Proof that eigenvalue of matrix product smaller than 1 Suppose that we are given an $M\times N$ complex matrix 
$\mathbf{A}$ and an $N\times N$ real diagonal matrix $\mathbf{D}$ with non-negative entries on the diagonal. Through numerical simulations, I found that the eigenvalues of $\mathbf{B}$, which is defined as $$\mathbf{B}=\mathbf{A}(\mathbf{A}^{\mathrm{H}}\mathbf{A} + \mathbf{D})^{-1}\mathbf{A}^{\mathrm{H}},$$ are no larger than $1$, where $(\cdot)^\mathrm{H}$ denotes matrix conjugate transpose. How can I prove that such an observation holds theoretically? Or is there any counter-example to show that this observation does not always hold?
I notice that $\mathbf{A}(\mathbf{A}^{\mathrm{H}}\mathbf{A} + \mathbf{D})^{-1}\mathbf{A}^{\mathrm{H}}$ shares the same non-zero eigenvalues as 
$(\mathbf{A}^{\mathrm{H}}\mathbf{A} + \mathbf{D})^{-1}\mathbf{A}^{\mathrm{H}}\mathbf{A}$. This motivates me to consider if I could approach this proof through an upper bound for the largest eigenvalue of matrix product, i.e., the product of $(\mathbf{A}^{\mathrm{H}}\mathbf{A} + \mathbf{D})^{-1}$ and $\mathbf{A}^{\mathrm{H}}\mathbf{A}$. However, so far I have gone nowhere. Any suggestion would be greatly appreciated.
 A: First, let us reduce this to the case $D=I$. Set $C:=AD^{-1/2}$. Then we have
$$
A^HA+D = D^{1/2}( D^{-1/2}A^HAD^{-1/2} + I)D^{1/2} = D^{1/2}(C^HC+I)D^{1/2}
$$
and
$$
A(A^HA+D)^{-1}A^H =A D^{-1/2}(C^HC+I)^{-1}D^{-1/2}A^H = C(C^HC+I)^{-1}C^H.
$$
The matrix $C^HC$ is Hermitian and positive semidefinite, hence is diagonalizable with real eigenvalues. 
Let $u,v$ be singular vectors of $C$ to singular value $\sigma\ge0$, i.e., $Cu = \sigma v$ and $C^Hv=\sigma u$. Then
$(C^HC+I)u=(\sigma^2+1)u$ implies $(C^HC+I)^{-1}u =(\sigma^2+1)^{-1}u$ and
$$
C(C^HC+I)^{-1}C^Hv=\frac{\sigma^2}{1+\sigma^2}v.
$$
Hence, $v$ is an eigenvector of $C(C^HC+I)C^H$ to the eigenvalue $\frac{\sigma^2}{1+\sigma^2}\le1$. As we can choose an orthonormal basis of such singular vectors $v$, the claim follows.
If you do not like to work with singular vectors, then we can do the following: Let $x$ be an eigenvector of $CC^H$ to the eigenvalue $\lambda\ge0$. Define $u:=(C^HC+I)^{-1}C^Hx$. Then it holds
$$
C(C^HC +I)u = CC^Hx = \lambda x,
$$
implying $Cu = \lambda x - CC^HCu$. Then we obtain
$$
x^HC(C^HC+I)^{-1}C^Hx=x^HCu = \lambda \|x\|^2-xCC^HCu = \lambda \|x\|^2-\lambda x^HCu,
$$
which implies $x^HCu = \frac\lambda{\lambda + 1} \|x\|^2$
and
$$
x^HC(C^HC+I)^{-1}C^Hx = \frac\lambda{\lambda + 1} \|x\|^2,
$$
hence $C(C^HC+I)^{-1}C^H$ has norm smaller than one.
