I'm having trouble proving the following.
$A^\top B A (A^\top B A)^{\dagger} A^\top B = A^\top B$
In the above, $A$ ($n \times m$) and $B$ ($n \times n$) are any matrices (i.e. possibly rank-deficient).
Edit: OK, I see it doesn't hold for general $B$. But what about the case where $A$ is still arbitrary, but $B$ is invertible?
I have made two attempts. First, there is the well-known identity $A^\top B A (A^\top B A)^{\dagger} A^\top B A = A^\top B A$. But this doesn't imply the result because $A$ may have linearly dependent rows.
Also, it is easy to show the result for $B=I$ by taking the SVD of $A$ and cancelling terms on the left-hand side. Sadly, this doesn't seem to generalize.
I apologize in advance if the question is trivial - I am a computer scientist and don't have much experience doing this sort of thing.