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:

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

(2) 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? Thanks!


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

$$ \lambda_M\left(R_1(R_1+R_2)^{-1}\left(R_1(R_1+R_2)^{-1}\right)^T\right) \le \lambda_M(R_1(R_1+R_2)^{-1})^2. $$

  • $\begingroup$ Thanks Bernhard! Yes, I was thinking of using Weyl's inequalities, and I constructed the following matrix $\mathbf{X} = (\mathbf{R}_1 + \mathbf{R}_2) (\mathbf{R}_1 + \mathbf{R}_2 +\mathbf{I}_M)^{-2} (\mathbf{R}_1 + \mathbf{R}_2) $. I can show that $\lambda_1(\mathbf{X}) \leq 1$. But the following matrix $\mathbf{G} = \mathbf{R}_1 (\mathbf{R}_1 + \mathbf{R}_2 +\mathbf{I}_M)^{-2} \mathbf{R}_2 + \mathbf{R}_2 (\mathbf{R}_1 + \mathbf{R}_2 +\mathbf{I}_M)^{-2} \mathbf{R}_1 + \mathbf{R}_2 (\mathbf{R}_1 + \mathbf{R}_2 +\mathbf{I}_M)^{-2} \mathbf{R}_2$ is not positive semidefinite... $\endgroup$ – Echooff Jul 28 '15 at 10:22
  • $\begingroup$ I extended the solution a little bit, maybe this helps... it's an interesting problem! $\endgroup$ – Bernhard Jul 28 '15 at 11:31
  • $\begingroup$ Thanks again for the references. I'm still looking into this problem. My covariance has a some special properties, e.g., the diagonal elements are equal. I'll update the question later. $\endgroup$ – Echooff Jul 28 '15 at 22:58

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