From (5) to (7), it seems difficult to directly evaluate **W**_{
i
} and **D**_{
j
} because they appear in both signal and noise terms. Therefore, before designing beamformers, three lemmas are derived to make **W**_{
i
} and **D**_{
j
} solvable.

### Lemma 1

Total relay transmission power can be expressed as a quadratic function of **w**:

\begin{array}{l}{P}_{\mathrm{r}}={P}_{1}{\mathbf{w}}^{H}\overline{\mathbf{H}}\mathbf{w}+{P}_{2}{\mathbf{w}}^{H}\mathbf{G}\mathbf{w}+{\sigma}_{v}^{2}{\mathbf{w}}^{H}\mathbf{w},\end{array}

(8)

where

\begin{array}{lcr}{\mathbf{w}}_{i}& =\hfill & \mathit{\text{vec}}\left({\mathbf{W}}_{i}\right),\phantom{\rule{1em}{0ex}}i=1,\dots ,L,\hfill \\ \mathbf{w}& =\hfill & {\left({\mathbf{w}}_{1}^{T},\dots ,{\mathbf{w}}_{L}^{T}\right)}^{T},\hfill \\ {\overline{H}}_{\mathit{\text{ik}}}& =\hfill & \left({\mathbf{h}}_{\mathit{\text{ik}}}^{T}\otimes {\mathbf{I}}_{N}\right),\hfill \\ \overline{H}& =\hfill & \sum _{k=1}^{M}{\overline{H}}_{1k}^{H}{\overline{H}}_{1k}\oplus \dots \oplus \sum _{k=1}^{M}{\overline{H}}_{\mathit{\text{Lk}}}^{H}{\overline{H}}_{\mathit{\text{Lk}}},\hfill \\ {\mathbf{G}}_{\mathit{\text{ik}}}& =\hfill & \left({\overline{g}}_{\mathit{\text{ik}}}^{T}\otimes {\mathbf{I}}_{N}\right),\hfill \\ \mathbf{G}& =\hfill & \sum _{k=1}^{M}{\mathbf{G}}_{1k}^{H}{\mathbf{G}}_{1k}\oplus \dots \oplus \sum _{k=1}^{M}{\mathbf{G}}_{\mathit{\text{Lk}}}^{H}{\mathbf{G}}_{\mathit{\text{Lk}}},\hfill \end{array}

are defined. The notation of **h**_{
ik
} and {\overline{\mathbf{g}}}_{\mathit{\text{ik}}} is given in the proof.

### Proof

From (5), the *k* th column of (**W**_{
i
}**H**_{
i
}) can be expressed as \left({\mathbf{h}}_{\mathit{\text{ik}}}^{T}\otimes {\mathbf{I}}_{N}\right){\mathbf{w}}_{i}, where **h**_{
ik
} denotes the *k* th column of matrix **H**_{
i
}. Therefore, the trace of \left({\mathbf{W}}_{i}{\mathbf{H}}_{i}{\mathbf{H}}_{i}^{H}{\mathbf{W}}_{i}^{H}\right) is given by the Frobenious norm of (**W**_{
i
}**H**_{
i
}), which can be computed as

\phantom{\rule{-14.0pt}{0ex}}\begin{array}{c}\text{tr}\left({\mathbf{W}}_{i}{\mathbf{H}}_{i}{\mathbf{H}}_{i}^{H}{\mathbf{W}}_{i}^{H}\right)={\mathbf{w}}_{i}^{H}\sum _{k=1}^{M}{\left({\mathbf{h}}_{\mathit{\text{ik}}}^{T}\otimes {\mathbf{I}}_{N}\right)}^{H}\left({\mathbf{h}}_{\mathit{\text{ik}}}^{T}\otimes {\mathbf{I}}_{N}\right){\mathbf{w}}_{i}.\end{array}

(9)

Expressing the trace of \left({\mathbf{W}}_{i}{\mathbf{G}}_{i}^{T}{\mathbf{G}}_{i}^{\ast}{\mathbf{W}}_{i}^{H}\right) in the similar way of (9) and substituting it and (9) into (5) yield

\begin{array}{lcr}{P}_{\mathrm{r}}& =\hfill & {P}_{1}\sum _{i=1}^{L}{\mathbf{w}}_{i}^{H}\sum _{k=1}^{M}{\left({\mathbf{h}}_{\mathit{\text{ik}}}^{T}\otimes {\mathbf{I}}_{N}\right)}^{H}\left({\mathbf{h}}_{\mathit{\text{ik}}}^{T}\otimes {\mathbf{I}}_{N}\right){\mathbf{w}}_{i}\hfill \\ +\phantom{\rule{2.77626pt}{0ex}}{P}_{2}\sum _{i=1}^{L}{\mathbf{w}}_{i}^{H}\sum _{k=1}^{M}{\left({\overline{g}}_{\mathit{\text{ik}}}^{T}\otimes {\mathbf{I}}_{N}\right)}^{H}\left({\overline{g}}_{\mathit{\text{ik}}}^{T}\otimes {\mathbf{I}}_{N}\right){\mathbf{w}}_{i}\hfill \\ +\phantom{\rule{2.77626pt}{0ex}}{\sigma}_{v}^{2}\sum _{i=1}^{L}{\mathbf{w}}_{i}^{H}{\mathbf{w}}_{i},\hfill \end{array}

where {\overline{g}}_{\mathit{\text{ik}}} denotes the *k* th column of matrix {\mathbf{G}}_{i}^{T}.

Using the definitions in *Lemma 1*, (8) can be derived.

SNR constraint is usually used in optimizing a relay network. For our problem, constraints on destination SNRs are expressed as

\begin{array}{l}{\text{SNR}}_{1}\ge {\gamma}_{1},\end{array}

(10a)

\begin{array}{l}{\text{SNR}}_{2}\ge {\gamma}_{2},\end{array}

(10b)

where *γ*_{1} and *γ*_{2} are required SNRs at transceiver 1 and transceiver 2, respectively. From (6) and (7), it is seen that (10) is related to **W**_{
i
} and **D**_{
j
} in a complicated form. In the rest of this section, two lemmas are derived to transform (10) into a manageable form.

The ZF constraint requires that

\begin{array}{l}{\mathbf{D}}_{1}\sum _{i=1}^{L}{\mathbf{H}}_{i}^{T}{\mathbf{W}}_{i}{\mathbf{G}}_{i}^{T}=\mathbf{I},\end{array}

(11a)

\begin{array}{l}{\mathbf{D}}_{2}\sum _{i=1}^{L}{\mathbf{G}}_{i}{\mathbf{W}}_{i}{\mathbf{H}}_{i}=\mathbf{I},\end{array}

(11b)

where **D**_{1} and **D**_{2} are defined as the left pseudoinverse of \sum _{i=1}^{L}{\mathbf{H}}_{i}^{T}{\mathbf{W}}_{i}{\mathbf{G}}_{i}^{T} and \sum _{i=1}^{L}{\mathbf{G}}_{i}{\mathbf{W}}_{i}{\mathbf{H}}_{i}, respectively.

### Lemma 2

From definitions of **D**_{1} and **D**_{2} given above, if {\mathbf{D}}_{1}^{H}{\mathbf{D}}_{1} and {\mathbf{D}}_{2}^{H}{\mathbf{D}}_{2} are diagonal matrices, we have

\begin{array}{l}\text{tr}\left({\mathbf{D}}_{1}^{H}{\mathbf{D}}_{1}\right)=\sum _{i=1}^{M}\frac{1}{\mid {\varphi}_{\mathit{\text{ii}}}{\mid}^{2}{\sigma}_{i}},\end{array}

(12a)

\begin{array}{l}\text{tr}\left({\mathbf{D}}_{2}^{H}{\mathbf{D}}_{2}\right)=\sum _{i=1}^{M}\frac{1}{\mid {\stackrel{~}{\varphi}}_{\mathit{\text{ii}}}{\mid}^{2}{\stackrel{~}{\sigma}}_{i}},\end{array}

(12b)

where the definitions of *ϕ*_{
ii
}, {\stackrel{~}{\varphi}}_{\mathit{\text{ii}}}, *σ*_{
i
}, and {\stackrel{~}{\sigma}}_{i} are given in the proof.

*Proof*.

It is straightforward to show that

\begin{array}{l}\phantom{\rule{-12.0pt}{0ex}}{\mathbf{D}}_{1}{\mathbf{D}}_{1}^{H}={\left(\sum _{i=1}^{L}{\mathbf{G}}_{i}^{\ast}{\mathbf{W}}_{i}^{H}{\mathbf{H}}_{i}^{\ast}\sum _{i=1}^{L}{\mathbf{H}}_{i}^{T}{\mathbf{W}}_{i}{\mathbf{G}}_{i}^{T}\right)}^{-1}.\end{array}

(13)

If we define \underset{\xcc\xb2}{H}={\left[{\mathbf{H}}_{1}^{T},\dots ,{\mathbf{H}}_{L}^{T}\right]}^{T}, \underset{\xcc\xb2}{W}={\mathbf{W}}_{1}\oplus \dots {\mathbf{W}}_{L}, and \underset{\xcc\xb2}{G}=\left[{\mathbf{G}}_{1},\dots ,{\mathbf{G}}_{L}\right], (13) can be expressed as

\begin{array}{l}{\mathbf{D}}_{1}{\mathbf{D}}_{1}^{H}={\left({\underset{\xcc\xb2}{\mathbf{G}}}^{\ast}{\underset{\xcc\xb2}{W}}^{H}{\underset{\xcc\xb2}{H}}^{\ast}{\underset{\xcc\xb2}{H}}^{T}\underset{\xcc\xb2}{W}\phantom{\rule{0.3em}{0ex}}{\underset{\xcc\xb2}{G}}^{T}\right)}^{-1}.\end{array}

(14)

Suppose that the eigendecomposition of {\underset{\xcc\xb2}{H}}^{\ast}{\underset{\xcc\xb2}{H}}^{T} is given by

\begin{array}{lcr}{\underset{\xcc\xb2}{H}}^{\ast}{\underset{\xcc\xb2}{H}}^{T}& =& \left(\phantom{\rule{0.3em}{0ex}}\begin{array}{cc}\mathbf{U}& {\mathbf{U}}_{\parallel}\end{array}\phantom{\rule{0.3em}{0ex}}\right)\left(\phantom{\rule{0.3em}{0ex}}\begin{array}{cc}\mathit{\Lambda}& \mathbf{0}\\ \mathbf{0}& \mathbf{0}\end{array}\phantom{\rule{0.3em}{0ex}}\right)\left(\begin{array}{c}{\mathbf{U}}^{H}\\ {\mathbf{U}}_{\parallel}^{H}\end{array}\right)\\ =\hfill & \mathbf{U}\mathit{\Lambda}{\mathbf{U}}^{H}.\hfill \end{array}

(15)

In (15), the diagonal of *Λ* consists of *M* nonzero eigenvalues. The matrix **U** consists of all the eigenvectors corresponding to these nonzero eigenvalues. **U**_{∥} consists of column vectors which are linearly dependent on columns of **U**. The dependence of eigenvectors is caused by rank deficiency of {\underset{\xcc\xb2}{H}}^{\ast}{\underset{\xcc\xb2}{H}}^{T} whose effective rank is *M*.

We define \overline{\underset{\xcc\xb2}{\mathbf{W}}}={\underset{\xcc\xb2}{W}\phantom{\rule{0.3em}{0ex}}\underset{\xcc\xb2}{G}}^{T}, and assume that \overline{\underset{\xcc\xb2}{\mathbf{W}}} can be represented by the complete orthogonal basis in the *NL*-dimensional space, where **U** is contained in the complete orthogonal basis, i.e.,

\begin{array}{l}\overline{\underset{\xcc\xb2}{\mathbf{W}}}=\left(\begin{array}{cc}\mathbf{U}& {\mathbf{U}}_{\perp}\end{array}\right)\left(\begin{array}{c}\mathit{\Phi}\\ {\mathit{\Phi}}_{\perp}\end{array}\right).\end{array}

(16)

In (16), \mathbf{U}\in {\mathrm{\xe2\u201e\u201a}}^{\mathit{\text{NL}}\times M}, \mathit{\Phi}\in {\mathrm{\xe2\u201e\u201a}}^{M\times M}, and **U**_{⊥} consist of *N*−*M* orthogonal basis of the *NL*-dimensional space, which can be obtained via Gram-Schmidt procedure based on **U**.

Substituting (15) and (16) into (14) yields

\begin{array}{lc}\phantom{\rule{-12.0pt}{0ex}}{\mathbf{D}}_{1}{\mathbf{D}}_{1}^{H}=& \left(\left(\begin{array}{cc}{\mathit{\Phi}}^{H}& {\mathit{\Phi}}_{\perp}^{H}\end{array}\right)\right.\left(\begin{array}{c}{\mathbf{U}}^{H}\\ {\mathbf{U}}_{\perp}^{H}\end{array}\right)\mathbf{U}\mathit{\Lambda}{\mathbf{U}}^{H}\hfill \\ \left(\begin{array}{cc}\mathbf{U}& {\mathbf{U}}_{\perp}\end{array}\right){\left(\right)close=")">\left(\begin{array}{c}\mathit{\Phi}\\ {\mathit{\Phi}}_{\perp}\end{array}\right)}^{}-1\hfill & ={\left({\mathit{\Phi}}^{H}\mathit{\Lambda}\mathit{\Phi}\right)}^{-1}.\end{array}\n

(17)

From (17), it is seen that

\begin{array}{l}\text{tr}|\left({\mathbf{D}}_{1}^{H}{\mathbf{D}}_{1}\right)=\text{tr}\left({\mathbf{D}}_{1}{\mathbf{D}}_{1}^{H}\right)\le \sum _{i=1}^{M}\frac{1}{\mid {\varphi}_{\mathit{\text{ii}}}{\mid}^{2}{\sigma}_{i}},\end{array}

(18)

where *ϕ*_{
ii
} and *σ*_{
i
} are the *i* th diagonal element of *Φ* and *Λ*, respectively.

Similarly, for **D**_{2}, we have

\begin{array}{l}\text{tr}\left({\mathbf{D}}_{2}^{H}{\mathbf{D}}_{2}\right)\le \sum _{i=1}^{M}\frac{1}{\mid {\stackrel{~}{\varphi}}_{\mathit{\text{ii}}}{\mid}^{2}{\stackrel{~}{\sigma}}_{i}},\end{array}

(19)

where {\stackrel{~}{\sigma}}_{i} is the *i* th nonzero eigenvalue of {\underset{\xcc\xb2}{G}}^{H}\underset{\xcc\xb2}{G}. {\stackrel{~}{\varphi}}_{\mathit{\text{ii}}} is the *i* th diagonal element of \stackrel{~}{\Phi}, where \underset{\xcc\xb2}{\mathit{\text{WH}}}=\left(\begin{array}{cc}\stackrel{~}{U}& {\stackrel{~}{U}}_{\perp}\end{array}\right)\left(\begin{array}{c}\stackrel{~}{\Phi}\\ {\stackrel{~}{\Phi}}_{\perp}\end{array}\right), and \stackrel{~}{U} consists of eigenvectors of {\underset{\xcc\xb2}{G}}^{H}\underset{\xcc\xb2}{G} corresponding to its nonzero eigenvalues.

### Lemma 3

Inequalities (10) can be relaxed as

\begin{array}{l}\frac{{\sigma}_{v}^{2}{\mathbf{w}}^{H}\mathbf{Hw}+M{\sigma}_{x}^{2}}{{\mathbf{w}}^{H}{\mathbf{d}}_{i}{\mathbf{d}}_{i}^{H}\mathbf{w}{\sigma}_{i}}\le \frac{1}{{\gamma}_{1}},\phantom{\rule{1em}{0ex}}\forall i=1,\dots ,M,\end{array}

(20a)

\begin{array}{l}\frac{{\sigma}_{v}^{2}{\mathbf{w}}^{H}\overline{G}\mathbf{w}+M{\sigma}_{y}^{2}}{{\mathbf{w}}^{H}{\stackrel{~}{\mathbf{d}}}_{i}{\stackrel{~}{\mathbf{d}}}_{i}^{H}\mathbf{w}{\stackrel{~}{\sigma}}_{i}}\le \frac{1}{{\gamma}_{2}},\phantom{\rule{1em}{0ex}}\forall i=1,\dots ,M,\end{array}

(20b)

\begin{array}{l}{\mathbf{w}}^{H}{\mathbf{e}}_{\mathit{\text{ij}}}^{\ast}=0,\phantom{\rule{1em}{0ex}}\forall i=1,\dots ,M,j\ne i,\end{array}

(20c)

\begin{array}{l}{\mathbf{w}}^{H}{\stackrel{~}{e}}_{\mathit{\text{ij}}}^{\ast}=0,\phantom{\rule{1em}{0ex}}\forall i=1,\dots ,M,j\ne i,\end{array}

(20d)

where

\begin{array}{lcr}{\mathbf{H}}_{\mathit{\text{ik}}}& =\hfill & {\left({\mathbf{I}}_{N}\otimes {\mathbf{h}}_{\mathit{\text{ik}}}\right)}^{\ast},\hfill \\ \mathbf{H}& =\hfill & \sum _{k=1}^{M}{\mathbf{H}}_{1k}{\mathbf{H}}_{1k}^{H}\oplus \dots \oplus \sum _{k=1}^{M}{\mathbf{H}}_{\mathit{\text{Lk}}}{\mathbf{H}}_{\mathit{\text{Lk}}}^{H},\hfill \\ {\overline{G}}_{\mathit{\text{ik}}}& =\hfill & {\left({\mathbf{I}}_{N}\otimes {\overline{g}}_{\mathit{\text{ik}}}\right)}^{\ast},\hfill \\ \overline{G}& =\hfill & \sum _{k=1}^{M}{\overline{G}}_{1k}{\overline{G}}_{1k}^{H}\oplus \dots \oplus \sum _{k=1}^{M}{\overline{G}}_{\mathit{\text{Lk}}}{\overline{G}}_{\mathit{\text{Lk}}}^{H},\hfill \\ {\mathbf{d}}_{i}& =\hfill & \mathbf{Q}\left(\mathbf{I}\otimes {\mathbf{u}}_{i}^{\ast}\right){\underset{\xcc\xb2}{\overline{g}}}_{i},\hfill \\ {\stackrel{~}{d}}_{i}& =\hfill & \mathbf{Q}\left(\mathbf{I}\otimes {\stackrel{~}{u}}_{i}^{\ast}\right){\underset{\xcc\xb2}{h}}_{i},\hfill \\ {\mathbf{e}}_{\mathit{\text{ij}}}& =\hfill & \mathbf{Q}\left(\mathbf{I}\otimes {\mathbf{u}}_{i}^{\ast}\right){\underset{\xcc\xb2}{\overline{g}}}_{j},\hfill \\ {\stackrel{~}{e}}_{\mathit{\text{ij}}}& =\hfill & \mathbf{Q}\left(\mathbf{I}\otimes {\stackrel{~}{u}}_{i}^{\ast}\right){\underset{\xcc\xb2}{h}}_{j},\hfill \end{array}

and **u**_{
i
}, {\stackrel{~}{u}}_{i}, {\underset{\xcc\xb2}{\overline{g}}}_{i}, {\underset{\xcc\xb2}{h}}_{i} and **Q** are defined in the following proof.

*Proof*.

With the ZF constraint, (6) and (7) can be simplified as

\phantom{\rule{-10.0pt}{0ex}}\begin{array}{c}{\text{SNR}}_{1}=\frac{{P}_{2}M}{\text{tr}\left({\mathbf{D}}_{1}\left({\sigma}_{v}^{2}\sum _{i=1}^{L}{\mathbf{H}}_{i}^{T}{\mathbf{W}}_{i}\sum _{i=1}^{L}{\mathbf{W}}_{i}^{H}{\mathbf{H}}_{i}^{\ast}+{\sigma}_{x}^{2}\mathbf{I}\right){\mathbf{D}}_{1}^{H}\right)}\end{array}

\phantom{\rule{-10.0pt}{0ex}}\begin{array}{c}{\text{SNR}}_{2}=\frac{{P}_{1}M}{\text{tr}\left({\mathbf{D}}_{2}\left({\sigma}_{v}^{2}\sum _{i=1}^{L}{\mathbf{G}}_{i}{\mathbf{W}}_{i}\sum _{i=1}^{L}{\mathbf{W}}_{i}^{H}{\mathbf{G}}_{i}^{H}+{\sigma}_{y}^{2}\mathbf{I}\right){\mathbf{D}}_{2}^{H}\right)}.\end{array}

From the property of tr(.), we may relax the inequality of SNR_{1} as

\begin{array}{lcr}{\text{SNR}}_{1}& =\hfill & \frac{{P}_{2}M}{\text{tr}\left({\mathbf{D}}_{1}\mathbf{B}{\mathbf{D}}_{1}^{H}\right)}\hfill \\ =\hfill & \frac{{P}_{2}M}{\text{tr}\left(\mathbf{B}{\mathbf{D}}_{1}^{H}{\mathbf{D}}_{1}\right)}\hfill \\ \ge \hfill & \frac{{P}_{2}M}{\text{tr}\left(\mathbf{B}\right)\text{tr}\left({\mathbf{D}}_{1}^{H}{\mathbf{D}}_{1}\right)},\hfill \end{array}

(21)

where \mathbf{B}=\left({\sigma}_{v}^{2}\sum _{i=1}^{L}{\mathbf{H}}_{i}^{T}{\mathbf{W}}_{i}\sum _{i=1}^{L}{\mathbf{W}}_{i}^{H}{\mathbf{H}}_{i}^{\ast}+{\sigma}_{x}^{2}\mathbf{I}\right).

Substituting (18) into (21) yields

\begin{array}{l}{\text{SNR}}_{1}\ge \frac{{P}_{2}M}{\text{tr}\left(\mathbf{B}\right)\sum _{i=1}^{M}\frac{1}{\mid {\varphi}_{\mathit{\text{ii}}}{\mid}^{2}{\sigma}_{i}}}.\end{array}

(22)

From (16) and the definition of \underset{\xcc\xb2}{\overline{W}}, we have

\begin{array}{lcr}\mathit{\Phi}& =\hfill & {\mathbf{U}}^{H}\underset{\xcc\xb2}{\overline{W}}={\mathbf{U}}^{H}\underset{\xcc\xb2}{W}\phantom{\rule{0.3em}{0ex}}{\underset{\xcc\xb2}{G}}^{T}\hfill \\ =\hfill & \left(\begin{array}{c}\text{vec}{\left(\underset{\xcc\xb2}{W}\right)}^{H}\left(\mathbf{I}\otimes {\mathbf{u}}_{1}^{\ast}\right)\\ \vdots \\ \text{vec}{\left(\underset{\xcc\xb2}{W}\right)}^{H}\left(\mathbf{I}\otimes {\mathbf{u}}_{M}^{\ast}\right)\end{array}\right){\underset{\xcc\xb2}{G}}^{T}.\hfill \end{array}

(23)

Therefore, the elements of *Φ* can be represented by

\begin{array}{l}{\varphi}_{\mathit{\text{ij}}}=\text{vec}{\left(\underset{\xcc\xb2}{W}\right)}^{H}\left(\mathbf{I}\otimes {\mathbf{u}}_{i}^{\ast}\right){\overline{\underset{\xcc\xb2}{\mathbf{g}}}}_{j},\end{array}

(24)

where **u**_{
i
} denotes the *i* th column of **U** and {\overline{\underset{\xcc\xb2}{\mathbf{g}}}}_{j}^{T} denotes the *j* th row of \underset{\xcc\xb2}{G}.

Similarly, for SNR_{2}, we have

\begin{array}{l}{\text{SNR}}_{2}\ge \frac{{P}_{1}M}{\text{tr}\left(\stackrel{~}{B}\right)\sum _{i=1}^{M}\frac{1}{\mid {\stackrel{~}{\varphi}}_{\mathit{\text{ii}}}{\mid}^{2}{\stackrel{~}{\sigma}}_{i}}},\end{array}

(25)

where \stackrel{~}{B}=\left({\sigma}_{v}^{2}\sum _{i=1}^{L}{\mathbf{G}}_{i}{\mathbf{W}}_{i}\sum _{i=1}^{L}{\mathbf{W}}_{i}^{H}{\mathbf{G}}_{i}^{H}+{\sigma}_{y}^{2}\mathbf{I}\right), and {\stackrel{~}{\varphi}}_{\mathit{\text{ij}}} can be expressed by

\begin{array}{l}{\stackrel{~}{\varphi}}_{\mathit{\text{ij}}}=\text{vec}{\left(\underset{\xcc\xb2}{W}\right)}^{H}\left(\mathbf{I}\otimes {\stackrel{~}{u}}_{i}^{\ast}\right){\underset{\xcc\xb2}{\mathbf{h}}}_{j},\end{array}

(26)

where {\stackrel{~}{u}}_{i} denotes the *i* th column of \stackrel{~}{U} and {\underset{\xcc\xb2}{\mathbf{h}}}_{j} denotes the *j* th column of \underset{\xcc\xb2}{H}. It is assumed that the eigendecomposition of {\underset{\xcc\xb2}{G}}^{H}\underset{\xcc\xb2}{G} is \stackrel{~}{U}\stackrel{~}{\Lambda}{\stackrel{~}{U}}^{H}.

Similar to *Lemma 1*, the trace of **B** and \stackrel{~}{B} can be expressed as

\begin{array}{lll}\text{tr}\left(\mathbf{B}\right)& =\phantom{\rule{2em}{0ex}}& {\sigma}_{v}^{2}{\mathbf{w}}^{H}\mathbf{Hw}+M{\sigma}_{x}^{2},\end{array}

(27a)

\begin{array}{lll}\text{tr}\left(\stackrel{~}{B}\right)& =\phantom{\rule{2em}{0ex}}& {\sigma}_{v}^{2}{\mathbf{w}}^{H}\overline{G}\mathbf{w}+M{\sigma}_{y}^{2},\end{array}

(27b)

where the definitions of **H** and \overline{G} are given in *Lemma 1*.

Substituting (24) and (27a) into (22) and (26), and (27b) into (25) yields

\begin{array}{l}\phantom{\rule{-19.0pt}{0ex}}{\text{SNR}}_{1}\ge \frac{{P}_{2}M}{\left({\sigma}_{v}^{2}{\mathbf{w}}^{H}\mathbf{H}\mathbf{w}+M{\sigma}_{x}^{2}\right)\sum _{i=1}^{M}\frac{1}{{\left|\text{vec}{\left(\underset{\xcc\xb2}{W}\right)}^{H}\left(\mathbf{I}\otimes {\mathbf{u}}_{i}^{\ast}\right){\overline{\underset{\xcc\xb2}{\mathbf{g}}}}_{i}\right|}^{2}{\sigma}_{i}}},\end{array}

(28a)

\begin{array}{l}\phantom{\rule{-19.0pt}{0ex}}{\text{SNR}}_{2}\ge \frac{{P}_{1}M}{\left({\sigma}_{v}^{2}{\mathbf{w}}^{H}\overline{G}\mathbf{w}+M{\sigma}_{y}^{2}\right)\sum _{i=1}^{M}\frac{1}{{\left|\text{vec}{\left(\underset{\xcc\xb2}{W}\right)}^{H}\left(\mathbf{I}\otimes {\stackrel{~}{u}}_{i}^{\ast}\right){\underset{\xcc\xb2}{\mathbf{h}}}_{i}\right|}^{2}{\stackrel{~}{\sigma}}_{i}}}.\end{array}

(28b)

From (28), (10) can be relaxed as

\begin{array}{l}\frac{{P}_{2}M}{{\gamma}_{1}}\ge \sum _{i=1}^{M}\frac{{\sigma}_{v}^{2}{\mathbf{w}}^{H}\mathbf{H}\mathbf{w}+M{\sigma}_{x}^{2}}{{\left|\text{vec}{\left(\underset{\xcc\xb2}{W}\right)}^{H}\left(\mathbf{I}\otimes {\mathbf{u}}_{i}^{\ast}\right){\overline{\underset{\xcc\xb2}{\mathbf{g}}}}_{i}\right|}^{2}{\sigma}_{i}},\end{array}

(29a)

\begin{array}{l}\frac{{P}_{1}M}{{\gamma}_{2}}\ge \sum _{i=1}^{M}\frac{{\sigma}_{v}^{2}{\mathbf{w}}^{H}\overline{G}\mathbf{w}+M{\sigma}_{y}^{2}}{{\left|\text{vec}{\left(\underset{\xcc\xb2}{W}\right)}^{H}\left(\mathbf{I}\otimes {\stackrel{~}{u}}_{i}^{\ast}\right){\underset{\xcc\xb2}{\mathbf{h}}}_{i}\right|}^{2}{\stackrel{~}{\sigma}}_{i}}.\end{array}

(29b)

If every term on the right side of (29a) and (29b) is smaller than \frac{{P}_{2}}{{\gamma}_{1}} and \frac{{P}_{1}}{{\gamma}_{2}}, respectively, i.e.,

\begin{array}{l}\phantom{\rule{-20.0pt}{0ex}}\frac{{P}_{2}}{{\gamma}_{1}}\ge \frac{{\sigma}_{v}^{2}{\mathbf{w}}^{H}\mathbf{H}\mathbf{w}+M{\sigma}_{x}^{2}}{{\left|\text{vec}{\left(\underset{\xcc\xb2}{W}\right)}^{H}\left(\mathbf{I}\otimes {\mathbf{u}}_{i}^{\ast}\right){\overline{\underset{\xcc\xb2}{\mathbf{g}}}}_{i}\right|}^{2}{\sigma}_{i}},\phantom{\rule{1em}{0ex}}\forall i=1,\dots ,M,\end{array}

(30a)

\begin{array}{l}\phantom{\rule{-20.0pt}{0ex}}\frac{{P}_{1}}{{\gamma}_{2}}\ge \frac{{\sigma}_{v}^{2}{\mathbf{w}}^{H}\overline{G}\mathbf{w}+M{\sigma}_{y}^{2}}{{\left|\text{vec}{\left(\underset{\xcc\xb2}{W}\right)}^{H}\left(\mathbf{I}\otimes {\stackrel{~}{u}}_{i}^{\ast}\right){\underset{\xcc\xb2}{\mathbf{h}}}_{i}\right|}^{2}{\stackrel{~}{\sigma}}_{i}},\phantom{\rule{1em}{0ex}}\forall i=1,\dots ,M,\end{array}

(30b)

(29) can be satisfied.

Because \underset{\xcc\xb2}{W} is block diagonal matrices, there are many zero elements in \text{vec}\left(\underset{\xcc\xb2}{W}\right), which do not contribute to the calculation of (30). Suppose **Q** is chosen such that

\begin{array}{l}\mathbf{w}=\mathbf{Q}\text{vec}\left(\underset{\xcc\xb2}{W}\right)\end{array}

(31)

holds.

To derive (30), we have make assumption that {\mathbf{D}}_{1}^{H}{\mathbf{D}}_{1} and {\mathbf{D}}_{2}^{H}{\mathbf{D}}_{2} should be diagonal. From (17), we may achieve this by forcing *Φ* to be a diagonal matrix. Therefore, the following equations should be satisfied:

\begin{array}{lll}{\varphi}_{\mathit{\text{ij}}}& =\phantom{\rule{2em}{0ex}}& \text{vec}{\left(\underset{\xcc\xb2}{W}\right)}^{H}\left(\mathbf{I}\otimes {\mathbf{u}}_{i}^{\ast}\right){\overline{\underset{\xcc\xb2}{\mathbf{g}}}}_{j}=0,\phantom{\rule{1em}{0ex}}\forall i\ne j,\end{array}

(32)

\begin{array}{lll}{\stackrel{~}{\varphi}}_{\mathit{\text{ij}}}& =\phantom{\rule{2em}{0ex}}& \text{vec}{\left(\underset{\xcc\xb2}{W}\right)}^{H}\left(\mathbf{I}\otimes {\stackrel{~}{u}}_{i}^{\ast}\right){\underset{\xcc\xb2}{\mathbf{h}}}_{j}=0,\phantom{\rule{1em}{0ex}}\forall i\ne \mathrm{j.}\end{array}

(33)

With (30) to (33) and definitions given in Lemma 1, (20) can be derived.