I'd like to find the inverse of a certain 2x2 block matrix. Since its structure is peculiar, the usual inverse formula cannot be applied. However, for the same reason, I think there is a way to elegantly compute it. Let $A$ be a $m\times n$ matrix, I am interested in computing the inverse of
$$M =\begin{bmatrix} I \quad A^T\\ A \quad 0 \end{bmatrix}$$
with $I$ the identity matrix of size $m$ and 0 the null square matrix of size $n$. Thus, is there a simple way of computing $M^{-1}$?
The block matrix inverse formula states that
$\begin{bmatrix} \mathbf{A} & \mathbf{B} \\ \mathbf{C} & \mathbf{D} \end{bmatrix}^{-1} = \begin{bmatrix} \mathbf{A}^{-1} + \mathbf{A}^{-1}\mathbf{B}\left(\mathbf{D} - \mathbf{CA}^{-1}\mathbf{B}\right)^{-1}\mathbf{CA}^{-1} & -\mathbf{A}^{-1}\mathbf{B}\left(\mathbf{D} - \mathbf{CA}^{-1}\mathbf{B}\right)^{-1} \\ -\left(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B}\right)^{-1}\mathbf{CA}^{-1} & \left(\mathbf{D} - \mathbf{CA}^{-1}\mathbf{B}\right)^{-1} \end{bmatrix}$
provided that $\mathbf{A}$ and $\mathbf{D}-\mathbf{C}\mathbf{A}^{-1}\mathbf{B}$ are both invertible.
Applying that formula yields
$$\begin{bmatrix}I & A^T \\ A & 0 \end{bmatrix}^{-1} = \begin{bmatrix} I - A^T(AA^T)^{-1}A & A^T(AA^T)^{-1}\\ (AA^T)^{-1}A& -(AA^T)^{-1}\end{bmatrix}$$
provided that $AA^T$ is invertible.