# pseudoinverse of a square block diagonal matrix with projected diagonal matirces on diagonal

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?

• If this matrix is invertible, then both of the diagonal blocks are invertible since the blocks represent restrictions of your matrix to an invariant subspace. Commented Aug 7, 2021 at 18:29

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!
• @user1767774 I'm unfortunately not that familiar with pseudoinverses, so I'm not sure. My first guess would be that $P\Sigma^{-1}P$ may do the trick (for the top left block matrix; for the bottom right it would be the same with $P$ replaced by $I-P$), but I'm having trouble verifying the necessary conditions. (Update: it's not correct.) Commented Aug 7, 2021 at 18:38