Course: 6.5 不形态主义 | The Way To Learn

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Let's introduce a new vector space that we haven't really studied before, the polynomial space.
::让我们介绍一个新的矢量空间, 我们以前没有真正研究过, 多元空间。

We say that $\begin{align*}\mathbb{P}_{n}(x) = \{a_{n}x^{n} + \cdots + a_{0} | a_{0},\cdots,a_{n} \in \mathbb{R}\}\end{align*}$
::我们说,Pn(x) anxna0a0a00,,anR}

So, let's take a function from
::所以,让我们从

$\begin{align*}\\ f:\mathbb{P}_{2} \to \mathbb{R}^{3} \\ f(ax^{2} + bx + c) = \begin{bmatrix}a\\b\\c\end{bmatrix}\end{align*}$
::f:P2-R3f( 轴2+bx+c) = [ abc]

Here's a property of this function:
::以下是此函数的属性 :

The function $\begin{align*}f\end{align*}$ is one-to-one, so if $\begin{align*}f(p_{1}(x)) = f(p_{2}(x))\end{align*}$ then $\begin{align*}p_{1}(x) = p_{2}(x)\end{align*}$
::函数 f 是一对一, 所以如果 f( p1 (x)) = f( p2(x) , 那么 p1 (x) = p2(x)

To prove this we see that if $\begin{align*}p_{1}(x) = a_{1}x^{2} + b_{1}x + c_{1} & \& & p_{2}(x) = a_{2}x^{2} + b_{2}x + c_{2}\end{align*}$
::要证明这一点, 我们可以看到, 如果 p1( x) =a1x2+b1x+c1&p2( x) =a2x2+b2x+c2, 则 p1( x) =a2x2+b2x+c2

if $\begin{align*}f(p_{1}(x)) = \begin{bmatrix}a_{1}\\b_{1}\\c_{1}\end{bmatrix} \\ f(p_{2}(x)) = \begin{bmatrix}a_{2}\\b_{2}\\c_{2}\end{bmatrix} \\ \begin{bmatrix}a_{1}\\b_{1}\\c_{1}\end{bmatrix} = \begin{bmatrix}a_{2}\\b_{2}\\c_{2}\end{bmatrix} \\ \to a_{1} = a_{2}, b_{1} = b_{2}, c_{1} = c_{2} \\ p_{1}(x) = a_{1}x^{2} + b_{1}x + c_{1} \\ = a_{2}x^{2} + b_{2}x + c_{2} \\ = p_{2}(x)\end{align*}$
::如果 f( p1 (x)) = [a1b1c1f( p2(x) ) = [a2b2c2][a1b1c1] =[a2b2c22] =[a2b2c2] a1=a2,b1=b2,c1=c2p1 (x) =a1x2+b1x+c1=a2x2+b2x+c2=p2(x)

Similarly, it is true that for every triplet $\begin{align*}\begin{bmatrix}a\\b\\c\end{bmatrix}\in\mathbb{R}^{3} \exists p(x) \in\mathbb{P}_{2}(x) \text{s.t.} f(p(x)) = \begin{bmatrix}a\\b\\c\end{bmatrix}\end{align*}$
::同样,对于每个三胞胎来说,的确都是[abc]R3p(x)P2(x)P2(x)s.t.f(p(x))=[abc]。

This property is called being onto.
::此属性被命名为“ 正在连接 ” 。

The function f is both one-to-one and onto and we say that that means that it is a bijection or an isomorphism and that both the domain and codomain are isomorphic.
::函数 f 既一对一,又一对一,我们说,这意味着它是一个双向或无形态,域和共域都是无形态的。

Another example of an isomorphism is the change of coordinates transformation.
::地貌变异的另一个例子是坐标变异的变化。

Let's say we have a vector space $\begin{align*}V = \mathbb{R}^{3}\end{align*}$ and we have the standard basis vectors, $\begin{align*}\Bigg\{\begin{bmatrix}1\\0\\0\end{bmatrix}, \begin{bmatrix}0\\1\\0\end{bmatrix}, \begin{bmatrix}0\\0\\1\end{bmatrix}\Bigg\}\end{align*}$ and then we have another basis $\begin{align*}\mathcal{B} = \Bigg\{\begin{bmatrix}2\\0\\-1\end{bmatrix},\begin{bmatrix}4\\-2\\1\end{bmatrix},\begin{bmatrix}1\\0\\-2\end{bmatrix}\Bigg\} \end{align*}$
::假设我们有一个矢量空间V=R3,我们有标准基矢量,{[100],[010],[001],[001],然后我们有另一个基点B[20-1],[4-21],[10-2]

The function or mapping
::函数或绘图

$\begin{align*}\vec{x} \to[\vec{x}]_{\mathcal{B}}\end{align*}$ is an isomorphism
::x[x]B是一个无形态

Let's look at what this mapping does:
::让我们看看这幅地图会做什么:

Let's for example take the vector in the standard basis to equal
::例如,让我们在标准基中以矢量为等

$\begin{align*}\vec{x} = \begin{bmatrix}-1\\2\\-3\end{bmatrix} \\ \vec{x} = -1\cdot\begin{bmatrix}1\\0\\0\end{bmatrix} + 2\cdot\begin{bmatrix}0\\1\\0\end{bmatrix} - 3\begin{bmatrix}0\\0\\1\end{bmatrix} \\ \vec{x} = -1\vec{e_{1}} + 2\vec{e_{2}} - 3\vec{e_{3}} \\ [\vec{x}]_{\mathcal{B}} = [-1\vec{e_{1}} + 2\vec{e_{2}} - 3\vec{e_{3}}]_{\mathcal{B}} \\ [\vec{x}]_{\mathcal{B}} = -1[\vec{e_{1}}]_{\mathcal{B}} + 2[\vec{e_{2}}]_{\mathcal{B}} - 3[\vec{e_{3}}]_{\mathcal{B}} \\ [\vec{e_{1}}]_{\mathcal{B}} = \begin{bmatrix}\frac{2}{3} \\ 0 \\ -\frac{1}{3}\end{bmatrix} \\ [\vec{e_{2}}]_{\mathcal{B}} = \begin{bmatrix}\frac{3}{2} \\ -\frac{1}{2} \\ -1\end{bmatrix} \\ [\vec{e_{3}}]_{\mathcal{B}} = \begin{bmatrix}\frac{1}{3} \\ 0 \\ -\frac{2}{3}\end{bmatrix} \\ P_{\mathcal{B}\leftarrow S} = \begin{bmatrix}\frac{2}{3} & \frac{3}{2} & \frac{1}{3} \\ 0 & -\frac{1}{2} & 0 \\-\frac{1}{3} & -1 & -\frac{2}{3}\end{bmatrix} \\ [\vec{x}]_{B} = P_{\mathcal{B}\leftarrow S}\vec{x} = \begin{bmatrix}\frac{4}{3} \\ -1 \\\frac{1}{3}\end{bmatrix}\end{align*}$
::{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{>{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{>{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{>{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{

Let's now prove that this is an isomorphism
::现在让我们证明这是一个无形态主义

First we can prove that this transformation is onto
::我们首先可以证明,这种转变已经进入了

We see that because the basis vectors of the basis B are linearly independent we can gather that every element of $\begin{align*}\mathbb{R}^{3}\end{align*}$ can be written as a linear combination of all of the vectors in B. To prove this, let's show that the matrix of the bases of B is invertible which is true, because the determinant is non-zero.
::我们可以看到,因为B基基的基矢量是线性独立的,所以我们可以收集到,R3的每一个元素都可以被写成B中所有矢量的线性组合。为了证明这一点,让我们证明B基的矩阵是不可逆的,这是真的,因为决定因素是非零。

Next, we check for the transformation being one-to-one.
::接下来,我们检查一对一的变换情况

We must show that
::我们必须表明,

$\begin{align*}[\vec{x_{1}}]_{\mathcal{B}} = [\vec{x_{2}}]_{\mathcal{B}} \to \vec{x_{1}} = \vec{x_{2}} \\ [\vec{x_{1}}]_{\mathcal{B}} = P_{\mathcal{B}\leftarrow S}\vec{x_{1}} \\ [\vec{x_{2}}]_{\mathcal{B}} = P_{\mathcal{B}\leftarrow S}\vec{x_{2}} \\ P_{\mathcal{B}\leftarrow S}\vec{x_{1}} = P_{\mathcal{B}\leftarrow S}\vec{x_{2}} \\ P_{\mathcal{B}\leftarrow S}^{-1}P_{\mathcal{B}\leftarrow S}\vec{x_{1}} = P_{\mathcal{B}\leftarrow S}^{-1}P_{\mathcal{B}\leftarrow S}\vec{x_{2}} \\ \vec{x_{1}} = \vec{x_{2}} \end{align*}$
::[x1}B=[x2}Bx1}x2}x2}[x1}B=BB=BB=BB=Sx1}[x2}]B=BB=Sx2}B=BB=Sx2}[x2}B=B=BB=Sx2}[x2}[x2}]B=B=PB=Sx2}[x2}]B=Sx1}[x1]B=B=B=BB=PB=BB*Sx1}Sx1}[x2}}]B=B=PB*Sx2}[x1]Sx1}[x2}}[x1]B}B=B=PB=PB=S_Sx2}[Sx2}[x2}B=S_BD*Sx1}[Sx1}[x1}B=S_B=S_B

Hence we see that this is an isomorphism.
::因此,我们看到这是一种无形态主义。

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