I am working on an Optimization problem and I need to show that
$$AB(B^TAB)^{-1} B^T = I_n$$
where $A$ is $n \times n$ and invertible, $B$ is $n \times k$ with rank $k$, and $k \le n$.
If $B$ is square, then this is a simple calcuation. For $B$ non-square, I have tried the following way (which I'm pretty sure is supposed to work, but there must be a slight flaw in my reasoning somewhere): Since $B$ represents an injection, it has a left inverse; therefore there must be a matrix $D$ such that $DB = I_k$ (and thus $B^T D^T = I_k$). Now, we claim that
$$(B^TAB)^{-1} = DA^{-1}D^T.$$
To prove this, we would like to show that $$(B^TAB) (DA^{-1}D^T) = I_k.$$
However, the multiplication doesn't simplify, since $D$ is a left inverse, not a right inverse.
Any help is greatly appreciated. Thank you!
If $k < n$ this is impossible, as the rank of the left side can't be more than $k$ (i.e. $B$ has rank at most $k$, and $\text{rank}(UV) \le \min(\text{rank}(U),\text{rank}(V))$.