Upper bound on the largest eigenvalue 
Let
$$A := W \left( P \Sigma P^{\top} + W^{\top} W \right)^{-1} W^{\top}$$
where $W \in \mathbb{R}^{n \times k}$ may either be fat or skinny, and it's possible $\text{Rank}(W) < \min(n,k)$. $P \in \mathbb{R}^{k \times k}$ is an orthonormal matrix, and $\Sigma \in \mathbb{R}^{k \times k}$ is a diagonal matrix with positive diagonal entries (so $P \Sigma P^{\top}$ is positive definite).
Show that the largest eigenvalue of $A$ is less than $1$.


I tried the Woodbury identity but failed. I also looked at this post, but its Cholesky decomposition does not seem to help much about upper bound on eigenvalue. And I did some numerical experiment and it seems to me that the eigenvalue must be bounded by $1$.
Any help would be greatly appreciated!
 A: We can write simply $A=W(S+W^TW)^{-1}W^T$, where $S$ is positive definite. Note that the matrix $A$ itself is positive semidefinite and hence its eigenvalues are real nonnegative.
Since this is true, the nonzero eigenvalues of $A$ are the same as nonzero eigenvalues of $B$, where
$$
B=(S+W^TW)^{-1}W^TW.
$$
Note that the eigenvectors can be chosen real as well.
Assume that $\lambda $ and $x$ is an eigenpair of $B$ so
$$\tag{1}
Bx=\lambda x 
\iff
(S+W^TW)^{-1}W^TWx=\lambda x.
$$
This gives $W^TWx=\lambda(S+W^TW)x$ and by pre-multiplying with $x^T$ we get
$$\tag{2}
(1-\lambda)x^TW^TWx=\lambda x^TSx.
$$
The only interesting case is when $\lambda>0$ (otherwise, there's nothing to prove). This implies that $Wx\neq 0$. Therefore, we have $x^TW^TWx>0$. Since we also have $x^TSx>0$, putting these facts together in (2), we get that $(1-\lambda)>0$ and hence $\lambda<1$.
A: Assume $W$ is full rank, then $W$ has a $QR$ factorization, i.e. $W = QR$ where $Q$ has orthonormal column and $R$ is an invertible upper triangular matrix. Hence it follows
\begin{align}
A = QR(P\Sigma P^T+R^TR)^{-1}R^T Q^T = Q(R^{-T}P\Sigma P^T  R^{-1}+I)^{-1}Q^T 
\end{align}
which means
\begin{align}
\|A \mathbf{x}\|^2 = \mathbf{x}^TA^TA\mathbf{x} = \mathbf{x}^TQ(R^{-T}P\Sigma P^T  R^{-1}+I)^{-2}Q^T\mathbf{x} \leq \mathbf{x}^TQQ^T\mathbf{x} = \|Q^T\mathbf{x}\|^2 \leq \|\mathbf{x}\|^2.
\end{align}
The rest is straightforward.
