7.3 异常基础

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  半正态基础

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  A set of orthogonal unit vectors is called an orthonormal set of vectors and if that set spans a subspace of  $\begin{array}{r}{\mathbb{R}}^n\end{array}$  then that set is known as an orthonormal basis.
  ::一组正对角单位矢量被称为一组正正态矢量,如果该矢量设定跨越Rn的子空间,则该集合被称为正正态基数。

   

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  We know that any orthonormal set that spans a subspace W is going to be linearly independent because any set of orthonormal  vectors only has the trivial coefficients when trying to find coefficients that make them equal scale and sum to equal the 0 vector. 
  ::我们知道,任何横跨一个子空间W的正方正态设置都将是线性独立的,因为任何一组正方正态矢量在试图找到使它们规模和总和等于0矢量的系数时,只有微小的系数。

   

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  We get this from the equations we found in the last lesson.
  ::我们从最后一堂课的方程式中 发现了这个

  ---

   

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  Let's look at an example:
  ::让我们举个例子:

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  Take the span of the  set of  vectors in  $\begin{array}{r}{\mathbb{R}}^3\end{array}$  given by:
  ::使用 R3 中的矢量范围, 由 :

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  $\begin{array}{r}\{\stackrel{\to }{v_1},\stackrel{\to }{v_2},\stackrel{\to }{v_3}\}=\{\left[\begin{array}{c}\frac{2}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}\\ 0\end{array}\right],\left[\begin{array}{c}\frac{-1}{\sqrt{6}}\\ \frac{2}{\sqrt{6}}\\ \frac{1}{\sqrt{6}}\end{array}\right],\left[\begin{array}{c}\frac{-2}{\sqrt{44}}\\ \frac{2}{\sqrt{44}}\\ \frac{-6}{\sqrt{44}}\end{array}\right]\}\end{array}$  
  ::{v1,v2,v3}[25150],[-162616],[-2444-644}}

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  and show that this is an orthonormal basis of  $\begin{array}{r}{\mathbb{R}}^3\end{array}$
  ::并显示这是 R3 的正态基数

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  Fir st, one can obviously see that these columns span the three dimensional reals, just check the column space on your own by inspection.
  ::首先,人们可以明显地看到,这些柱子横跨三维真实体,只需检查一下你自己的柱子空间。

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  Next, we prove that these  vectors are orthonormal.
  ::接下来,我们证明这些矢量是正正态的。

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  First, we see that each of these vectors have magnitude 1:
  ::首先,我们看到这些矢量中的每一种具有1级:

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  $\begin{array}{r}||\stackrel{\to }{v_1}||=\sqrt{(\frac{2}{\sqrt{5}}{)}^2+(\frac{1}{\sqrt{5}}{)}^2+0}\\ ||\stackrel{\to }{v_1}||=\sqrt{\frac{4}{5}+\frac{1}{5}}\\ ||\stackrel{\to }{v_1}||=\sqrt{1}\\ ||\stackrel{\to }{v_1}||=1\\ ||\stackrel{\to }{v_2}||=\sqrt{(-\frac{1}{\sqrt{6}}{)}^2+(\frac{2}{\sqrt{6}}{)}^2+(\frac{1}{\sqrt{6}}{)}^2}\\ ||\stackrel{\to }{v_2}||=\sqrt{\frac{1}{6}+\frac{4}{6}+\frac{1}{6}}\\ ||\stackrel{\to }{v_2}||=\sqrt{1}\\ ||\stackrel{\to }{v_2}||=1\\ ||\stackrel{\to }{v_3}||=\sqrt{(-\frac{2}{\sqrt{44}}{)}^2+(\frac{2}{\sqrt{44}}{)}^2+(-\frac{6}{\sqrt{44}}{)}^2}\\ ||\stackrel{\to }{v_3}||=\sqrt{\frac{4}{44}+\frac{4}{44}+\frac{36}{44}}\\ ||\stackrel{\to }{v_3}||=\sqrt{1}\\ ||\stackrel{\to }{v_3}||=1\end{array}$  
  :25)2+(15)2+(15)2+(0)1+(45)+(15)1+(15)4+(15)1+(5)1+(1)1+(1)1+(16)2+(26)2+(16)2+(6)2+(16)+(6)2+(16)+(6)2+(6)+(6)2+(1)1}(1)+(2)2+(2)2+(2)2+(2)2+(2)+(4)2+(4)2+(5)+(4)2+(4)+(4)44)+(44)+(4)4)+(4)4)+(4)4}(4)+(4)

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  Next, checking the  vectors dot products we see that this is an orthonormal 
  ::接下来,检查矢量点产品 我们看到这是一种正正态

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  $\begin{array}{r}\stackrel{\to }{v_1}\cdot \stackrel{\to }{v_2}=\frac{-2}{\sqrt{30}}+\frac{2}{\sqrt{30}}+0=0\\ \stackrel{\to }{v_1}\cdot \stackrel{\to }{v_3}=\frac{-10}{\sqrt{220}}+\frac{10}{\sqrt{220}}=0\\ \stackrel{\to }{v_2}\cdot \stackrel{\to }{v_3}=\frac{2}{\sqrt{264}}+\frac{4}{\sqrt{264}}-\frac{6}{\sqrt{264}}=0\end{array}$  
  ::v1v2230+230+0=0v1v310220+10220=0v2v32264+4264-6264=0

  ---

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  Here are some theorems and properties to consider about orthonormal sets of vectors and matrices with orthonormal columns:
  ::以下是一些理论和属性, 用于考虑有正态矢量和带有正态柱的矩阵的正态矢量和矩阵:

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  Theorem: A matrix  $\begin{array}{r}U\end{array}$  with  $\begin{array}{r}m\times n\end{array}$  dimensions has orthonormal columns if and only if  $\begin{array}{r}{U}^{T}U=I\end{array}$
  ::定理: 带有 mxn 维度的矩阵U , 只有当 UTU=I 时, 才会有正异矩形列

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  Proof: We will prove just one case, however, the proof is easily generalizable. In this proof, assume that  $\begin{array}{r}U\end{array}$  is a  $\begin{array}{r}3\times 3\end{array}$  matrix 
  ::证据:我们只证明一个案例,不过,证据很容易普及。在这个证据中,假设U是一个 3x3 矩阵。

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  We can write the matrix  $\begin{array}{r}U=\left[\begin{array}{ccc}\stackrel{\to }{u_1}& \stackrel{\to }{u_2}& \stackrel{\to }{u_3}\end{array}\right]\end{array}$  which means that we can write the matrix  $\begin{array}{r}{U}^T\end{array}$  in the simplified form  $\begin{array}{r}{U}^T=\left[\begin{array}{c}{\stackrel{\to }{u_1}}^T\\ {\stackrel{\to }{u_2}}^T\\ {\stackrel{\to }{u_3}}^T\end{array}\right]\end{array}$  which just follows directly from  $\begin{array}{r}{U}^T\end{array}$  being the transpose of the matrix  $\begin{array}{r}U.\end{array}$
  ::我们可以写表U=[u1]u2u3],这意味着我们可以以简化表UT=[u1]Tu2Tu3T]的形式写表UT,这直接源于UT的转换。

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  Taking the product 
  ::获取产品

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  $\begin{array}{r}{U}^{T}U=\left[\begin{array}{c}{\stackrel{\to }{u_1}}^T\\ {\stackrel{\to }{u_2}}^T\\ {\stackrel{\to }{u_3}}^T\end{array}\right]\cdot \left[\begin{array}{ccc}\stackrel{\to }{u_1}& \stackrel{\to }{u_2}& \stackrel{\to }{u_3}\end{array}\right]\\ U^{T}U=\left[\begin{array}{ccc}{\stackrel{\to }{u_1}}^T\cdot \stackrel{\to }{u_1}& {\stackrel{\to }{u_1}}^T\cdot \stackrel{\to }{u_2}& {\stackrel{\to }{u_1}}^T\cdot \stackrel{\to }{u_3}\\ {\stackrel{\to }{u_2}}^T\cdot \stackrel{\to }{u_1}& {\stackrel{\to }{u_2}}^T\cdot \stackrel{\to }{u_2}& {\stackrel{\to }{u_2}}^T\cdot \stackrel{\to }{u_3}\\ {\stackrel{\to }{u_3}}^T\cdot \stackrel{\to }{u_1}& \stackrel{\to }{u_3^T}\cdot \stackrel{\to }{u_2}& {\stackrel{\to }{u_3}}^T\cdot \stackrel{\to }{u_3}\end{array}\right]\end{array}$
  ::UTU=[u1]TU2[u2]u3]UTU=[u1]Tu1[u1]u1]u1][u1]Tu2}u1}u2}u1}[u1]Tu2}[u2]u2}u2}u2}u2}u2}Tu3}u3}Tu1}u1}u1}u2}T2}u3}T

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  Now we know that from the last lesson that two vectors are orthogonal if and only if their dot product is 0.  We also know that the dot product of a vector of itself is 1 if and only if that vector has unit magnitude because the dot product is just the magnitude squared and the magnitude cannot be negative, hence this statement is true.
  ::现在,我们从最后的教训中知道,两个矢量是正向的,如果并且只有它们的点产值为0。我们也知道,一个矢量本身的点产值是1,如果并且只有该矢量具有单位数量,因为点产值是正方形的,而这个数量不能是负的,所以这个声明是真实的。

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  This then proves that the matrix product is the identity if and only if these vectors are orthonormal.
  ::然后,这证明矩阵产品是特性,如果而且只有在这些矢量是正正态的情况下。

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  Our next theorem teaches us some important properties about matrices with orthonormal columns.
  ::我们的下一个定理告诉我们一些重要的属性 有关带有正态柱的矩阵。

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  Theorem: If  $\begin{array}{r}U\end{array}$  is a matrix with orthonormal column vectors and dimensions  $\begin{array}{r}m\times n\end{array}$  then given any two vectors  $\begin{array}{r}\stackrel{\to }{x},\stackrel{\to }{y}\in {\mathbb{R}}^n\end{array}$  we have that multiplication by the matrix  $\begin{array}{r}U\end{array}$  is length preserving, preserve the value of the dot product and also preserve orthogonality.
  ::论理: 如果 U 是带有正态柱矢量和尺寸 mxn 的矩阵, 那么给任何两个矢量 x, yRn 我们的矩阵U的乘法是长度保存, 保存点产值, 并保存正数性 。

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  In other words:
  ::换言之:

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  $\begin{array}{r}1.||U\cdot \stackrel{\to }{x}||=||\stackrel{\to }{x}||\\ 2.(U\cdot \stackrel{\to }{x})\cdot (U\cdot \stackrel{\to }{y})=\stackrel{\to }{x}\cdot \stackrel{\to }{y}\\ 3.(U\cdot \stackrel{\to }{x})\cdot (U\cdot \stackrel{\to }{y})=0\text{ if and only if }\stackrel{\to }{x}\cdot \stackrel{\to }{y}=0\end{array}$
  ::{{{{{{{{{{{{}}}}{{{{}}{{{}}{{{}}{{{}}{{{}}{{{}}{{{}}{{{}}{{{{{}}{{}{{{}}{{{}}{{}}{{{}}{{}}{{}}}{{{}}{{}}{{}}{{{}}{{{}}}{{}}}{{}}}{}}{}{}}{}{}{{{}}{}{}{}}{}{}{}{}}{}{}{}{}{}{}{}{}{}{}{}{{}{}{{{}{}{}{{}{>{{>{}{{}{}{}}{>{}{>{{}}{}}}{{{{{{{{}}{}}}}}{{{{{{{{{{{{{{}}}{{{{{{{{{{{{}}}}}}}}}{{{{{{{{{{{{{{{}}}}}}}}}{,但只有{,但,但只要{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{,只要{{{{{{,只要{{{{{{{{{{{,只要{{{{{{{{{{,但{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}{,只要{,只要{{{,但只有{,但只有{{{,只要{{{{{{{{{{{

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  We can see that the third property follows easily from the second property.
  ::我们可以看到,第三处财产很容易从第二处财产继承。

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  Again, for the proof of this theorem I will just show that this is true for a 3 x 3 matrix and generalizing this proof is not that difficult, I just want to give a concrete example.
  ::再一次,为了证明这个理论,我只想表明,3x3矩阵的情况就是这样,而概括这一证据并不那么困难,我只想举一个具体的例子。

   

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  Proof: If  $\begin{array}{r}\stackrel{\to }{x}\in {\mathbb{R}}^3\text{ and }U=\left[\begin{array}{ccc}\stackrel{\to }{u_1}& \stackrel{\to }{u_2}& \stackrel{\to }{u_3}\end{array}\right]\end{array}$  we have that 
  ::证明: 如果 xR3 和 U=[u1>u2}u3}我们得到了

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  $\begin{array}{r}U\cdot \stackrel{\to }{x}=x_1\cdot \stackrel{\to }{u_1}+x_2\cdot \stackrel{\to }{u_2}+x_3\cdot \stackrel{\to }{u_3}\end{array}$
  ::Uxx1u1x22x33

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  When calculating the magnitude of this vector we get that because scalar multiples of orthogonal vectors keep the vectors orthogonal that
  ::当计算此矢量的大小时,我们得到这一点,因为正向矢量的弧倍数使矢量的正向量保持了正向量。

   

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  $\begin{array}{r}||U\cdot \stackrel{\to }{x}|{|}^2\\ =||x_1\stackrel{\to }{u_1}+x_2\stackrel{\to }{u_2}+x_3\stackrel{\to }{u_3}|{|}^2\\ =||x_1\stackrel{\to }{u_1}|{|}^2+||x_2\stackrel{\to }{u_2}|{|}^2+||x_3\stackrel{\to }{u_3}|{|}^2\\ =x_1^2||\stackrel{\to }{u_1}|{|}^2+x_2^2||\stackrel{\to }{u_2}|{|}^2+x_3^2||\stackrel{\to }{u_3}|{|}^2\\ =x_1^2+x_2^2+x_3^2\\ ||U\cdot \stackrel{\to }{x}|{|}^2=x_1^2+x_2^2+x_3^2\\ ||U\cdot \stackrel{\to }{x}||=\sqrt{x_1^2+x_2^2+x_3^2}=||\stackrel{\to }{x}||\\ ||U\cdot \stackrel{\to }{x}||=||\stackrel{\to }{x}||\end{array}$
  ::22222222222223322222222222222222222222x1222X22212x22+x22+x322x1122222222222222224xxxxxx1

   

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  Hence multiplication by a matrix with orthonormal columns is length preserving.
  ::由带有正态列的矩阵乘以时长保持长度。

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  Next, let's show that  $\begin{array}{r}(U\cdot \stackrel{\to }{x})\cdot (U\cdot \stackrel{\to }{y})=\stackrel{\to }{x}\cdot \stackrel{\to }{y}\end{array}$
  ::接下来,让我们展示一下(Ux) (U)()()()()()()()()()()()()())()()()()()。

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  $\begin{array}{r}U\cdot \stackrel{\to }{x}=\stackrel{\to }{u_1}{x}_1+\stackrel{\to }{u_2}{x}_2+\stackrel{\to }{u_3}{x}_3\\ U\cdot \stackrel{\to }{y}=\stackrel{\to }{u_1}{y}_1+\stackrel{\to }{u_2}\stackrel{\to }{y_2}+\stackrel{\to }{u_3}{y}_3\\ (U\cdot \stackrel{\to }{x})\cdot (U\cdot \stackrel{\to }{y})=(\stackrel{\to }{u_1}\cdot \stackrel{\to }{u_1})x_1{y}_1+(\stackrel{\to }{u_2}\cdot \stackrel{\to }{u_2})x_2{y}_2+(\stackrel{\to }{u_3}\cdot \stackrel{\to }{u_3})x_3{y}_3\end{array}$
  ::Uáxu1x1+u2_x2+u2_x2+u3_x3U}{y1+u2_y2}{{{{{{{{}}}}}{{{}}}{{{}}{{}}{}{}{}{}}{}}{}}x1}1+(u2}{}}x2}}}x2}2}+(u3}}}{}x3}}}}}x3}}}}}}}{}}{}}{}}}{}}{}{}}}{}{}{}}{}}{}{}{}}{}{}{}{}{}{}{}{}{}{}{}{{}{}{}{}{}{}{{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}}}}{}}}}{{{{}}}}}{{{{{{{{{}}}{{{{{{}}}}}}}}}}}}{{{{{{{{{{{{{{{{{{}}}}}}}}}}}{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{}}}}}{{{{{{{{{{{{{{{{{{{{}}}}}{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{

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  Now,  because the columns of U are orthonormal we know that they have unit magnitude and thus their dot products equal 1.
  ::现在,因为U的柱体是正方形 我们知道它们有单位数量 所以它们的点产品等于1

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  Hence,
  ::因此,

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  $\begin{array}{r}(U\cdot \stackrel{\to }{x})\cdot (U\cdot \stackrel{\to }{y})=x_1{y}_1+x_2{y}_2+x_3{y}_3\\ \stackrel{\to }{x}\cdot \stackrel{\to }{y}=x_1{y}_1+x_2{y}_2+x_3{y}_3\\ (U\cdot \stackrel{\to }{x})\cdot (U\cdot \stackrel{\to }{y})=\stackrel{\to }{x}\cdot \stackrel{\to }{y}\end{array}$
  :Ux) (Uy) =x1y1+x2y2+x3y3x (yx1y1+x2y2+x3y3)(Ux__) (Uy__) =xy__

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  In proving this, we also prove property three.
  ::在证明这一点时,我们也证明财产三。

  ---

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  Now, let's look at an example to show that our theorems actually hold.
  ::现在,让我们举个例子来说明 我们的定理确实有效。

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  Take the matrix  $\begin{array}{r}U\end{array}$  of dimensions 3 by 2 and the vector  $\begin{array}{r}\stackrel{\to }{x}\in {\mathbb{R}}^2\end{array}$
  ::将3x2号维度和矢量 xR2 的矩阵表 U 和矢量 xR2 制成矩阵表U

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  Let  $\begin{array}{r}U=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{2}{3}\\ \frac{1}{\sqrt{2}}& \frac{-2}{3}\\ 0& \frac{1}{3}\end{array}\right],\stackrel{\to }{x}=\left[\begin{array}{c}\sqrt{2}\\ 3\end{array}\right]\end{array}$
  ::Let=U[122312-23013],x_[23]

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  $\begin{array}{r}U\end{array}$  has orthonormal columns because
  ::U有正正异常列栏,因为

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    $\begin{array}{r}\sqrt{(\frac{1}{\sqrt{2}}{)}^2+(\frac{1}{\sqrt{2}}{)}^2+0^1}=\sqrt{\frac{1}{2}+\frac{1}{2}}=1\\ \text{ and }\\ \sqrt{(\frac{2}{3}{)}^2+(\frac{-2}{3}{)}^2+(\frac{1}{3}{)}^2}=\sqrt{\frac{4}{9}+\frac{4}{9}+\frac{1}{9}}=1\end{array}$
  :12)2+(12)2+01=12+12=1和(23)2+(-23)2+(13)2+(13)2=49+49+19=1

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  and similarly,
  ::同样,

  $\begin{array}{r}(\frac{1}{\sqrt{2}})(\frac{2}{3})+(\frac{1}{\sqrt{2}})(-\frac{2}{3})+(0)\cdot (\frac{1}{3})=\frac{\sqrt{2}}{3}-\frac{\sqrt{2}}{3}+0=0\end{array}$

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  Hence the columns are orthonormal. 
  ::因此,各栏是正态的。

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  Now we try and calculate  $\begin{array}{r}{U}^{T}U\end{array}$  and check that it is  $\begin{array}{r}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}$
  ::现在我们尝试计算UTU 并检查它是否为 [1001]

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  $\begin{array}{r}{U}^{T}U=\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0\\ \frac{2}{3}& -\frac{2}{3}& \frac{1}{3}\end{array}\right]\cdot \left[\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{2}{3}\\ \frac{1}{\sqrt{2}}& -\frac{2}{3}\\ 0& \frac{1}{3}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{1}{2}+\frac{1}{2}& \frac{\sqrt{2}}{3}-\frac{\sqrt{2}}{3}\\ \frac{\sqrt{2}}{3}-\frac{\sqrt{2}}{3}& \frac{4}{9}+\frac{4}{9}+\frac{1}{9}\end{array}\right]\\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}$
  ::UTU = [12202323-2313] = [122312-23013] = [12+1223-2323-2323-2349+49+19] = [1001]

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  Now, on your own verify the rest of the properties and for properties 2 and 3 of our second theorem use the vector  $\begin{array}{r}\stackrel{\to }{y}=\left[\begin{array}{c}1\\ \sqrt{7}\end{array}\right]\end{array}$ . 
  ::现在,由你来验证其余的属性 以及第二个理论的属性 2 和 3 使用矢量 y 。 [17] 。

  ---

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  Now, from here on out we are going to define a matrix with orthonormal columns as an orthogonal matrix (it really should be called an orthonormal matrix, but I don't know why mathematicians use the nomenclature that they do). Some people also define orthogonal matrices as ones whose transpose equal their inverse.
  ::现在,从现在开始,我们将定义一个带有正方形列的矩阵,作为正方形矩阵(这应该真正被称为正方形矩阵,但我不知道数学家为什么使用他们使用的术语。 一些人也把正方形矩阵定义为其反向转换等于正方形的矩阵。

   

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  On your own, prove that both these definitions are the same (it's quite trivial).
  ::证明这两个定义都是一样的(这相当微不足道)。

  ---

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  Here are some valuable videos to watch to help solidify your understanding of this topic:
  ::以下是一些有价值的影片, 帮助巩固你们对这个议题的理解: