are equal to 1/2, hence $\mbox{tr}(\gamma^S)=d-1$.

For mixed state $\rho$ on a $d$-dimensional Hilbert space, we have $\mbox{tr}[\gamma(\rho)] = d - \mbox(\rho^2)$, and the same for $\gamma^S(\rho)$. The covariance matrix (symmetric and non-symmetric) also satisfies the concavity property, i.e., if $\rho = \sum_k p_k \rho_k$ is a convex combination of states $\rho_k$, then $$\gamma(\rho) \geq \sum_k p_k \gamma(\rho_k).$$

### Covariance matrix criterion for separability

**Theorem 7**[@cmc1; @cmc2]\
*Let $\rho_{AB}$ be a separable state on $\mathcal{H}_A \otimes \mathcal{H}_B$ and $A_k$ and $B_k$ be the orthonormal observables on the Hilbert spaces $\mathcal{H}_A$ and $\mathcal{H}_B$, with the latter having dimensions $d_A$ and $d_B$ respectively. Define $\{ M_k \} =\{ A_k \otimes I, I \otimes B_k\}$ as discussed previously. Then there exist pure states ${\ensuremath{|\psi_k\rangle}}{\ensuremath{\langle\psi_k|}}$ for $A$, ${\ensuremath{|\phi_k\rangle}}{\ensuremath{\langle\phi_k|}}$ for $B$, and convex weights $p_k$ such that if we define $\kappa_A = \sum_k p_k \gamma({\ensuremath{|\psi_k\rangle}}{\ensuremath{\langle\psi_k|}}, \{ A_{k'} \})$ and $\kappa_B = \sum_k p_k \gamma({\ensuremath{|\phi_k\rangle}}{\ensuremath{\langle\phi_k|}}, \{ B_{k'} \})$, the inequality $$\gamma(\rho_{AB},\{M_k\}) \geq \kappa_A \oplus \kappa_B \Leftrightarrow
\begin{pmatrix}
\mathcal{A} && \mathcal{X} \\
\mathcal{X}^T && \mathcal{B}
\end{pmatrix}
\geq 
\begin{pmatrix}
\kappa_A && 0 \\
0 && \kappa_B
\end{pmatrix}$$ holds.*\
Proof:\
Let us assume $\rho_{AB}$ be a separable state with the following pure state decomposition, $$\rho_{AB} = \sum_k p_k ({\ensuremath{|\psi_k\rangle}}{\ensuremath{\langle\psi_k|}} 