \text{[same]}} $$ We then compute $$ P^{-1} = \pmatrix{\frac{1}{a+bm + cn} K_m + \frac{1}{a + cn}(I_m - K_m)\\ & \frac{1}{a+bm} K_m + \frac{1}{a}(I_m - K_m) \\ & & \ddots \\ &&& \frac{1}{a+bm} K_m + \frac{1}{a}(I_m - K_m)} \\ = K_n \otimes (\frac{1}{a+bm + cn} K_m + \frac{1}{a + cn}(I_m - K_m)) + (I_n - K_n) \otimes (\frac{1}{a+bm} K_m + \frac{1}{a}(I_m - K_m)) $$ From there, we may compute $M^{-1} = (F_n \otimes F_m) P^{-1}(F_n \otimes F_m)^*$ to find that $$ M^{-1} = \frac{J_n}{n} \otimes (\frac{1}{a+bm + cn} \frac{J_m}{m} + \frac{1}{a + cn}(I_m - \frac{J_m}{m})) + (I_n - \frac{J_n}{n}) \otimes (\frac{1}{a+bm} \frac{J_m}{m} + \frac{1}{a}(I_m - \frac{J_m}{m})) $$