Inverse of the sum of two orthogonal projections I am trying to find out, if there is a formula for finding the inverse of the sum of two orthogonal projections. So basically my questions is:
If $\left[\mathbf{A},\mathbf{B}\right]$ is full rank, then
$\left(\mathbf{A}\left(\mathbf{A}^{\mathrm{H}}\mathbf{A}\right)^{-1}\mathbf{A}^{\mathrm{H}} + 
\mathbf{B}\left(\mathbf{B}^{\mathrm{H}}\mathbf{B}\right)^{-1}\mathbf{B}^{\mathrm{H}}\right)^{-1}$ = ?  
 A: A projection splits your vector space (which I guess here is $\mathbb{R}^n$), and expresses it as a direct sum of the image and the kernel. Obviously, everything in the kernel is sent to $0$. On the image, the projection acts by the identity.
So now consider the sum of two projections, $P$ and $Q$. 
We can divide up the vector space as a direct sum of four subspaces: $\text{im}(P)\cap\text{im}(Q)$, $\text{im}(P)\cap\text{ker}(Q)$, $\text{ker}(P)\cap \text{im}(Q)$ and $\text{ker}(P)\cap\text{ker}(Q)$.
As we want to find the inverse, we're assuming that $P+Q$ has no kernel, so the last of these four is trivial, and we only have three parts to deal with.
On $\text{ker}(P)\cap \text{im}(Q)$ the map $P+Q$ acts as the identity. The same is true of $\text{im}(P)\cap\text{ker}(Q)$. So the inverse on those subspaces is the identity.
On $\text{im}(P)\cap\text{im}(Q)$ the map $P+Q$ acts as twice the identity, so the inverse there is half the identity.
All that's left then is to apply the change of basis needed to get between a basis that matches the direct sum to the one you are interested in.
