课程： 5.2 电源和电源空间

章节大纲

选择节常规

折叠展开
常规

全部折叠展开全部
- 选择活动新闻通告
  
  新闻通告讨论区
选择节电源和电源空间

折叠展开
电源和电源空间
In the last lesson we reviewed eigenvalues and how they are the values that a linear transformation takes on whenever its output vector is a scalar multiple of its input vector. However, an eigenvector is the vector that corresponds to an eigenvalue and when the transformation is applied we get a scalar multiple of that vector.
::在最后一堂课中,我们考察了电子元值,以及线性变换在输出矢量是其输入矢量的弧数乘数时所使用的数值。然而,电子元变换是对应电子元值的矢量,在应用这种变换时,我们得到该矢量的弧数乘数。

When we looked at the geometry of the linear transformation we were able to see which vectors just became scalar multiples of themselves, but did not have a clearcut algebraic way to solve for their value. The importance of having a good method to solve for these comes up when we are applying linear algebra to something like image processing where we would have extremely high dimensional matrices that cannot be visualized very well whereas in our 2d and 3d cases it is pretty easy.
::当我们查看线性变换的几何学时,我们可以看到哪些矢量本身就变成了天体的倍数,但并没有清晰的代数法来解决它们的价值问题。当我们应用线性代数处理图像处理等图像处理时,我们发现哪些矢量刚刚变成了天体的天体倍数,但却没有清晰的代数法来解决它们的价值问题。当我们应用线性代数处理图像处理时,找到一个解决这些矢量的好方法的重要性就浮现了出来,在我们的2D和3D案例中,我们将会有非常高的维量矩阵,无法被视觉化得非常好,而非常容易。

Going back to when we learned about the null space of a matrix, denoted by $\begin{align*}Nul(A)\end{align*}$ we saw that it was defined to be $\begin{align*}\text{Nul}(A) := \big\{\vec{x}\mid A\cdot\vec{x} = \vec{0}\big\}\end{align*}$ , so the null space of a matrix is the set of vectors such that when multiplied by that matrix we get the zero vector. In the next chapter we will study this a bit more in-depth when we talk about linear transformations across different vector spaces.
::回到我们学到了Nul(A)所描述的矩阵空格时,我们看到它被定义为Nul(A):xAx0},所以矩阵空格是矢量的一组,这样当乘以该矩阵时,我们就会得到零矢量。在下一章,当我们谈论不同矢量空间的线性转换时,我们将对此进行更深入的研究。

So the eigenvectors are the vectors in the equation $\begin{align*}A\cdot\vec{x} = \lambda\cdot\vec{x}\end{align*}$ and by subtracting from both sides the lambda times the identity matrix portion we see that $\begin{align*}\vec{x} \in \text{ Nul}(A-\lambda\cdot I)\end{align*}$ and the eigenvector is just the basis of this null space. And then we define the eigenspace, denoted by $\begin{align*}E_{\lambda}\end{align*}$ to be $\begin{align*}E_{\lambda} := \text{ Nul}(A-\lambda\cdot I) := \big\{\vec{x} \mid (A-\lambda\cdot I)\vec{x} = \vec{0}\big\}\end{align*}$ and each eigenspace is defined separately with respect to each eigenvalue.
::等式中的矢量是Axxx, 通过从两侧减去 xxxx 和 x 乘以身份矩阵部分,我们看到 xnul(AI) 和 genvictor 仅仅是这个空格的基础。然后我们定义了 egenspace, 以 E 表示为 E: = Nul(AI): x{(AI) xx0} , 并且每个电子空间就每个电子价值分别定义。

So let's look at a numerical example, let $\begin{align*}A=\begin{bmatrix}2 & 1\\1 & 2\end{bmatrix}\end{align*}$ . Using techniques from the last lesson you can verify that the eigenvalues are $\begin{align*}\lambda = 1 \text{ and } \lambda = 3\end{align*}$ . Now let's find the eigenvectors and more generally, the eigenspace for $\begin{align*}\lambda = 3.\end{align*}$
::让我们看看一个数字示例,让我们来看看A=[2112]。使用上一个课的技术,你可以验证 igenvalues 是%1 和%3。现在让我们找到 igenvalues , 更广义地说, igenspace for%3 。

$\begin{align*}E_{\lambda = 3} = \text{ Nul}\Bigg(\begin{bmatrix}2&1\\1&2\end{bmatrix} - 3\cdot I\Bigg) = \text{ Nul}\Bigg(\begin{bmatrix}-1&1\\1&-1\end{bmatrix}\Bigg)\end{align*}$
::E3=Nul([[2112]-3I]=Nul([-111-1])

So we want to solve $\begin{align*}\begin{bmatrix}-1&1\\1&-1\end{bmatrix}\cdot\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix} = \vec{0}\end{align*}$
::所以我们想解决[-111-1] {[x1x2]_0]_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

This is just equivalent to solving the simultaneous system of linear equations generated by multiplying this matrix with this vector:

$\begin{align*}\begin{bmatrix}-x_{1}+x_{2}\\x_{1}-x_{2}\end{bmatrix} = \vec{0} \text{ which is just } x_{1} = x_{2} \text{ making } \vec{x} = \begin{bmatrix}x_{1}\\x_{1}\end{bmatrix} = x_{1}\cdot\begin{bmatrix}1\\1\end{bmatrix}\end{align*}$
::[- x1+x2x1- x2] @ @ @%0, 它只是 x1=x2 制造 *x=[x1x1] =x1}=x1}[11]

::这只相当于用此矢量乘以此矩阵而同时产生的线性方程式系统 : [- x1+x2x1 - x1 - x2] @ @%0, 它只是 x1=x2 making *x=[x1x1]=x1]=x1}[11]

Now our eigenvector is $\begin{align*}\vec{x} = \begin{bmatrix}1\\1\end{bmatrix}\end{align*}$ and our eigenspace is all linear combinations of our eigenvectors which is just $\begin{align*}E_{\lambda = 3} = \big\{a\vec{x}\mid a \in \mathbb{R}\big\}\end{align*}$ .
::现在,我们的向导是x=[11],我们的向导是我们向导的所有线性组合, 也就是E3axaxa{R}。

Now we can repeat the exact same process to find the eigenspace for our other eigenvalue.
::现在我们可以重复同样的过程来为我们的另外一种基因价值找到一个基因空间。

$\begin{align*}E_{\lambda = 1} = \text{Nul}\bigg(\begin{bmatrix}2&1\\1&2\end{bmatrix} - I\bigg) = \text{Nul}\bigg(\begin{bmatrix}1&1\\1&1\end{bmatrix}\bigg)\end{align*}$
::E1=Nul([2112]-I)=Nul([1111])

So now we want to solve the equation for all $\begin{align*}x_{1}\text{ and }x_{2}\end{align*}$ such that $\begin{align*}\begin{bmatrix}1 & 1\\1&1\end{bmatrix}\vec{x} = \begin{bmatrix}1&1\\1&1\end{bmatrix}\cdot\begin{bmatrix}x_{1} \\ x_{2}\end{bmatrix} = \begin{bmatrix}x_{1} + x_{2}\\x_{1} + x_{2}\end{bmatrix} = \vec{0}.\end{align*}$
::所以我们现在要解析所有 x1 和 x2 的方程, 这样 [1111] x=[1111] {x=[1111]}}}}[x1x2]=[x1+x2x1+x2]_#0] 。

This is true if and only if $\begin{align*}x_{1} = -x_{2}\end{align*}$ thus our eigenspace is $\begin{align*}\text{Span}\bigg\{\begin{bmatrix}1\\-1\end{bmatrix}\bigg\}\end{align*}$ .
::如果且只有在 x1 x2 的情况下, 我们的等离子空间是 Span{ [1- 1]} , 才会发生这种情况。

Here are some extra resources to help develop your understanding of eigenvectors and eigenspaces:
::以下是一些额外的资源,

章节大纲

常规

电源和电源空间