Let $\Sigma$ be a $2\times 2$ covariance matrix and ${\bf h}$ a vector of complex values entries.
$$A= \log(1+ {\bf h}^* \Sigma {\bf h} )$$
$$\Sigma = \begin{bmatrix} 1-|\rho_1|^2 & \rho_3 - \rho_1 \rho_2^* \\ \rho_3^* - \rho_1^*\rho_2 & 1-|\rho_2|^2 \\ \end{bmatrix}$$
$$0<|\rho_i|<1$$
Someone mentioned that I can upperbound the above by $$A\leq \log(1+ {\bf h}^* {\bf I} {\bf h} )$$
${\bf I}$ being the identity matrix. Does anyone know why this is the case?
Can one say that the following inequality is true
$$\Sigma \leq \frac{1}{2}{\bf I}$$ Thank you. Please let me know if I can provide more details.
I would ask that "someone" for more details first why this should be true.
Forget about the logarithm; if its arguments are well-defined (positive), the inequality holds (assuming it should hold for any $h$) if and only if $\Sigma\preceq I$. But take, e.g., for $\rho_1=\rho_2=\rho_3=1/4$ and $h=[1,1]^T$, $$ h^*\Sigma h=\frac{5}{4}\not\leq\frac{4}{4}=1=h^*h. $$ In fact, if $\rho_1=\rho_2=\rho_3=\rho$ (assuming for simplicity that $0<\rho<1$), $$ \Lambda(\Sigma)=\{1-\rho,(1-\rho)(1+2\rho)\}. $$ Choosing a positive $\rho$ smaller than $1/2$ (e.g., $\rho=1/4$ as in the first example) makes the second eigenvalue larger than one.