For reference, I am linking the Zhou textbook pdf available online. I want to understand an equation appearing in Lemma 5.3 (on page 122 of the book).
In its proposition, it has the following identity:
For transfer functions $P(s) = P$ and $K(s) = K$,
$$\begin{bmatrix}I & -K \\ -P & I\end{bmatrix}^{-1} = \begin{bmatrix}I + K(I-PK)^{-1}P & K(I-PK)^{-1} \\ (I-PK)^{-1}P & (I-PK)^{-1}\end{bmatrix}.$$
It does not seem to be derived from standard matrix inverse as it deals with transfer functions. How do I compute the inverse of transfer function matrix such as the one above? Is there a general formula for such?
The block matrix inversion formula states that if a square matrix $X$ can be partitioned into four blocks $$\mathbf{X} = \begin{bmatrix} \mathbf{A} & \mathbf{B} \\ \mathbf{C} & \mathbf{D} \end{bmatrix},$$ then provided that $\mathbf{A}$ and $\mathbf{D}-\mathbf{C}\mathbf{A}^{-1}\mathbf{B}$ are both invertible, the inverse of $\mathbf{X}$ is given by $$\mathbf{X}^{-1} = \begin{bmatrix} \mathbf{A} & \mathbf{B} \\ \mathbf{C} & \mathbf{D} \end{bmatrix}^{-1} = \begin{bmatrix} \mathbf{A}^{-1} + \mathbf{A}^{-1}\mathbf{B}\left(\mathbf{D} - \mathbf{CA}^{-1}\mathbf{B}\right)^{-1}\mathbf{CA}^{-1} & -\mathbf{A}^{-1}\mathbf{B}\left(\mathbf{D} - \mathbf{CA}^{-1}\mathbf{B}\right)^{-1} \\ -\left(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B}\right)^{-1}\mathbf{CA}^{-1} & \left(\mathbf{D} - \mathbf{CA}^{-1}\mathbf{B}\right)^{-1} \end{bmatrix}.$$
The specific formula you stated is a special case where $\mathbf{A}$ and $\mathbf{D}$ are both identity matrices.