Possible eigenvalues of matrix $C = A(B^TA)^{-1}B^T$ We also have the conditions that $A$ and $B$ are rectangular matrices such that $B^TA$ is square and invertible.
Thus, we can assume that $A$ and $B$ have dimensions $p \times n$ and hence $B^TA$ has dimensions $n \times n$.

So, what I have done so far is that since $$C \times C = A(B^TA)^{-1}B^TA(B^TA)^{-1}B^T = A(B^TA)^{-1}B^T = C$$
we have $$ C^2 = C$$
So, by the fact that eigenvalues are roots of the characteristic polynomial, possible eigenvalues can be $0$ or $1$.
But isn't there a problem for the eigenvalue $0$ as if $Cx = 0$ for some non-zero vector $x$ then multiplying $B^T$ from the left (this can be done as $C$ has dimensions $p \times p$ and $B^T$ has dimensions $n \times p$) gives $B^TCx = 0$ which means $$B^TA(B^TA)^{-1}B^Tx = 0 \implies B^TAy = 0$$ for some non-zero $y$, so this is contradicting that $B^TA$ is invertible.

So does this mean that only possible eigenvalue is $1$?
Or is it possible that $y$ can be a zero vector so I can't conclude anything about that?
Edit: I think that $y$ can be the zero vector because in that case, we just need $B^Tx$ to be zero as $B^TA$ is invertible. $B^Tx = 0$ is possible for a non-zero $x$, isn't it?
 A: $B^T x=0$ is possible for nonzero $x$. A simple example comes from taking 
$$
A=B=\begin{pmatrix} 1 & 0 \\
0 & 1 \\
0 & 0
\end{pmatrix}.
$$
This meets all the desired properties and the matrix $C$ has zero as an eigenvalue.
Also note that if $A$ and $B$ are $p \times n$ and $ B^T A$ is invertible, then we must have $n \leq p$, since $A$, hence $B^T A$, has rank at most $\min (p,n)$ and $B^T A$ is $n \times n$. Furthermore, the rank of $C$ is at most $n$, since the rank of $A$ is at most $n$. It follows that $C$ always zero as an eigenvalue with geometric multiplicity at least $p-n$. 
A: To supplement Eric’s answer, one way that this expression comes up is in the construction of a projector onto the eigenspace corresponding to a non-defective eigenvalue, i.e., one with algebraic and geometric multiplicity both equal to some $r>0$. The columns of $A$ consist of $r$ linearly-independent right eigenvectors, while the columns of $B$ consist of left eigenvectors. If the eigenspace is not the entire ambient space, then these matrices will be rank-deficient, and in particular $B^T$ will have a nontrivial null space, and so will the projector.
