论文中的一处推导

x H diag { w w H } x x H w w H x \frac {\mathbf{x}^{H}\text{diag}\left\{\mathbf{w}\mathbf{w}^{H}\right\}\mathbf{x}}{\mathbf{x}^{H}\mathbf{w}\mathbf{w}^{H}\mathbf{x}} xHwwHxxHdiag{wwH}x?

要将给定的表达式化简为形如 $\frac{\Vert \mathbf{a}{\Vert }^2}{\Vert \mathbf{b}{\Vert }^2}$ 的形式，我们可以考虑定义新的向量 $\mathbf{a}$ 和 $\mathbf{b}$，使得原始的分子和分母分别变为 $\Vert \mathbf{a}{\Vert }^2$ 和 $\Vert \mathbf{b}{\Vert }^2$。

给定的表达式是：

[
x H diag { w w H } x x H w w H x \frac {\mathbf{x}^{H}\text{diag}\left\{\mathbf{w}\mathbf{w}^{H}\right\}\mathbf{x}}{\mathbf{x}^{H}\mathbf{w}\mathbf{w}^{H}\mathbf{x}} xHwwHxxHdiag{wwH}x?
]

首先，令 $\mathbf{a}={\mathbf{w}}^H\mathbf{x}$，这样我们就有 $\Vert \mathbf{a}{\Vert }^2={\mathbf{x}}^H\mathbf{w}{\mathbf{w}}^H\mathbf{x}$。

然后，令 $\mathbf{b}=\mathbf{w}$，这样我们就有 $\Vert \mathbf{b}{\Vert }^2={\mathbf{w}}^H\mathbf{w}$。

现在，我们可以将原始的表达式用 $\mathbf{a}$ 和 $\mathbf{b}$ 重新表示：

[
x H diag { w w H } x x H w w H x = x H diag { b b H } x ∥ a ∥ 2 \frac {\mathbf{x}^{H}\text{diag}\left\{\mathbf{w}\mathbf{w}^{H}\right\}\mathbf{x}}{\mathbf{x}^{H}\mathbf{w}\mathbf{w}^{H}\mathbf{x}} = \frac {\mathbf{x}^{H}\text{diag}\left\{\mathbf{b}\mathbf{b}^{H}\right\}\mathbf{x}}{\|\mathbf{a}\|^2} xHwwHxxHdiag{wwH}x?=∥a∥2xHdiag{bbH}x?
]

最后，我们可以进一步简化为：

[
f ( w , Θ ) = x H diag { b b H } x ∥ a ∥ 2 = ∥ b ∥ 2 ∥ x ∥ 2 ∥ a ∥ 2 = ∥ w ∥ 2 ∥ x ∥ 2 ∥ w H x ∥ 2 , \mathbf{f}\left(\mathbf{w},\mathbf{\Theta}\right) = \frac {\mathbf{x}^{H}\text{diag}\left\{\mathbf{b}\mathbf{b}^{H}\right\}\mathbf{x}}{\|\mathbf{a}\|^2} = \frac {\|\mathbf{b}\|^2 \|\mathbf{x}\|^2}{\|\mathbf{a}\|^2} = \frac {\|\mathbf{w}\|^2 \|\mathbf{x}\|^2}{\|\mathbf{w}^{H}\mathbf{x}\|^2}, f(w,Θ)=∥a∥2xHdiag{bbH}x?=∥a∥2∥b∥2∥x∥2?=∥wHx∥2∥w∥2∥x∥2?,

where

$\mathbf{x}={\mathbf{t}}_{1T}{\mathbf{t}}_{1R}^H{\mathbf{\Theta }}^H{\mathbf{s}}_T{\mathbf{s}}_R^H{\mathbf{\Theta }}^H{\mathbf{r}}_{2T}$
]

or

[
f ( w , Θ 1 , Θ 2 ) = x H diag { b b H } x ∥ a ∥ 2 = ∥ b ∥ 2 ∥ x ∥ 2 ∥ a ∥ 2 = ∥ w ∥ 2 ∥ x ∥ 2 ∥ w H x ∥ 2 , \mathbf{f}\left(\mathbf{w},\mathbf{\Theta}_1,\mathbf{\Theta}_2\right) = \frac {\mathbf{x}^{H}\text{diag}\left\{\mathbf{b}\mathbf{b}^{H}\right\}\mathbf{x}}{\|\mathbf{a}\|^2} = \frac {\|\mathbf{b}\|^2 \|\mathbf{x}\|^2}{\|\mathbf{a}\|^2} = \frac {\|\mathbf{w}\|^2 \|\mathbf{x}\|^2}{\|\mathbf{w}^{H}\mathbf{x}\|^2}, f(w,Θ1?,Θ2?)=∥a∥2xHdiag{bbH}x?=∥a∥2∥b∥2∥x∥2?=∥wHx∥2∥w∥2∥x∥2?,

where

$\mathbf{x}={\mathbf{t}}_{1T}{\mathbf{t}}_{1R}^H{\mathbf{\Theta }}_1^H{\mathbf{s}}_T{\mathbf{s}}_R^H{\mathbf{\Theta }}_2^H{\mathbf{r}}_{2T}=\mathbf{h}$
]

?(P7)? min ? w , Θ 1 , Θ 2 ∥ w ∥ 2 ∥ t 1 T t 1 R H Θ 1 H s T s R H Θ 2 H r 2 T ∥ 2 ∥ w H t 1 T t 1 R H Θ 1 H s T s R H Θ 2 H r 2 T ∥ 2 ?s.t.? ( 1 + β T ) tr ? ( w w H ) ≤ P t ,? Θ 1 = diag ? { θ 11 , ? ? , θ 1 L } , ∣ θ 1 i ∣ = 1 , i ∈ { 1 , ? ? , L } , Θ 2 = diag ? { θ 21 , ? ? , θ 2 L } , ∣ θ 2 i ∣ = 1 , i ∈ { 1 , ? ? , L } . \begin{align} \text { (P7) } \quad \min _{\mathbf{w}, \boldsymbol{\Theta}_1, \boldsymbol{\Theta}_2} & \frac {\|\mathbf{w}\|^2 \|\mathbf{t}_{1T}\mathbf{t}_{1R}^{\mathrm{H}}\boldsymbol{\Theta}_{1}^{\mathrm{H}}\mathbf{s}_{T}\mathbf{s}_{R}^{\mathrm{H}}\boldsymbol{\Theta}_{2}^{\mathrm{H}}\mathbf{r}_{2T}\|^2}{\|\mathbf{w}^{\mathrm{H}}\mathbf{t}_{1T}\mathbf{t}_{1R}^{\mathrm{H}}\boldsymbol{\Theta}_{1}^{\mathrm{H}}\mathbf{s}_{T}\mathbf{s}_{R}^{\mathrm{H}}\boldsymbol{\Theta}_{2}^{\mathrm{H}}\mathbf{r}_{2T}\|^2} \\ \text { s.t. } &\left( 1 + \beta_T \right) \operatorname{tr}\left(\mathbf{w w}^{\mathrm{H}}\right) \leq P_t \text {, } \\ & \boldsymbol{\Theta}_1=\operatorname{diag}\left\{\theta_{11}, \cdots, \theta_{1L}\right\}, \\ & \left|\theta_{1i}\right|=1, \quad i \in\{1, \cdots, L\} , \\ & \boldsymbol{\Theta}_2=\operatorname{diag}\left\{\theta_{21}, \cdots, \theta_{2L}\right\}, \\ & \left|\theta_{2i}\right|=1, \quad i \in\{1, \cdots, L\} . \\ \end{align} ?(P7)?w,Θ1?,Θ2?min??s.t.??∥wHt1T?t1RH?Θ1H?sT?sRH?Θ2H?r2T?∥2∥w∥2∥t1T?t1RH?Θ1H?sT?sRH?Θ2H?r2T?∥2?(1+βT?)tr(wwH)≤Pt?,?Θ1?=diag{θ11?,?,θ1L?},∣θ1i?∣=1,i∈{1,?,L},Θ2?=diag{θ21?,?,θ2L?},∣θ2i?∣=1,i∈{1,?,L}.??

?(P7)? min ? w , Θ ∥ w ∥ 2 ∥ t 1 T t 1 R H Θ H s T s R H Θ H r 2 T ∥ 2 ∥ w H t 1 T t 1 R H Θ H s T s R H Θ H r 2 T ∥ 2 ?s.t.? ( 1 + β T ) tr ? ( w w H ) ≤ P t ,? Θ = diag ? { θ 1 , ? ? , θ L } , ∣ θ i ∣ = 1 , i ∈ { 1 , ? ? , L } . \begin{align} \text { (P7) } \quad \min _{\mathbf{w}, \boldsymbol{\Theta}} & \frac {\|\mathbf{w}\|^2 \|\mathbf{t}_{1T}\mathbf{t}_{1R}^{\mathrm{H}}\boldsymbol{\Theta}^{\mathrm{H}}\mathbf{s}_{T}\mathbf{s}_{R}^{\mathrm{H}}\boldsymbol{\Theta}^{\mathrm{H}}\mathbf{r}_{2T}\|^2}{\|\mathbf{w}^{\mathrm{H}}\mathbf{t}_{1T}\mathbf{t}_{1R}^{\mathrm{H}}\boldsymbol{\Theta}^{\mathrm{H}}\mathbf{s}_{T}\mathbf{s}_{R}^{\mathrm{H}}\boldsymbol{\Theta}^{\mathrm{H}}\mathbf{r}_{2T}\|^2} \\ \text { s.t. } &\left( 1 + \beta_T \right) \operatorname{tr}\left(\mathbf{w w}^{\mathrm{H}}\right) \leq P_t \text {, } \\ & \boldsymbol{\Theta}=\operatorname{diag}\left\{\theta_{1}, \cdots, \theta_{L}\right\}, \\ & \left|\theta_{i}\right|=1, \quad i \in\{1, \cdots, L\} . \\ \end{align} ?(P7)?w,Θmin??s.t.??∥wHt1T?t1RH?ΘHsT?sRH?ΘHr2T?∥2∥w∥2∥t1T?t1RH?ΘHsT?sRH?ΘHr2T?∥2?(1+βT?)tr(wwH)≤Pt?,?Θ=diag{θ1?,?,θL?},∣θi?∣=1,i∈{1,?,L}.??

其实这里的norm应该改成abs

$\mathbf{w}=\sqrt{\frac{P_t}{1+{\beta }_T}}$