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notes-linear-algebra's Introduction

线性代数笔记

简介

作者:子实

线性代数笔记,使用 jupyter notebook (ipython notebook) 编写展示。

Github 加载 .ipynb 的速度较慢,建议在 Nbviewer 中查看该项目。


目录

来自MIT课程线性代数的笔记,可以在麻省理工公开课:线性代数观看。

笔记格式借鉴Jin Li机器学习笔记

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notes-linear-algebra's Issues

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In chapter 18, the 8th property of determinant :

当且仅当A可逆时,有det A=0

det A should not equal to 0 when A is reversible.

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ch03

逆(方阵) 小节,第 3 段

观察这个方阵,我们如果用另一个矩阵乘$A$,则得到的结果矩阵中的每一列应该都是【$\begin{bmatrix}1\2\end{bmatrix}$的倍数】,所以我们不可能从$AB$的乘积中得到单位矩阵$I$。

应该是 $\begin{bmatrix}1\\3\end{bmatrix}$的倍数

ch17

Gram-Schmidt正交化法 小节,倒数第 2 段

* 单位化,$q_1=\frac{1}{\sqrt 3}\begin{bmatrix}1\1\1\end{bmatrix},\quad 【q_2=\frac{1}{\sqrt 2}\begin{bmatrix}1\0\2\end{bmatrix}$】...

应该是 q_2=\frac{1}{\sqrt 2}\begin{bmatrix}0\\-1\\1\end{bmatrix}$

第一讲中的式子错误

方程组的矩阵型式不对。
例如:
线性方程组如下:

2x - y   = 0 
-x +2y   = 3 

矩阵型式:

[2    -1 ]   [ x ]      [0]
                       =
[-1   2  ]   [ y ]      [3]

但是文中却是:xy

lect 18 3.a

"3. a. $\begin{vmatrix}ta&tb\\tc&td\end{vmatrix}=t\begin{vmatrix}a&b\\c&d\end{vmatrix}$。\n",

tc&td should be c&d

lect11 subspace dimensions

求并(intersect):S∪U=D,dim(S∪U)=9
求交(sum): S∩U=M,dim(S∩U)=3

单词错了
交集 intersection: S∩U=D, dim(S∩U)=3
?:S∪U or S+U
这里教授强调union不是一个subspace,所以改用了加号combination。这里是不是说明一下会更好。

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第二步,我们希望在第三更方程中消去$z$项

这里是不是应该改成

第二步,我们希望在第三个方程中消去$y$项

Report Mistake: Lec18的笔记的第3点

由于issue里面的markdown暂时不支持数学公式,我就只能这样写了:
第3.a行列式性质里面有一个小错误

3.a.
| ta tb |
| tc td | 写错了

应该是:
| ta tb |
| c d |

screenshot

第三章 错误

现在来看看什么矩阵有逆,设A=[1237],我们来求A−1。[1237][acbd]=[1001],使用列向量线性组合的**,我们可以说A乘以A−1的第j列,能够得到I的第j列,这时我会得到一个关于列的方程组。

接下来介绍高斯-若尔当(Gauss-Jordan)方法,该方法可以一次处理所有的方程:

这个方程组为⎧⎩⎨⎪⎪⎪⎪[1237][ab]=[10][1237][cd]=[01],我们想要同时解这两个方程;

按 “A乘以A−1的第j列,能够得到I的第j列” 这个方程组为⎧⎩⎨⎪⎪⎪⎪[1237][ac]=[10][1237][bd]=[01]

我看了视频“[1237][acbd]=[1001]” 应该改为 [1237][abcd]=[1001] ab为第一列 cd为第二列

report mistakes

  • chapter 08
    The prof. takes $b_1=1,b_2=5,b_3=6 $ for an example, so
    $
    \begin{eqnarray*}
    x_1 & + & 2x_3 & = & 1 \
    & & 2x_3 & = & 3 \
    \end{eqnarray*}
    $
    $
    \begin{eqnarray*}
    x_1 & = & -2 \
    x_3 & = & \frac{3}{2} \
    \end{eqnarray*}
    $

  • chapter 16

$Pb$将会把向量投影在$A$的行空间中

Should be
$Pb$将会把向量投影在$A$的列空间中

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