Inverting a Block-Toeplitz matrix with the Sherman-Morrison formula

Question

Suppose we are given the following Block-Toeplitz matrix: \begin{eqnarray} T=\left(\begin{matrix} A & 0 & ... & 0\\ B & A & ... & \vdots\\ \vdots & \ddots & \ddots &\vdots\\ 0 & ... & B & A \end{matrix}\right) \end{eqnarray} where each $B,A$ are matrices of dimension $N\times N$. Note that we only have blocks $A$ in the diagonal, as well as blocks $B$ on the lower-diagonal; the rest of blocks are $N\times N$ matrices with zero entries. Our aim is to calculate the inverse of such matrix $T^{-1}$. One can decompose this matrix as a sum of a circulant matrix and a matrix with a single block in the upper-right corner, so that: \begin{eqnarray} T=\left(\begin{matrix} A & 0 & ... & B\\ B & A & ... & \vdots\\ \vdots & \ddots & \ddots &\vdots\\ 0 & ... & B & A \end{matrix}\right) + \left(\begin{matrix} 0 & 0 & ... & -B\\ 0 & 0 & ... & \vdots\\ \vdots & \ddots & \ddots &\vdots\\ 0 & ... & 0 & 0 \end{matrix}\right)=C + D \end{eqnarray} Now, my question is if one can, in this very concrete case, apply the Shermann and Morrison lemma (https://en.wikipedia.org/wiki/Sherman%E2%80%93Morrison_formula ) here, so that we can make: \begin{eqnarray} (C+D)^{-1} = C^{-1} - \frac{1}{1+\text{Tr}(DC^{-1})}C^{-1}.D.C^{-1} \end{eqnarray} since the inverse of $C^{-1}$ has a closed form because $C$ is block-circulant matrix. Is the above statement correct? Note that $D$ is not invertible, but if I understood correctly, this is not a requirement for the formula to be correct.

Answer 1

The Shermann Morrison formula only applies if the update matrix (which you refer to as $D$) has rank $1$.

However, we can do what you're trying to do using the Woodbury matrix identity. In particular, we can write $$ D = \pmatrix{-B\\0\\ \vdots \\ 0} I_N \pmatrix{0 & \cdots & 0 & I_N} = UIV, $$ where $I$ denotes the identity matrix. With that, we have $$ (C + D)^{-1} = (C + UI_NV)^{-1}= C^{-1} - C^{-1}U \left(I_N + VC^{-1}U \right)^{-1} VC^{-1}. $$