课程： 2.6 不可逆矩阵定理

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选择节不可逆矩阵定理

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不可逆矩阵定理
For the past couple of chapters we have talked about a variety of topics whether it be linear independence, solving systems of equations or pivot columns and its relation to parametric vector form. However, all of these concepts are related to inverting matrices. In this lesson I will introduce a theorem that we will build on throughout the entirety of the course called the invertible matrix theorem. This theorem is a series of equivalent statements that essentially tells us when a matrix is or isn't invertible and all of these statements are true or they are all false.
::对于过去的两章,我们谈论了各种议题,无论是线性独立,解决方程式或轴柱系统及其与参数矢量表的关系。然而,所有这些概念都与反转矩阵有关。在这个教训中,我将提出一个我们在整个过程中将发扬的理论,即所谓的不可翻转矩阵定理。这个定理是一系列相当的表述,基本上告诉我们一个矩阵是或不是不可翻转的,所有这些陈述都是真实的,或者都是虚假的。

Before we talk about this, go back to section 1.3 and look at the final question where it discusses linear combination and span of vectors in $\begin{align*}\mathbb{R}^{n}\end{align*}$ . The conclusion you should have drawn is that $\begin{align*}n\end{align*}$ linearly independent vectors span $\begin{align*}\mathbb{R}^{n}\end{align*}$ . This will be relevant in our discussion of invertible matrices in this lesson .
::在我们讨论这个问题之前,请回到第1.3节,看看它讨论Rn矢量的线性组合和范围的最后问题。你本应该得出的结论是,线性独立的矢量横跨Rn。这对于我们讨论这一教训中不可倒置的矩阵很重要。

Theorem: The Invertible Matrix Theorem
::定理: 不可逆矩阵定理

(Only applies to square matrices as those are the only types of matrices that one can invert)
:仅适用于平方矩阵,因为这些是唯一可以倒转的矩阵类型)

An $\begin{align*}n \times n\end{align*}$ matrix is invertible if and only if:
::nxn 矩阵在以下条件下是不可倒置的:

The columns span $\begin{align*}\mathbb{R}^{n}\end{align*}$
::横跨Rn的列

For every vector $\begin{align*}\vec{b}\in\mathbb{R}^{n}\end{align*}$ there is a unique solution to $\begin{align*}A\vec{x} = \vec{b}\end{align*}$ for the vector $\begin{align*}\vec{x}\end{align*}$
::对于每个矢量 bRn , 对于矢量 x, Axb 有一种独特的解决方案。

The columns are linearly independent
::列线是线性独立的

There are $\begin{align*}n\end{align*}$ pivot columns
::有正轴柱

The absolute value of the determinant is greater than 0
::决定因素的绝对值大于0

The matrix can be row reduced to the identity matrix
::矩阵可以减到身份矩阵

The homogeneous system Ax = 0 has only the trivial solution
::平质系统 Ax = 0 只有一个微小的溶液

There are about a dozen more we will add to this list by the end of the course as we gain more knowledge about matrices and linear transformations.
::随着我们对矩阵和线性变换的更多了解,在课程结束时,我们将增加大约十几个人。

Now, let's give a proof of what we have so far. In order to prove this, we have to that everything implies the matrix is invertible and everything is equivalent.
::现在,让我们来证明一下我们目前的情况。为了证明这一点,我们必须证明,一切意味着矩阵是不可逆的,一切是等效的。

Going in the first direction, if A is invertible then $\begin{align*}A\vec{x}=\vec{0}\end{align*}$ has the trivial solution, because $\begin{align*}\vec{x}=A^{-1}\vec{0}=\vec{0}\end{align*}$ and that is the only solution. If that is the only solution then the columns are linearly independent because there is no linear dependence relation other than the trivial linear dependence relation. Because these n columns are linearly independent it spans $\begin{align*}\mathbb{R}^{n}\end{align*}$ and has a solution for every vector $\begin{align*}\vec{b}\in\mathbb{R}^{n}\end{align*}$ and a unique solution, because $\begin{align*}\vec{x}=A^{-1}\vec{b}\end{align*}$ . Because there is a unique solution then the matrix on the left side of the augment can be reduced to the identity matrix, because every entry has a solution and you would get $\begin{align*}x_{1}=a,x_{2}=b\end{align*}$ and so on, where the form if you put it in an augmented coefficient matrix leaves you with the identity on the left of the augment and a vector on the right. And if that is true then there are n pivot columns because there is a leading one in each column. Now for this direction, the last statement we have top prove is that the absolute value of the determinant is greater than 0.
::向第一个方向前进, 如果 A 是不可倒转的, 那么 Ax0 具有小的解决方案, 因为 xA- 100, 而这是唯一的解决方案。如果这是唯一的解决方案, 那么列将线性独立, 因为除了线性依赖关系之外, 没有线性依赖关系。因为这些 n 列是线性独立的, 它横跨 Rn , 并且对每个矢量 bRn 都有一个解决方案, 因为 xA- 1b 。因为有一个独特的解决方案, 那么增加值左侧的矩阵可以缩到身份矩阵中, 因为每个条目都有解决方案, 而您会得到 x1=a, x2=b 等, 如果将表格放入一个增强的系数矩阵中, 则该表格会保持增长值左侧的特性和右边的矢量。如果确实如此, 则有 n pivot 列, 因为每列中有一个导列。现在, 我们最后的语句会证明这个方向的绝对值大于 0 。

We have already shown that if a matrix is invertible then the columns are linearly independent. In an earlier lecture we showed that if the columns were linearly dependent of a matrix then the determinant would be 0. Recall we showed a brute force numerical proof for the two and three dimensional case. When we went to study higher dimensions we thought about what it meant geometrically in that the columns could only span at most some n-1 dimensional subspace of $\begin{align*}\mathbb{R}^{n}\end{align*}$ and that the determinant calculates the n dimensional analogue of area in that space and that the n-1 dimensional analogue would make it 0. We got this from seeing that the volume of a plane and the area of a line and the length of a point were all 0. Look here for a better visual:
::我们已经表明,如果矩阵是垂直的,那么列就是线性独立的。在早先的一次演讲中,我们表明,如果列的线性依附于一个矩阵,那么决定因素将是0。回顾我们为两维和三维案例展示了粗力的数值证明。当我们去研究更高维时,我们从几何角度思考了它的含义,即柱子最多只能跨越Rn的某个 n-1 维次空间,而决定因素计算出该空间的区域的 n 维类比,而 n-1 维类比则会使它成为0。我们从看到一平面的体积和一条线的面积以及一个点的长度都是0,我们从中了解到这一点。看这里看一个更好的视觉:

So essentially, the logic is that if the columns are linearly dependent then the matrix determinant is 0. So taking the contrapositive we get that if the columns are linearly independent then the matrix has a nonzero determinant.
::因此,基本上,逻辑是,如果柱子线性依赖线性,矩阵的决定因素是0。用相反的参数,我们得到的是,如果柱子线性独立,则矩阵有一个非零的决定因素。

We already showed that if the matrix is invertible then the columns are linearly independent and by the transitivity of logic we get that because linear independence of columns implies nonzero determinant, a matrix being invertible implies nonzero determinant.
::我们已经表明,如果矩阵是不可倒置的,那么各列是线性独立的,根据逻辑的中性,我们理解,因为各列的线性独立性意味着非零决定因素,一个不可倒置的矩阵意味着非零决定因素。

Now, we have to prove the other way. We have to show that if all of these cases hold, so does matrix invertibility.
::现在,我们必须证明相反的方法。我们必须证明,如果所有这些案例都存在,矩阵也是不可置疑的。

If the determinant is nonzero then the matrix is invertible. The formula for inverting a matrix only holds the possibility of not existing if the determinant is 0 because we assume every entry is a real number. So if the determinant is nonzero then the matrix is invertible.
::如果决定因素为非零,则矩阵是不可倒置的。如果决定因素为0,则颠倒矩阵的公式只能存在不存在的可能性,因为我们假设每个条目都是真实数字。所以如果决定因素为非零,则矩阵是不可倒置的。

Now, we have already shown, that we just articulated above, that when ever the columns are linearly independent, then the determinant is nonzero and as we just proved before this whenever the determinant is nonzero the matrix is invertible. Hence, when the columns of a matrix are linearly independent the matrix is invertible.
::我们已经表明,我们刚才已经说过,当纵列线性独立时,决定因素是非零的,正如我们以前证明的那样,当决定因素是非零时,矩阵是不可倒置的。因此,当一个矩阵的纵列线性独立时,矩阵是不可倒置的。

Now, if there are n pivot columns then the columns are linearly independent. If there are n pivot columns then we have the identity matrix to the left of the augmented matrix and the columns of the identity only have the trivial linear dependence relation and span $\begin{align*}\mathbb{R}^{n}\end{align*}$
::现在, 如果有正对角列, 则列是线性独立的。如果有正对角列, 那么在扩大的矩阵左侧有身份矩阵, 身份的列只有小线性依赖关系, 横跨 Rn 。

Now, if the columns span $\begin{align*}\mathbb{R}^{n}\end{align*}$ then the matrix is invertible because the columns spanning $\begin{align*}\mathbb{R}^{n}\end{align*}$ are row equivalent to the identity matrix and then there are n pivot columns. Hence we have proven the invertibility for n pivot columns and spanning $\begin{align*}\mathbb{R}^{n}\end{align*}$ .
::现在,如果列横跨Rn,则矩阵是不可倒置的,因为横跨Rn的列相当于身份矩阵的行,然后是正角列。因此,我们已经证明了正角列和横跨Rn的列是不可倒置的。

Trivially, it follows that if the matrix is row equivalent to the identity matrix then the columns span $\begin{align*}\mathbb{R}^{n}\end{align*}$ and obviously there are n pivot columns. In the end, we get that the matrix is invertible because the columns are linearly independent.
::从三方面来看,如果矩阵的行与身份矩阵相等,则列横跨Rn,显然有正角列。最后,我们发现矩阵是不可倒置的,因为列是线性独立的。

Lastly we have to prove that the homogeneous system of equations of $\begin{align*}A\vec{x}=\vec{0}\end{align*}$ has only the trivial and more so there is one and only one solution to $\begin{align*}A\vec{x}=\vec{b}\end{align*}$ for every $\begin{align*}\vec{b}\in\mathbb{R}^{n}\end{align*}$ .
::最后,我们必须证明,Ax*0等方程式的同质系统只是微不足道的,而且更多,因此,对于每个bRn来说,Ax*b只有一种解决办法。

Now, if there is one and only one solution to $\begin{align*}A\vec{x}=\vec{b}\end{align*}$ then the columns span $\begin{align*}\mathbb{R}^{n}\end{align*}$ , because for every $\begin{align*}\vec{b}\end{align*}$ you can grab a solution and then the columns span $\begin{align*}\mathbb{R}^{n}\end{align*}$ so they are row equivalent to the identity matrix.
::现在,如果Axb只有一个解决方案,那么柱子横跨Rn,因为对于每一个b,你可以抓住一个解决方案,然后柱子横跨Rn,所以它们相当于身份矩阵的行。

Lastly, because we proved this in general for $\begin{align*}\vec{b}\end{align*}$ we have proved this for the 0 vector and we are done proving the invertible matrix theorem, for now...
::最后,由于我们为0矢量证明了这一点, 现在已经证明了这一点, 而现在我们证明的是不可置疑的矩阵定理, 目前...

章节大纲

常规

不可逆矩阵定理