I need to take inverse of this matrix:
\begin{equation}
i \mathbf{I}_8-\begin{pmatrix}
(a+ib-c)\sigma\otimes\sigma & a \sigma\otimes\sigma \\
a \sigma\otimes\sigma & (a-ib-c) \sigma\otimes\sigma
\end{pmatrix}
\end{equation}
Here $\mathbf{I}_8$ is $8*8$ identity matrix and, $\sigma$ is Pauli matrix such that $\sigma\otimes\sigma= \sum_{i=1}^3 \sigma^i\otimes \sigma^i=\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & -1 & 2 & 0 \\
0 & 2 & -1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)$.
I would like to have the final result in terms of $\sigma\otimes\sigma$ and Identity matrix. I tried brute force inverse of the resulting $8*8$ matrix but that way I could not write the final result in terms of $\sigma\otimes\sigma$. I was wondering if there are any identities that can help me simplfy this calculation.
Best Answer
You may use the properties of the Schur complement
\begin{equation} \left(\begin{matrix} A & B \\ C & D \end{matrix}\right)^{-1} = \left(\begin{matrix} A^{-1} + A^{-1} B S^{-1} C A^{-1} & -A^{-1} B S^{-1} \\ -S^{-1} C A^{-1} & S^{-1} \end{matrix}\right)= \left(\begin{matrix} I_n & -A^{-1} B \\ 0 & I_m \end{matrix}\right) \left(\begin{matrix} A^{-1} & 0 \\ 0 & S^{-1} \end{matrix}\right) \left(\begin{matrix} I_n & 0 \\ -C A^{-1} & I_m \end{matrix}\right), \end{equation}
with the Schur matrix
\begin{equation} S = D - C A^{-1} B \end{equation}
and the inverse of an outer product
\begin{equation} (E \otimes F)^{-1} = E^{-1} \otimes F^{-1}. \end{equation}
Note that you still have to invert the submatrices $A, B, C, D$.
Regards