Largest eigenvalue of a Hermitian Toeplitz matrix I have two Toeplitz positive semi-definite Hermitian matrices $\mathbf{R}_1, \mathbf{R}_2 \in \mathbb{C}^{M \times M}$. They are in fact covariance matrices satisfing the following conditions:

*

*${\mathop{\rm diag}\nolimits}\{\mathbf{R}_1\} = d_1 \mathbf{I}_M$ and ${\mathop{\rm diag}\nolimits}\{\mathbf{R}_2\} = d_2 \mathbf{I}_M$, where $d_1$ and $d_2$ are real numbers.


*The off-diagonal entries of the covariance matrix are complex with absolute value no larger than the diagonal entires. In other words, the ij-th element of $\mathbf{R}$, namely $r_{ij}, \forall i\neq j$, satisfies $|r_{ij}| \leq d$, where $d$ is the diagonal element(s) of $\mathbf{R}$.
I'm interested in the largest eigenvalue (or spectral norm) of the following matrix:
$$\mathbf{R}_1 (\mathbf{R}_1 + \mathbf{R}_2 +\mathbf{I}_M)^{-2} \mathbf{R}_1,$$
where $\mathbf{I}_M$ is the identity matrix. I tried with Matlab and observed that the largest eigenvalue is always smaller than $1$. However I couldn't prove it. Is there anyone who can show me the way?

A first step could be the following: A corollary to Weyl's eigenvalue inequalities (see, e.g., Horn & Johnson, second edition, Corollary 4.3.15), is that the largest eigenvalue of a sum of matrices is as least as large as the sum of the largest eigenvalue of one of the matrices and the smallest eigenvalue of the other matrix, i.e.,
$$ \lambda_M(R_1+R_2)\ge\lambda_M(R_1)+\lambda_1(R_2) $$
Since the matrices are covariance matrices, they are positive semidefinite, and hence $\lambda_1(R_2)\ge0$.
Note that you can easily omit the identity matrix by just adding it to the matrix $R_2$. Note further that, since all matrices involved are symmetric,
$$ R_1(R_1+R_2)^{-2}R_1 = R_1(R_1+R_2)^{-1}(R_1+R_2)^{-1}R_1 = R_1(R_1+R_2)^{-1}\left(R_1(R_1+R_2)^{-1}\right)^T. $$
Hence, for the moment it might suffice to determine the largest eigenvalue of $R_1(R_1+R_2)^{-1}$ and then use the Corollary to Gelfand's formula to bound
$$ \lambda_M\left(R_1(R_1+R_2)^{-1}\left(R_1(R_1+R_2)^{-1}\right)^T\right) \le \lambda_M(R_1(R_1+R_2)^{-1})^2. $$
