# Energy Efficiency Optimization in Massive MIMO Secure Multicast Transmission
## Appendix A Proof of Theorem 1  
We can observe that the upper bound of the eavesdropper transmit rate \(R_{\text{ev,ub}}\) in (15) is irrelevant to the off-diagonal elements of \(\mathbf{V}^{H}\mathbf{Q}\mathbf{V}\). Then, we define \(\tilde{\mathbf{Q}}=\mathbf{V}^{H}\mathbf{Q}\mathbf{V}\). The multicast rate of UTs \(R_{\text{se}}\) in (20) can be expressed as follows,  
\[R_{\text{se}}=\min_{k}\quad\mathsf{E}\left\{\log\det\left\{\mathbf{I}_{N_{t}} +\mathbf{G}_{k}\tilde{\mathbf{Q}}\mathbf{G}_{k}^{H}\right\}\right\}-R_{\text{ ev,ub}}(\tilde{\mathbf{Q}}), \tag{10}\]  
Following a similar line of reasoning of the proof in [49], define \(\mathbf{D}_{r}\in\mathbb{R}^{N_{r}\times N_{r}}\) as a diagonal matrix whose diagonal entries are all \(1\) except the \((r,r)\)th entry being \(-1\). Please note that the matrix \(\mathbf{D}_{r}\) is an unitary matrix, which means \(\mathbf{G}_{k}\mathbf{D}_{r}\) has the same distribution as \(\mathbf{G}_{k}\)[49], thus  
\[R_{\text{se}}\left(\mathbf{D}_{r}\tilde{\mathbf{Q}}\mathbf{D}_ {r}^{H}\right)\] \[= \min_{k}\quad\mathsf{E}\left\{\log\det\left\{\mathbf{I}_{N_{r}} +\mathbf{G}_{k}\mathbf{D}_{r}\tilde{\mathbf{Q}}\mathbf{D}_{r}^{H}\mathbf{G}_{ k}^{H}\right\}\right\}-R_{\text{ev,ub}}(\mathbf{D}_{r}\tilde{\mathbf{Q}} \mathbf{D}_{r}^{H})\] \[= \min_{k}\quad\mathsf{E}\left\{\log\det\left\{\mathbf{I}_{N_{t}} +\mathbf{G}_{k}\tilde{\mathbf{Q}}\mathbf{G}_{k}^{H}\right\}\right\}-R_{\text{ ev,ub}}(\tilde{\mathbf{Q}})\] \[= R_{\text{se}}\left(\tilde{\mathbf{Q}}\right). \tag{11}\]  
Here, \(\tilde{\mathbf{Q}}\) is equal to \(\mathbf{D}_{r}\tilde{\mathbf{Q}}\mathbf{D}_{r}^{H}\) expect for the off-diagonal entries in the \(r\)th row and the \(r\)th column, whose signs are reverse. We want to prove that nulling all the off-diagonal entries of \(\tilde{\mathbf{Q}}\) in the \(r\)-th row and the \(r\)-th column will not decrease the transmit rate with respect to \(\tilde{\mathbf{Q}}\). We invoke [PERSON]'s inequality to prove it.  
\[R_{\text{se}}\left(\frac{1}{2}\left(\tilde{\mathbf{Q}}+\mathbf{D} _{r}\tilde{\mathbf{Q}}\mathbf{D}_{r}^{H}\right)\right)\] \[= \min_{k}\quad\mathsf{E}\left\{\log\det\left\{\mathbf{I}_{N_{r}} +\frac{1}{2}\left(\mathbf{G}_{k}\left(\tilde{\mathbf{Q}}+\mathbf{D}_{r}\tilde {\mathbf{Q}}\mathbf{D}_{r}^{H}\right)\mathbf{G}_{k}^{H}\right)\right\}\right\}-R _{\text{ev,ub}}(\frac{1}{2}\left(\tilde{\mathbf{Q}}+\mathbf{D}_{r}\tilde{ \mathbf{Q}}\mathbf{D}_{r}^{H}\right))\] \[\geq \min_{k}\left(\frac{1}{2}\mathsf{E}\left\{\log\det\left\{\mathbf{ I}_{N_{t}}+\mathbf{G}_{k}\tilde{\mathbf{Q}}\mathbf{G}_{k}^{H}\right\}\right\} \right)-\frac{1}{2}R_{\text{ev,ub}}(\tilde{\mathbf{Q}})\] \[+ \min_{k}\left(\frac{1}{2}\mathsf{E}\left\{\log\det\left\{\mathbf{ I}_{N_{t}}+\mathbf{G}_{k}\mathbf{D}_{r}\tilde{\mathbf{Q}}\mathbf{D}_{r}^{H}\mathbf{G}_{ k}^{H}\right\}\right\}\right)-\frac{1}{2}R_{\text{ev,ub}}(\mathbf{D}_{r}\tilde{ \mathbf{Q}}\mathbf{D}_{r}^{H})\] \[= \frac{1}{2}R_{\text{se}}\left(\tilde{\mathbf{Q}}\right)+\frac{1}{2 }R_{\text{se}}\left(\mathbf{D}_{r}\tilde{\mathbf{Q}}\mathbf{D}_{r}^{H}\right)\] \[= R_{\text{se}}\left(\tilde{\mathbf{Q}}\right). \tag{12}\]We repeat the \(r\) for 1 to \(N_{r}\), and find that the numerator of the problem is maximized when \(\tilde{\mathbf{Q}}\) is diagonal. Meanwhile, the consumed power \(P\triangleq\zeta\text{tr}\left\{\mathbf{Q}\right\}+MP_{\text{c}}+P_{\text{s}}\) is irrelevant to the off-diagonal elements of \(\mathbf{V}^{H}\mathbf{Q}\mathbf{V}\). Thus, the objective problem in (20) is maximized when \(\tilde{\mathbf{Q}}=\mathbf{V}^{H}\mathbf{Q}\mathbf{V}=\mathbf{V}^{H}\mathbf{ \Phi}\Lambda\mathbf{\Phi}^{H}\mathbf{V}\) is diagonal. This concludes the proof.