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  Sometimes when you are dealing with linear algebra in real life, you'll get an inconsistent system of linear equations. Our former selves in chapters one and two would have just given up, but with our new skills we can try and approximate a solution for an inconsistent system.
  ::有时当你在现实生活中处理线性代数时,你会得到一个不一致的线性方程系统。我们在第一章和第二章中的前一自我可能刚刚放弃了,但是有了我们的新技能,我们可以尝试并大致地为不一致的系统找到一个解决方案。

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  Say we want to solve the system  $\begin{array}{r}A\stackrel{\to }{x}=\stackrel{\to }{b}\end{array}$
  ::说我们要破解系统 Axb

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  However, if  $\begin{array}{r}\stackrel{\to }{b}\notin \text{Col}(A)\end{array}$  then we can't find the simple solution of  $\begin{array}{r}\stackrel{\to }{x}=A^{-1}\stackrel{\to }{b}\end{array}$ . 
  ::然而,如果bCol(A),那么我们无法找到 xA-1b的简单解决方案。

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  To approximate  a  solution, we take what is called  a least-squares solution.
  ::为了接近一个解决办法,我们采取所谓的最平方解决办法。

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  Definition: A least squares solution given a matrix  $\begin{array}{r}A\in {\mathbb{R}}^{m\times n},\stackrel{\to }{b}\in {\mathbb{R}}^m\end{array}$  of the equation 
  ::定义: 给方程式矩阵 ARmxn, bRm 给出的最小方格解决方案

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  $\begin{array}{r}A\stackrel{\to }{x}=\stackrel{\to }{b}\end{array}$  is a vector  $\begin{array}{r}\hat{x}\in {\mathbb{R}}^n\end{array}$  such that
  ::Axb是矢量 xRn ,

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  $$\begin{array}{r}||\stackrel{\to }{b}-A\hat{x}||\le ||\stackrel{\to }{b}-A\stackrel{\to }{x}||\\ \mathrm{\forall }\stackrel{\to }{x}\in {\mathbb{R}}^n\end{array}$$

  
  ::

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  Now, let's look at finding the optimal least squares solution. 
  ::现在,让我们来看看 找到最佳的最小方形解决方案。

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  Theorem: The optimal least-squares solution (just referred to as the least-squares solution) of  $\begin{array}{r}A\stackrel{\to }{x}=\stackrel{\to }{b}\end{array}$  is the set of solutions to the normal equations 
  ::理论:Axb 的最佳最小平方解决办法(称为最小平方解决办法)是正常方程的一套解决办法

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  $$\begin{array}{r}{A}^{T}A=A^T\stackrel{\to }{b}\end{array}$$

  
  ::ATA=ATB

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  This theorem is only valid if  $\begin{array}{r}A\in {\mathbb{R}}^{m\times n}\text{ and }\stackrel{\to }{b}\in {\mathbb{R}}^m\end{array}$ . 
  ::此定理只有在 ARmn 和 bRm 的情况下才有效 。

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  The key step here is to write an orthogonal decomposition of  $\begin{array}{r}\stackrel{\to }{b}\end{array}$ .  Yo u can go back and learn about  this in  lesson's 2 and 3 of this chapter.
  ::这里的关键步骤是写下b的正对分解。 您可以在本章第2和第3课中回去学习这一点 。

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  Here is an informal proof as to why this makes sense:
  ::以下是一个非正式的证据,说明为什么这样做是有道理的:

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  S uppose that  $\begin{array}{r}\hat{x}\end{array}$  satisfies the normal equations described above, so
  ::假设 x__ 满足上述正常方程,所以

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  $$\begin{array}{r}{A}^{T}A\hat{x}=A^T\stackrel{\to }{b}\end{array}$$

  Then we have that
  ::ATAXATB,然后我们有了

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  $$\begin{array}{r}{A}^{T}A\hat{x}-A^T\stackrel{\to }{b}=\stackrel{\to }{0}\\ A^T(A\hat{x}-\stackrel{\to }{b})=\stackrel{\to }{0}\end{array}$$

  Thus,  $\begin{array}{r}A\hat{x}-\stackrel{\to }{b}\end{array}$  is orthogonal to the rows of  $\begin{array}{r}{A}^T\end{array}$   which means that it is also orthogonal to the columns of  $\begin{array}{r}A\end{array}$ . 
  ::ATAxAT0AT( Axb) = 0Thus, AxbOAT( Axb) 与 AT 的行对齐, 这意味着对 A 列也对齐。

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  Now, let's write the orthogonal decomposition of $\begin{array}{r}\stackrel{\to }{b}\end{array}$ . We get that the column space of  $\begin{array}{r}A\end{array}$  is orthogonal to  $\begin{array}{r}A\hat{x}-\stackrel{\to }{b}\end{array}$  so we can decompose  $\begin{array}{r}\stackrel{\to }{b}\end{array}$  into
  ::现在,让我们来写下bQQ的正心分解。 我们发现 A 的列空间与 AxBQQQQ 的正弦空间是正弦的, 这样我们就可以分解 bQQ 。

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  $$\begin{array}{r}\stackrel{\to }{b}=A\hat{x}+(\stackrel{\to }{b}-A\hat{x})\end{array}$$

  Hence, we have that  $\begin{array}{r}A\hat{x}\end{array}$  will be orthogonal to the projection of $\begin{array}{r}\stackrel{\to }{b}\end{array}$  onto $\begin{array}{r}\text{Col}(A)\end{array}$ . This then means that

  $$\begin{array}{r}A\hat{x}={\text{proj}}_{\text{Col }A}\stackrel{\to }{b}\end{array}$$

  T his proves that  $\begin{array}{r}\hat{x}\end{array}$  is the least squares solution. By the definition of the projection of  $\begin{array}{r}\stackrel{\to }{b}\end{array}$  onto $\begin{array}{r}\text{Col}(A)\end{array}$ , we get that that is the closes vector in the column space of A to b and hence makes  $\begin{array}{r}\hat{x}\end{array}$  that minimal solution we were looking for.
  ::báx (bAx}) 根據此, 我們有Ax 會對射 b 投射到 Col( A) 。 這意味著 Ax projcol AbThis 證明 x 是最小的平方溶液 。 根据 投射 b 投射到 Col( A) 的定義, 我們得到的是 A至 b 列空間的關閉矢量, 因此 x 使得我們所尋找的 最小的溶液 。

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  Now, we have proved that if a vector is a solution to the normal equations then it is a least squares solution.
  ::现在,我们已经证明,如果矢量是正常方程式的解决方案,那么它就是最小方形的解决方案。

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  We want to show that this works in both directions. We are now going to prove that if  we have  $\begin{array}{r}\hat{x}\end{array}$  as  the least squares solution, then it satisfies the normal equations described in the theorem written above.
  ::我们要证明这双向运作。我们现在要证明,如果我们有x作为最小方块的解决方案,那么它就满足了上面所写理论中描述的正常方程式。

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  From now on, let's just call  $\begin{array}{r}\hat{b}={\text{proj}}_{\text{Col}(A)}\stackrel{\to }{b}\end{array}$ . So, if  $\begin{array}{r}\hat{x}\end{array}$  satisfies the equation  $\begin{array}{r}A\hat{x}=\hat{b}\end{array}$  then by orthogonal decomposition we can take that  $\begin{array}{r}\hat{b}\end{array}$  has the property that  $\begin{array}{r}\stackrel{\to }{b}-\hat{b}\end{array}$  is orthogonal to  $\begin{array}{r}\text{Col}(A)\end{array}$ . 
  ::从现在开始,我们打给bprojCol(A)b。所以,如果 x 满足了Axb的方程式, 然后通过正方形分解, 我们可以假设b 具有bb 向Col(A) 正方形的属性 。

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  Now b ecause  $\begin{array}{r}\hat{b}=A\hat{x}\end{array}$  we have that  $\begin{array}{r}\stackrel{\to }{b}-A\hat{x}\end{array}$  is orthogonal to $\begin{array}{r}\text{Col}(A)\end{array}$ .
  ::现在,因为bAx我们得到的bAx是直角到Col( A) 。

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  This means that for any column of  $\begin{array}{r}A\end{array}$ , say  $\begin{array}{r}\stackrel{\to }{a_i}\end{array}$ , we get that
  ::这意味着,对于A的任何一列, 说Ai,我们明白

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    $\begin{array}{r}\stackrel{\to }{a_i}\cdot (\stackrel{\to }{b}-A\hat{x})=\stackrel{\to }{0}\\ {\stackrel{\to }{a_i}}^T(\stackrel{\to }{b}-A\hat{x})=\stackrel{\to }{0}\end{array}$  
  ::

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  T his means that every row of  $\begin{array}{r}{A}^T\end{array}$  is orthogonal to  $\begin{array}{r}\stackrel{\to }{b}-A\hat{x}\end{array}$  and thus
  ::这意味着 AT 每行的 AT 都与 bAx 垂直, 因此

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  $$\begin{array}{r}{A}^T(\stackrel{\to }{b}-A\hat{x})=\stackrel{\to }{0}\\ A^T\stackrel{\to }{b}-A^{T}A\hat{x}=\stackrel{\to }{0}\\ A^{T}A\hat{x}=A^T\stackrel{\to }{b}\end{array}$$

    and thus we have that the least squares solution to
  ::AT(b) AT(Ax) = 0AT(AT) AT(ATAT) AT(ATA) = 0 AT(ATA) AT(ATA) = 0 AT(ATA) AT(AT)

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  $$\begin{array}{r}A\stackrel{\to }{x}=\stackrel{\to }{b}\end{array}$$

  satisfies the normal equations
  ::满足正常方程

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  $$\begin{array}{r}{A}^{T}A\stackrel{\to }{x}=A^T\stackrel{\to }{b}\end{array}$$

  and we can simplify it to
  ::我们可以把它简化到

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    $\begin{array}{r}\hat{x}=(A^{T}A{)}^{-1}{A}^T\stackrel{\to }{b}\end{array}$  
  ::x(ATA)-1ATb

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  Before I continue, here are some videos to help you understand this more. After we will go over an example:
  ::在我继续之前,这里有一些视频可以帮助你理解更多。

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  Let's look at  an example now.
  ::让我们现在举一个例子。

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  Example:
  ::示例:

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  Let's find the least squares solution for the system of equations given by the matrix-vector equation
  ::让我们为矩阵- 矢量方程给出的方程系统找到最小方格的解决方案

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  $$\begin{array}{r}A\stackrel{\to }{x}=\stackrel{\to }{b}\\ A=\left[\begin{array}{cc}4& 0\\ 0& 2\\ 1& 1\end{array}\right]\\ \stackrel{\to }{b}=\left[\begin{array}{c}2\\ 0\\ 1\end{array}\right]\end{array}$$

  First we must confirm that  $\begin{array}{r}\stackrel{\to }{b}\notin \text{Col}(A)\end{array}$ . 
  ::AxbA=[400211]bb[201] 首先我们必须确认 bcol(A)。

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  $\begin{array}{r}\text{Col}(A)=\text{Span}\{\left[\begin{array}{c}4\\ 0\\ 1\end{array}\right],\left[\begin{array}{c}0\\ 2\\ 1\end{array}\right]\}\\ \text{Col}(A)=c_1\cdot \left[\begin{array}{c}4\\ 0\\ 1\end{array}\right]+c_2\cdot \left[\begin{array}{c}0\\ 2\\ 1\end{array}\right]\\ =\left[\begin{array}{c}4c_1\\ 2c_2\\ c_1+c_2\end{array}\right]\end{array}$
  ::Col(A) = Span{[401],[021] ol(A) = c1} [401] co(A) = c1} [401] + c2} [021] = [4c12c2c1+c2]

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  The only way that  $\begin{array}{r}\stackrel{\to }{b}\end{array}$  could be in  $\begin{array}{r}\text{Col}(A)\end{array}$  is if  $\begin{array}{r}{c}_1=2,c_2=1\end{array}$  (which would take care of the first two entries of the column space). However, if that would be true then the third entry of the linear combination would be  $\begin{array}{r}3\ne 1\end{array}$ , so  $\begin{array}{r}\stackrel{\to }{b}\notin \text{Col}(A)\end{array}$ .   
  ::在 Col( A) 中, b 的唯一方法就是 c1=2, c2=1 (这将照顾列空间的前两个条目 ) 。 但是, 如果这是正确的的话, 那么线性组合的第三个条目将是 31, 所以 bcol( A) 。

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  We can get that
  ::我们能办到的

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  $$\begin{array}{r}\hat{x}=(A^{T}A{)}^{-1}{A}^T\stackrel{\to }{b}\\ \hat{x}=\frac{1}{84}\left[\begin{array}{cc}5& -1\\ -1& 17\end{array}\right]\cdot \left[\begin{array}{c}19\\ 11\end{array}\right]\\ \hat{x}=\left[\begin{array}{c}1\\ 2\end{array}\right]\\ \text{We get this because}\\ A^{T}A=\left[\begin{array}{ccc}4& 0& 1\\ 0& 2& 1\end{array}\right]\cdot \left[\begin{array}{cc}4& 0\\ 0& 2\\ 1& 1\end{array}\right]\\ A^{T}A=\left[\begin{array}{cc}17& 1\\ 1& 5\end{array}\right]\\ A^T\stackrel{\to }{b}=\left[\begin{array}{ccc}4& 0& 1\\ 0& 2& 1\end{array}\right]\cdot \left[\begin{array}{c}2\\ 0\\ 11\end{array}\right]\\ A^T\stackrel{\to }{b}=\left[\begin{array}{c}19\\ 11\end{array}\right]\\ (A^{T}A{)}^{-1}=\frac{1}{84}\left[\begin{array}{cc}5& -1\\ -1& 17\end{array}\right]\end{array}$$

  
  ::X(ATA)-1ATbx184[5-1-1-117][1911]x[12]我们理解这一点,因为ATAT=[401021][400211]ATA=[17115]ATb[401021]_[2011]ATb[1911](ATA)-1=184[5-1-1]

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  Now, the last theorem I'll show is that if  $\begin{array}{r}A\in {\mathbb{R}}^{m\times n}\end{array}$  with linearly independent columns such it has a QR factorization, t hen the least-squares solution of the equation  $\begin{array}{r}A\stackrel{\to }{x}=\stackrel{\to }{b}\end{array}$  is
  ::现在,我将展示的最后一个理论是 如果ARmxn有线性独立列的ARmxn 具有QR乘数化, 那么方程式Axb的最小平方解法就是

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  $$\begin{array}{r}\hat{x}=R^{-1}{Q}^T\stackrel{\to }{b}\end{array}$$

   
  ::xR-1QTb

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