$$\nonumber
I_{sA} = \frac{1}{2}\log_2\det\left(\mathbf{I}_{n_T} +\frac{\rho}{n_T}\left[%
\begin{array}{cccc}
  \lambda_1 & 0 & \alpha_1 & \alpha_2 \\
  0 & \lambda_1 & -\alpha_2^* & \alpha_1^* \\
  \alpha_1^* & -\alpha_2 & \lambda_2 & 0 \\
  \alpha_2^* & \alpha_1 & 0 & \lambda_2 \\
\end{array}%
\right]\right),$$ where $\lambda_i=\sum_{j=1}^{n_T}|h_{ij}|^2$, $\alpha_1=h_{11}h_{21}^*+h_{12}h_{22}^*+h_{13}h_{23}^*+h_{14}h_{24}^*$, and $\alpha_2=-h_{11}h_{22}+h_{12}h_{21}-h_{13}h_{24}^*+h_{14}h_{23}$. Using Fischer’s inequality $$\nonumber
    \det\left(\left[%
\begin{array}{cc}
  \mathbf{A} & \mathbf{B^H} \\         %Oliver B-->B^H
  \mathbf{B} & \mathbf{D} \\           %Oliver C-->B
\end{array}%
\right]\right)\leq \det(\mathbf{A})\det(\mathbf{D})$$ yields $$\nonumber
I_{sA}\leq \log_2\left(\left(1+\frac{\rho}{n_T}\lambda_1\right)\left(1+\frac{\rho}{n_T}\lambda_2\right)\right).$$ By using the arithmetic-geometric inequality, we arrive at $$\nonumber
I_{sA}\leq 2\log_2\left(1+\frac{\rho}{2n_T}||\mathbf{H}||^2\right).$$ This upper bound equals to twice the rate of a full code rate OSTBC for $n_T=4$ transmit and $n_R=2$ receive antennas with a power penalty of $3$ dB. In this particular case a more precise statement can be made due to the following strict form of Fischer’s inequality [@hen.wun-itg05]

\[lem.strictFischer\] Let $\mathbf{P}=\left[
     \begin{array}{cc}
        \mathbf{A} & \mathbf{B^H} \\
        \mathbf{B} & \mathbf{D} \\
     \end{array}%
   \right]$ ($\mathbf{A},\mathbf{D}$ square, nonempty) be positive definite. Then $$\text{$\mathbf{B}$ has full rank}
      \quad\Rightarrow\quad
     \det\mathbf{P} < (\det \mathbf{A}) (\det \mathbf{D})$$

Let $\mathbf{R}\succ0$ denote positive definiteness, and $\mathbf{R}\succ \mathbf{S}$ defined by $(\mathbf{R}-\mathbf{S})\succ 0$. Then [@HornJohnson 7.7.6] $\mathbf{P}\succ0 \Leftrightarrow
    (\mathbf{A}\succ0, \mathbf{D}\succ \mathbf{BA}^{-1}\mathbf{B}^H)$. Thus for arbitrary $\mathbf{B}$ holds $\mathbf{D}-(\mathbf{D}-\mathbf{BA}^{-1}\mathbf{B}^H)=\mathbf{BA}^{-1}\mathbf{B}^H\succeq0$ and becomes strict if $\mathbf{B}$ has full rank. Since $\big(0\prec \mathbf{S}\prec \mathbf{R} 