Eigenvalues of $(A'A)^{-1}(A'BA)$ Let $\mathbf A$ be a $n\times r$ matrix with $\mathbf A'\mathbf A$ invertible and, for some $u<n$ let
$$\mathbf B = \begin{bmatrix}\mathbf I_u & \mathbf 0 \\ \mathbf 0 & \mathbf 0\end{bmatrix}_{n\times n},$$
where $\mathbf I_u$ is the $u\times u$ identity matrix.

What can be said about the eigenvalues of $(\mathbf A'\mathbf A)^{-1}(\mathbf A'\mathbf B\mathbf A)$?

I suspect that the eigenvalues always belong to $[-1,1]$, but don't know how start. Any hints would be very welcome.
 A: First consider that $BA$ is just $A$ with the bottom $n-u$ rows zeroed out.
Now let 
$$
C=
\begin{bmatrix}
0 & 0 \\
0 & I_{n-u}
\end{bmatrix}
$$
Consider that $(A'A)^{-1}(A'BA) + (A'A)^{-1}(A'CA) = I$ by linearity.  
Hope that's a helpful start!
A: We can show that the eigenvalues of $M = (A'A)^{-1}(A'BA)$ are always in $[0,1]$.   We also know that the number of zero-eigenvalues will always be $(n-u)$.

Claim: $M$ has real, non-negative eigenvalues

Proof: We find that the similar matrix
$$
(A'A)^{1/2}M(A'A)^{-1/2} = 
(A'A)^{1/2}(A'A)^{-1}(A'BA)(A'A)^{-1/2} = 
(A'A)^{-1/2}(A'BA)(A'A)^{-1/2}
$$
is positive semidefinite. 

Claim: The eigenvalues of $M$ are smaller than $1$.

Proof: Suppose that $A$ and $B$ are such that the resulting $M$ has an eigenvalue greater than $1$.  Note that by Badam's observation, $I - M$ is similar to a $\tilde M$ resulting from some other pair $\tilde A,\tilde B$ (or, more directly: $M$ is similar to a positive semidefinite matrix).  However, this implies that the eigenvalues of $I - M$ are non-negative.  This implies that the eigenvalues of $M$ are $\leq 1$, as desired.

Given a vector $y$, $My$ is the solution $x$ to the equation
$$
A'(Ax) = A'B(Ay)
$$
which is the least squares solution to the equation $Ax = BAy$.
We can also write $M = A^+BA$, where $A^+$ denotes the pseudoinverse.
