# Robust Transceiver Design for IRS-Assisted Cascaded MIMO Communication Systems
## 3 Achievable Rate (Full-Duplex Relay)
### Algorithm Description
We have

\[\mathbf{H}_{R\uparrow}\mathbf{\Theta}\mathbf{H}_{IR}= \left(\sum_{r_{1}=1}^{K_{r}}\sum_{r_{2}=1}^{M}h_{RIr_{1},r_{2}} \left|r_{1}\right\rangle\!\!\left\langle r_{2}\right|\right)\left(\sum_{r_{3} =1}^{M}\theta_{r_{3},r_{3}}\left|r_{3}\right\rangle\!\!\left\langle r_{3} \right|\right)\left(\sum_{r_{4}=1}^{M}\!\!\sum_{r_{5}=1}^{K_{t}}h_{IRr_{4},r_ {5}}\left|r_{4}\right\rangle\!\!\left\langle r_{5}\right|\right) \tag{17}\] \[=\sum_{r_{1}=r_{2}=r_{3}=r_{4}}^{K_{r}}\!\!\sum_{r_{3}=r_{4}}^{M }\!\!\sum_{r_{5}=1}^{K_{t}}h_{RIr_{1},r_{2}}\theta_{r_{3},r_{3}}h_{IRr_{4},r_ {5}}\!\!\left|r_{1}\right\rangle\!\!\left\langle r_{2}\right|\!\!\left\langle r _{3}\right|\!\!\left\langle r_{4}\right\rangle\!\!\left\langle r_{5}\right|\] (18) \[=\sum_{r_{1}=1}^{K_{r}}\sum_{r_{3}=1}^{M}\sum_{r_{5}=1}^{K_{t}}h_{ RIr_{1},r_{3}}\theta_{r_{3},r_{3}}h_{IRr_{3},r_{5}}\!\!\left|r_{1}\right\rangle \!\!\left\langle r_{5}\right|\,. \tag{19}\]

Now, after turning the matrix representation into the vector representation we obtain

\[Vec(\mathbf{H}_{R\uparrow}\mathbf{\Theta}\mathbf{H}_{IR})= \sum_{r_{1}=1}^{K_{r}}\sum_{r_{3}=1}^{M}\sum_{r_{5}=1}^{K_{t}}h_{ RIr_{1},r_{3}}\theta_{r_{3},r_{3}}h_{IRr_{3},r_{5}}\!\!\left|r_{1},r_{5} \right\rangle \tag{20}\] \[= \sum_{r_{1}=1}^{K_{r}}\sum_{r_{3}=1}^{M}\sum_{r_{4}=1}^{M}\sum_{r _{5}=1}^{K_{t}}h_{RIr_{1},r_{4}}\theta_{r_{3},r_{3}}h_{IRr_{4},r_{5}}\!\!\left| r_{1},r_{5}\right\rangle\left\langle r_{4}\right|\!\!\left\langle r_{3}\right\rangle \tag{21}\]Figure 7:0 Figure 7:1

As a result, we have, \(||\hat{\mathbf{H}}_{r}+\mathbf{H}_{RI}\mathbf{\Theta}\mathbf{H}_{IR}||_{F}^{2} =||Vec(\hat{\mathbf{H}}_{r})+(\mathbf{H}_{IR}*\mathbf{H}_{RI}^{T})Vec(\mathbf{ \Theta})||_{2}^{2}\), and the proof is complete. ## Appendix B Proof of Theorem 1

Before stating the proof, first we introduce the following definitions. **Definition A1**. : _For a vector **a**, we denote vector \(\textbf{a}^{\perp}\) which has the same components as **a** except that they are sorted in a decreasing order._

**Definition A2**. : _Vector **a** is said to be majorized by vector **b** and denoted by \(\textbf{a}\prec\textbf{b}\) if:_

Figure 7:2 Figure 7:3

_where \(a_{i}^{\downarrow}\) is the \(i\)'th component of \(\textbf{a}^{\downarrow}\), \(N_{r}\) is the number of vector components and \(K\leq N\). If the last equality does not hold, **a** is said to be weakly majorized by **b** and denoted by **a**\(\prec_{w}\textbf{b}\)._

**Definition A3**. : _Vector **a** is said to be multiplicatively majorized by vector **b** and denoted by **a**\(\prec_{\times}\textbf{b}\) if:_

Figure 7:4 Figure 7:5

_In addition, it is easy to check_

Figure 7:6

To begin with, we know that for \(n\times m\) matrix \(\textbf{A}\) and \(m\times n\) matrix \(\textbf{B}\) we have \(\lambda_{i}(\textbf{A}\textbf{B})=\lambda_{i}(\textbf{B}\textbf{A}),\ \forall i \in\{1,\cdots,\min(m,n)\}\). In addition, the only difference between eigenvalues of \(\textbf{B}\textbf{A}\) and \(\textbf{A}\textbf{B}\) are the number of eigenvalues 0. Thus, non-zero eigenvalues of \(\hat{\mathbf{H}}_{r}\mathbf{Q}_{r}\hat{\mathbf{H}}_{r}^{H}\) and \(\mathbf{Q}_{r}\hat{\mathbf{H}}_{r}^{H}\hat{\mathbf{H}}_{r}\) and also \(\mathbf{H}_{1}\mathbf{Q}_{s}\mathbf{H}_{1}^{H}\) and \(\mathbf{Q}_{s}\mathbf{H}_{1}^{H}\mathbf{H}_{1}\) are equal, respectively. Notice that all \(\mathbf{Q}_{s}\), \(\mathbf{H}_{1}^{H}\mathbf{H}_{1}\), \(\mathbf{Q}_{r}\) and \(\hat{\mathbf{H}}_{r}^{H}\hat{\mathbf{H}}_{r}\) are square matrices. For \(\mathbf{H}_{1}^{H}\mathbf{H}_{1}\) and \(\bar{\mathbf{H}}_{r}^{H}\bar{\mathbf{H}}_{r}\) we define \(\lambda_{i}(\mathbf{H}_{1}^{H}\mathbf{H}_{1})=\sigma_{i}^{2}(\mathbf{H}_{1})\) and \(\lambda_{i}(\bar{\mathbf{H}}_{r}^{H}\bar{\mathbf{H}}_{r})=\sigma_{i}^{2}(\bar{ \mathbf{H}}_{r})\), respectively. As discussed in Remark 1, the equality \(\lambda_{i}(\mathbf{Q}_{s}\mathbf{H}_{1}^{H}\mathbf{H}_{1})=\lambda_{\rho(i)}( \mathbf{Q}_{s})\lambda_{i}(\mathbf{H}_{1}^{H}\mathbf{H}_{1})\) does not hold in general.