7.7 内产产品空间

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  From lesson 1 in this chapter, we learned about inner products which help us define our notions of length and distance in vector spaces. In this lesson, we are going to revisit that notion of an inner product and talk about spaces which have inner products called inner product spaces.
  ::从本章第1课中,我们学到了内产物,这些内产物帮助我们定义矢量空间的长度和距离概念。 在这个教训中,我们将重新审视内产物的概念,并谈论内产物的内产物空间。

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   Recall the definition of an inner product.
  ::回顾内部产品的定义。

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  Definition: An inner product in a vector space  $\begin{array}{r}V\end{array}$  is a function of two  vectors in the vector space, denoted by  $\begin{array}{r}⟨\stackrel{\to }{u},\stackrel{\to }{v}⟩\end{array}$  . Now, the inner product has the following properties:
  ::定义:矢量空间V中的一种内产物是矢量空间中两种矢量的函数,代号为u,v。现在,内产物具有以下特性:

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  $\begin{array}{r}1.⟨\stackrel{\to }{u},\stackrel{\to }{v}⟩=⟨\stackrel{\to }{v},\stackrel{\to }{u}⟩\\ 2.⟨\stackrel{\to }{u}+\stackrel{\to }{v},\stackrel{\to }{w}⟩=⟨\stackrel{\to }{u},\stackrel{\to }{w}⟩+⟨\stackrel{\to }{v},\stackrel{\to }{w}⟩\\ 3.⟨c\stackrel{\to }{u},\stackrel{\to }{v}⟩=c⟨\stackrel{\to }{u},\stackrel{\to }{v}⟩\\ 4.⟨\stackrel{\to }{u},\stackrel{\to }{u}⟩\ge 0\text{ and }⟨\stackrel{\to }{u},\stackrel{\to }{u}⟩=0\text{ if and only if }\stackrel{\to }{u}=\stackrel{\to }{0}\end{array}$
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}哦! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}哦! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}哦! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}哦!

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  Now, we already have examples of inner products in lesson 6.1, so in case you need a refresher go back to that lesson.
  ::现在,我们已经在6.1课中有了内部产品的例子, 所以,如果你需要再补习一下, 回到那个教训。

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  Now, because we  have an inner product, we can extend the Gram-Schmidt process to any inner product space. 
  ::现在,因为我们有一个内部产品, 我们可以把Gram-Schmidt工艺 扩展到任何内部产品空间。

  Similarly, we can extend the least squares problem to any inner product space as well as inequalities such as the Cauchy-Schwarz inequality and the triangle inequality.

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  Now, let's look at a  two examples of different inner products and inner product spaces and try to apply some of the generalizations of those theorems above.

  Take  $\begin{array}{r}{\mathbb{P}}_4\end{array}$  to be the vector space with inner product defined as

  $$\begin{array}{r}⟨p(x),q(x)⟩=p(-2)q(-2)+\cdots +p(2)q(2)\end{array}$$

  where the numbers in between are also integers.

  Now, let's list out all of the values in a given polynomial evaluated at those indices in some vector in  $\begin{array}{r}{\mathbb{R}}^5\end{array}$  

  So let's say that we want to find an orthonormal basis for the polynomials  $\begin{array}{r}\{1,t,t^2\}\end{array}$  

  $$\begin{array}{r}1=\left[\begin{array}{c}1\\ 1\\ 1\\ 1\\ 1\end{array}\right],t=\left[\begin{array}{c}-2\\ -1\\ 0\\ 1\\ 2\end{array}\right],t^2=\left[\begin{array}{c}4\\ 1\\ 0\\ 1\\ 4\end{array}\right]\end{array}$$

   Looking at this we see that the dot product in  $\begin{array}{r}{\mathbb{R}}^5\end{array}$  is the same as this inner product in  $\begin{array}{r}{\mathbb{P}}_4\end{array}$  when given the special representation of these vectors.

  Let,  $\begin{array}{r}\stackrel{\to }{v_1}=\left[\begin{array}{c}1\\ 1\\ 1\\ 1\\ 1\end{array}\right]\end{array}$  and then

  $\begin{array}{r}\stackrel{\to }{v_2}=\left[\begin{array}{c}-2\\ -1\\ 0\\ 1\\ 2\end{array}\right]\end{array}$  because we'd end up getting that 

  So now let's try and make the third vector orthogonal to those first two.

  $$\begin{array}{r}\stackrel{\to }{v_3}=(4,1,0,1,4)-\frac{(1,1,1,1,1)(4,1,0,1,4)}{(1,1,1,1,1)(1,1,1,1,1)}(1,1,1,1,1)-\frac{(-2,-1,0,1,2)(4,1,0,1,4)}{(-2,-1,0,1,2)(-2,-1,0,1,2)}(-2,-1,0,1,2)\\ \stackrel{\to }{v_3}=(4,1,0,1,4)-(2,2,2,2,2)-0(-2,0,1,2)\\ \stackrel{\to }{v_3}=(2,-1,-2,-1,2)\\ \stackrel{\to }{v_3}=t^2-2\end{array}$$

   

  Thus, normalizing all of these vectors we get that

  $$\begin{array}{r}\frac{\stackrel{\to }{v_1}}{\sqrt{5}},\frac{\stackrel{\to }{v_2}}{\sqrt{10}},\frac{\stackrel{\to }{v_3}}{\sqrt{14}}\\ p_0(x)=\frac{1}{\sqrt{5}}\\ p_1(x)=\frac{t}{\sqrt{10}}\\ p_2(x)=\frac{t^2-2}{\sqrt{14}}\end{array}$$

   

  Doing our second example, let's try and find a best approximation applying the least squares method.

  Take the same vector space and that same orthogonal basis that we found  and approximate the  polynomial closest to  $\begin{array}{r}q(x)=5-\frac{1}{2}{t}^4\end{array}$  that is in that basis.

  Now, we get that

  $$\begin{array}{r}⟨q(x),p_0(x)⟩=8\\ ⟨p_0(x),p_0(x)⟩=5\\ ⟨q(x),p_1(x)⟩=0\\ ⟨q(x),p_2(x)⟩=-31\\ ⟨p_2(x),p_2(x)⟩=14\end{array}$$

  So, in order to find the polynomial closes to this basis we have to take the projection of  

  $\begin{array}{r}q(x)\text{ onto }\text{Span}\{p_0(x),p_1(x),p_2(x)\}\end{array}$

  We get that this vector is

  $$\begin{array}{r}\frac{⟨q(x),p_0(x)⟩}{⟨p_0(x),p_0(x)⟩}{p}_0(x)+\frac{⟨q(x),p_1(x)⟩}{⟨p_1(x),p_1(x)⟩}{p}_1(x)+\frac{⟨q(x),p_2(x)⟩}{⟨p_2(x),p_2(x)⟩}{p}_2(x)\\ \frac{8}{5}+\frac{-31}{14}(t^2-2)\\ \frac{8}{5}+\frac{-31t^2+62}{14}\\ \frac{422-155t^2}{70}\end{array}$$

   

  Now, let's look at a picture of this equation versus  $\begin{array}{r}q(x)\end{array}$  to get

   

   

   

  
  

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  This video is a review of the rest of the chapter:
  ::本录像回顾本章的其余部分: