$\hat{Z}_K$.

An Algorithm
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Covariance Matrix Estimation:
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The optimization problem (2.4) can be written as:\
$$\begin{aligned}
\hat{R}_K=\operatorname*{arg\,min}_{R=R^T|(\lambda, \gamma) \in \hat{\mathscr{S}}^{K}_1 }~f(R),\end{aligned}$$ where $$\begin{aligned}
f(R)=||R-K||^2_F +  \lambda \|{R}^-\|_1  + \gamma\sum_{i=1}^{p} \{\sigma_i(R)-\bar{\sigma}(R)\}^2.\end{aligned}$$ Note that $\sum_{i=1}^{p} \{\sigma_i(R)-\bar{\sigma}(R)\}^2=tr(R^2)-2~tr(R)+p$, where we have used the constraint $tr(R)=p$. Therefore, $$\begin{aligned}
f(R) & = & \|R-K\|_F^2+\lambda \|{R}^-\|_1 +\gamma~ tr(R^2)-2~ \gamma~tr(R) +p\\
&  = & tr(R^2 (1+\gamma))-2tr\{R(K+\gamma I)\}+ tr(K^TK) +\lambda ~\|{R}^-\|_1 +p\\
&  = & (1+\gamma)\{tr(R^2)-2/(1+\gamma) tr\{R(K+\gamma I)\}+ (1/(1+\gamma))tr(K^TK)\} \\
&   &~~~~+\lambda ~\|{R}^-\|_1 +p \\
&  = & (1+\gamma)\{\|R- (K+\gamma I)/(1+\gamma)\|^2_F+ (1/(1+\gamma))tr(K^TK)\} \\
 &   & ~~~~+\lambda ~\|{R}^-\|_1 +p.\end{aligned}$$ The solution of (4.1) is soft thresholding estimator and it is given by: $$\begin{aligned}
\hat{R}_K =\frac{1}{1+\gamma}~\text{sign}(K)*\text{pmax}\{\text{abs}(K+\gamma ~I)-\frac{\lambda}{2},0\}\end{aligned}$$ with $(\hat{R}_{K})_{ii}=(K_{ii}+\gamma)/(1+\gamma)$, $pmax(A,b)_{ij}:=max(A_{ij},b)$ is elementwise max function for each entry of the matrix $A$. Note that for each $(\lambda,\gamma) \in \hat{\mathscr{S}}^{K}_1$, $\hat{R}_{K}$ is positive definite.\
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[**Choice of $\lambda$ and $\gamma$:**]{} For a given value of $\gamma$, we can find the value of $\lambda $ satisfying: $$\begin{aligned}
\sigma_{min}\{(K+\gamma I)-\frac{\lambda}{2}*sign(K+\gamma I)\}>0\end{aligned}$$ which can be simplified to $$\begin{aligned}
\lambda < \frac{\sigma_{min}(K+\gamma I)}{C_{12}~\sigma_{max}(\text{sign}(K))}. \end{aligned}$$ For some $C_{12} \ge 0.5$. Such choice of $(\lambda,\gamma)\in \hat{\mathscr{S}}_1^{{K}}$, and the estimator $\hat{R}_{K}$ is positive definite. Smaller values of $C_{12}$ yeild a solution which is more sparse but may not be 