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线性代数笔记
"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
请问公式的图片是用什么生成的呢
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
谢谢整理,笔记很简洁,但有些地方省略后阅读起来有点难受。。
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
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
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。
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
?
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}$
。
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
如标题
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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