Consider the following block-diagonal matrix:
\begin{pmatrix} P \Sigma P & 0\\ 0 & (I-P) \Sigma (I-P) \end{pmatrix}
Where $P$ is an orthogonal projection matrix, and $\Sigma$ is a diagonal matrix. The individual diagonal elements $P \Sigma P$ and $(I-P) \Sigma (I-P)$ are generally not invertible (unless P doesn't reduce the rank), however, the entire matrix seems to be invertible (validated numerically). Is there a simple expression for its inverse?
Update: @Carl_Schildkraut has shown the matrix is not invertible. Is there a simple expression for the pseudo inverse?
A block matrix $$\begin{pmatrix}A&0\\0&B\end{pmatrix}$$ with $A$ and $B$ square is invertible if and only if $A$ and $B$ are invertible -- otherwise, the rank is less than full (alternatively, its determinant is $\det(A)\det(B)$, which is $0$ if one of $A$ or $B$ fails to be invertible). Its inverse is $$\begin{pmatrix}A^{-1}&0\\0&B^{-1}\end{pmatrix}.$$ However, in your example, it is impossible for both components to be invertible. This would require both $P$ and $I-P$ to be invertible, which means that $P$ cannot have an eigenvalue of $0$ or $1$. However, these are the only possible eigenvalues of a projection matrix!