Suppose $\bf A$ and $\bf B$ are $n \times m$ matrices with $n\ge m$, suppose $\bf AA^\top+BB^\top$ is invertible, and consider the matrix
$$\bf C = A^\top(AA^\top+BB^\top)^{-1}A.$$
Conjecture:
The matrix $\bf C$ has at least $n-m$ eigenvalues equal to $1$ and all other eigenvalues between $0$ and $1$
I am having a hard time proving this, so any help would be much appreciated. I have verified it numerically using the simple Matlab code bellow:
n = 7;
m = 5;
A = randn(n,m);
B = randn(n,m);
C = A'*((A*A'+B*B')^(-1))*A;
disp(eig(C))
$rank(AA^T)\leq m,rank(BB^T)\leq m$ implies that the symmetric $\geq 0$ matrices $AA^T,BB^T$ have at least $n-m$ zero eigenvalues, the other eigenvalues being $\geq 0$. Note also that $(AA^T+BB^T)^{-1}$ is symmetric $>0$ and, consequently, (1) the symmetric matrix $C$ is $\geq 0$.
On the other hand, (2) $spectrum(AA^T(AA^T+BB^T)^{-1})=spectrum(C)\cup \{0_1,\cdots,0_{n-m}\}$.
Since (3) $AA^T(AA^T+BB^T)^{-1}=I_n-BB^T(AA^T+BB^T)^{-1}$ and $rank(BB^T(AA^T+BB^T)^{-1})\leq m$, we deduce that $AA^T(AA^T+BB^T)^{-1}$ has at least $n-m$ eigenvalues equal to $1$ and, according to (2), $C$ too.
Since $BB^T(AA^T+BB^T)^{-1}$ is the product of a symmetrix matrix $\geq 0$ and a symmetric matrix $>0$, it has only $\geq 0$ eigenvalues and all the eigenvalues of $AA^T(AA^T+BB^T)^{-1}$ are $\leq 1$.
According to (1),(2), $spectrum(C)\subset [0,1]$.