利用矩阵的迹巧妙解决矩阵的求导问题

题目引入：

在维纳滤波的过程中需要利用以前的信息的线性组合来预测当前值，同时要满足最小均方误差准则。另外，在线性MIMO接收机的设计过程中，有以下信道模型：

$$y=Hx+n$$

为了根据接收到的信号y，恢复出发送的信号x，需要对发送向量做一个估计，假设采用最小均方误差准则（MMSE），存在系数矩阵A使得：

$$\hat{x}_{MMSE}=Ay$$

均方误差为：

$$e = E||\hat{x}_{MMSE}-x||^2=E(Ay-x)^T(Ay-x)$$

现在的问题是如何得到矩阵A，使得上述均方误差达到最小？

很多教材上对此大都未曾提及，这里介绍了已知巧妙简单易懂的方法来解决上述问题。首先介绍一些关于矩阵迹的先验知识。

矩阵的迹就是矩阵对角元素之和。有以下性质：

$$tr(AB)=tr(BA)$$

$$\frac{{\partial tr\left( {AB} \right)}}{{\partial A}} = {B^T}$$

根据以上两个性质基本上就可以进行接下来的操作。首先利用上述性质计算如下导数：
$$\frac{{\partial tr\left( {AB{A^T}C} \right)}}{{\partial A}}$$

对于不同位置含有两个矩阵A，该如何操作呢？这里先引入微积分中对$x^2$的分步求导过程：
$$\frac{{d{x^2}}}{{dx}} = \frac{{dxx}}{{dx}} = x\frac{{dx}}{{dx}} + x\frac{{dx}}{{dx}} = 2x$$

同样地，
\[\begin{array}{l}

\frac{{\partial tr\left( {AB{A^T}C} \right)}}{{\partial A}} = \frac{{\partial tr\left( {{A^T}CAB} \right)}}{{\partial A}}\\

= {\left( {B{A^T}C} \right)^T} + CAB\\

= {C^T}A{B^T} + CAB

\end{array}\]

接下来，我们来解决前面提出的问题。
\[e = E\left[ {{{\left( {Ay - x} \right)}^T}\left( {Ay - x} \right)} \right]\]

上式对A求导：
\[\begin{array}{l}

\frac{{de}}{{dA}} = \frac{d}{{dA}}E\left[ {{{\left( {Ay - x} \right)}^T}\left( {Ay - x} \right)} \right]\\

= E\left[ {\frac{d}{{dA}}{{\left( {Ay - x} \right)}^T}\left( {Ay - x} \right)} \right]

\end{array}\]

待求期望的部分是
\[\begin{array}{l}

\frac{d}{{dA}}{\left( {Ay - x} \right)^T}\left( {Ay - x} \right) = \frac{d}{{dA}}tr\left( {{{\left( {Ay - x} \right)}^T}\left( {Ay - x} \right)} \right)\\

= \frac{d}{{dA}}tr\left( {{y^T}{A^T}Ay - {y^T}{A^T}x - {x^T}Ay + {x^T}x} \right)\\

= \frac{d}{{dA}}tr\left( {{y^T}{A^T}Ay - {y^T}{A^T}x - {x^T}Ay} \right) + \frac{d}{{dA}}tr\left( {{x^T}x} \right)\\

= \frac{d}{{dA}}tr\left( {{y^T}{A^T}Ay - {y^T}{A^T}x - {x^T}Ay} \right)\\

= \frac{d}{{dA}}tr\left( {Ay{y^T}{A^T} - {A^T}x{y^T} - Ay{x^T}} \right)\\

= {\left( {y{y^T}{A^T}} \right)^T} + Ay{y^T} - x{y^T} - {\left( {y{x^T}} \right)^T}\\

= 2Ay{y^T} - 2x{y^T}

\end{array}\]

令上式为零就可得到MMSE下的解。
\[\begin{array}{l}

\frac{{de}}{{dA}} = E\left[ {\frac{d}{{dA}}{{\left( {Ay - x} \right)}^T}\left( {Ay - x} \right)} \right]\\

= E\left( {2Ay{y^T} - 2x{y^T}} \right) = 0

\end{array}\]

假设
\begin{array}{l}

E\left[ {x{x^T}} \right] = I\\

E\left[ {n{x^T}} \right] = 0\\

E\left[ {n{n^T}} \right] = {\sigma ^2}I

\end{array}

则
\[\begin{array}{l}

\frac{{de}}{{dA}}{\rm{ = }}E\left( {2Ay{y^T} - 2x{y^T}} \right){\rm{ = }}0\\

AE\left( {y{y^T}} \right) - E\left( {x{y^T}} \right) = 0\\

E\left( {y{y^T}} \right) = E\left( {\left( {Hx + n} \right){{\left( {Hx + n} \right)}^T}} \right)\\

= E\left( {Hx{x^T}{H^T} + Hx{n^T} + n{x^T}{H^T} + n{n^T}} \right)\\

= HE\left( {x{x^T}} \right){H^T} + E\left( {n{n^T}} \right)\\

= HI{H^T} + {\sigma ^2}I\\

= H{H^T} + {\sigma ^2}I

\end{array}\]

从而有
\[\begin{array}{l}

AE\left( {y{y^T}} \right) = E\left( {x{y^T}} \right) = E\left( {x{{\left( {Hx + n} \right)}^T}} \right) = {H^T}\\

A\left( {H{H^T} + {\sigma ^2}I} \right) = {H^T}\\

A = {H^T}{\left( {H{H^T} + {\sigma ^2}I} \right)^{ - 1}}

\end{array}\]

至此就求得了矩阵A，从而得到了MIMO接收机的发送信号估计公式：
\[\begin{array}{l}

{{\hat x}_{MMSE}} = Ay\\

= {H^T}{\left( {H{H^T} + {\sigma ^2}I} \right)^{ - 1}}y

\end{array}\]

利用上述方法简介明了地推出了MMSE下MIMO接收机发送信号估计公式。利用矩阵的迹的性质还可以简便地推出最小二乘法的表达式。读者可以尝试。这在博客http://blog.csdn.net/acdreamers/article/details/44662633得到了实现。