## Appendix A Proof of Theorem 1
Let \(\mathbf{P}^{\star}=\mathbf{U}_{P}\mathbf{D}_{P}^{\star}\mathbf{V}_{P}^{H}\) be the singular-value decomposition (SVD) of \(\mathbf{P}^{\star}\), where the main diagonal elements of the rectangular diagonal matrix \(\mathbf{D}_{P}^{\star}\) consist of the singular values of \(\mathbf{P}^{\star}\), and the column vectors of \(\mathbf{U}_{P}\) and \(\mathbf{V}_{P}\) are the left and right singular vectors of \(\mathbf{P}^{\star}\), respectively. Then, the diagonal structure of \(\mathbf{U}_{Q}^{H}\mathbf{P}^{\star H}\mathbf{P}^{\star}\mathbf{U}_{Q}\) can be simply obtained with the choice of \(\mathbf{V}_{P}=\mathbf{U}_{Q}\). Multiplying (A3c) on the right-hand side by \((\tilde{\mathbf{T}}+\tilde{\mathbf{T}}^{\star})\) and using the condition \((\tilde{\mathbf{T}}+\tilde{\mathbf{T}}^{\star})\mathbf{\Gamma}_{T}=\mathbf{0}\) in (A3e), one can obtain that  
\[=J(\mathbf{P},\mathbf{\tilde{T}},\mathbf{\tilde{R}},\mathbf{ \tilde{Q}})+\mu_{P}\Big{[}\text{Tr}(\mathbf{P}\mathbf{P}^{H})-P_{T}\Big{]}\]2 \[=J(\mathbf{P},\mathbf{\tilde{T}},\mathbf{\tilde{R}},\mathbf{ \tilde{Q}})+\mu_{P}\Big{[}\text{Tr}(\mathbf{P}\mathbf{P}^{H})-P_{T}\Big{]}\]3 (A8)  
Also, multiplying (A3a) on the left-hand side by \(\mathbf{P}^{\star H}\), we have  
\[=J(\mathbf{P},\mathbf{\tilde{T}},\mathbf{\tilde{R}},\mathbf{ \tilde{Q}})+\mu_{P}\Big{[}\text{Tr}(\mathbf{P}\mathbf{P}^{H})-P_{T}\Big{]}\]4 (A9)  
By substituting (A9) into (A8), the condition (A8) can be rewritten as  
\[=J(\mathbf{P},\mathbf{\tilde{T}},\mathbf{\tilde{R}},\mathbf{ \tilde{Q}})+\mu_{P}\Big{[}\text{Tr}(\mathbf{P}\mathbf{P}^{H})-P_{T}\Big{]}\]5 (A10)  
Since the right-hand side of (A10) is Hermitian, the left-hand side of (A10) is also Hermitian. From this, we have \((\tilde{\mathbf{T}}+\tilde{\mathbf{T}}^{\star})\tilde{\mathbf{T}}^{\star}=\tilde{ \mathbf{T}}^{\star}(\tilde{\mathbf{T}}+\tilde{\mathbf{T}}^{\star})\), or equivalently, \(\tilde{\mathbf{T}}^{\star}\tilde{\mathbf{T}}=\tilde{\mathbf{T}}\tilde{\mathbf{T}}^ {\star}\). From Lemma A1, it follows that \(\tilde{\mathbf{T}}\) and \(\tilde{\mathbf{T}}^{\star}\) share the same eigenvectors. Let \(\tilde{\mathbf{T}}=\mathbf{U}_{T}\mathbf{\Lambda}_{T}\mathbf{U}_{T}^{H}\) be the EVD of \(\hat{\mathbf{T}}\). Then, the structure of \(\tilde{\mathbf{T}}^{\star}\) is obtained as \(\tilde{\mathbf{T}}^{\star}=\mathbf{U}_{T}\mathbf{D}_{T}^{\star}\mathbf{U}_{T}^ {H}\), where \(\mathbf{D}_{T}^{\star}\) is an \(M_{t}\times M_{t}\) real diagonal matrix whose diagonal elements consist of the eigenvalues of \(\tilde{\mathbf{T}}^{\star}\). By using \(\tilde{\mathbf{T}}^{\star}=\mathbf{U}_{T}\mathbf{D}_{T}^{\star}\mathbf{U}_{T}^ {H}\), (A10) can be rewritten as  
\[=J(\mathbf{P},\mathbf{\tilde{T}},\mathbf{\tilde{R}},\mathbf{ \tilde{Q}})+\mu_{P}\Big{[}\text{Tr}(\mathbf{P}\mathbf{P}^{H})-P_{T}\Big{]}\]6 (A11)  
Since the left-hand side of (A11) is diagonal, \(\mathbf{U}_{T}^{H}\big{(}\mathbf{Z}^{-1}-\mu_{P}\mathbf{P}^{\star}\mathbf{P}^ {\star H}\big{)}\mathbf{U}_{T}\) should be diagonal. With some manipulations, it can be shown that the diagonal structure of \(\mathbf{U}_{T}^{H}\big{(}\mathbf{Z}^{-1}-\mu_{P}\mathbf{P}^{\star}\mathbf{P}^ {\star H}\big{)}\mathbf{U}_{T}\) can be obtained when \(\mathbf{U}_{T}^{H}\mathbf{U}_{P}\mathbf{D}_{P}^{\star}\) is a rectangular diagonal matrix. This can be achieved by simply setting \(\mathbf{U}_{P}=\mathbf{U}_{T}\). Therefore, the optimal training structure is obtained as \(\mathbf{P}^{\star}=\mathbf{U}_{T}\mathbf{D}_{P}^{\star}\mathbf{U}_{Q}^{H}\).