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#LyX 2.3 created this file. For more info see http://www.lyx.org/
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\end_header
\begin_body
\begin_layout Section
向量的内积
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
向量的内积
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
在前面章节中, 我们研究了向量的线性运算, 并利用它讨论向量之间的线性关系, 但尚未涉及到向量的度量性质.
\end_layout
\begin_layout Standard
在空间解析几何中, 向量
\begin_inset Formula $\overrightarrow{x}=\left\{ x_{1},x_{2},x_{3}\right\} $
\end_inset
和
\begin_inset Formula $\overrightarrow{y}=\left\{ y_{1},y_{2},y_{3}\right\} $
\end_inset
的长度与夹角等度量性质可以通过两个向量的数量积
\begin_inset Formula
\[
\overrightarrow{x}\cdot\overrightarrow{y}=|\overrightarrow{x}||\overrightarrow{y}|\cos(\overrightarrow{x},\overrightarrow{y})
\]
\end_inset
来表示, 且在直角坐标系中, 有
\begin_inset Formula
\[
\begin{gathered}\overrightarrow{x}\cdot\overrightarrow{y}=x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3},\\
|\overrightarrow{x}|=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}.
\end{gathered}
\]
\end_inset
本节中, 我们要将数量积的概念推广到
\begin_inset Formula $n$
\end_inset
维向量空间中, 引入内积的概念.
\end_layout
\end_deeper
\begin_layout Subsection
内积及其性质
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
内积及其性质
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
设有
\begin_inset Formula $n$
\end_inset
维向量
\begin_inset Formula
\[
x=\begin{bmatrix}x_{1}\\
x_{2}\\
\vdots\\
x_{n}
\end{bmatrix},\quad y=\begin{bmatrix}y_{1}\\
y_{2}\\
\vdots\\
y_{n}
\end{bmatrix}
\]
\end_inset
称
\series bold
\begin_inset Formula $[x,y]$
\end_inset
为向量
\begin_inset Formula $x$
\end_inset
与
\begin_inset Formula $y$
\end_inset
的内积
\series default
.
内积是两个向量之间的一种运算, 其结果是一个实数, 按矩阵的记法可表示为
\begin_inset Formula
\[
[x,y]=x^{T}y=\begin{bmatrix}x_{1} & x_{2} & \cdots & x_{n}\end{bmatrix}\begin{bmatrix}y_{1}\\
y_{2}\\
\vdots\\
y_{n}
\end{bmatrix}.
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
计算内积
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设有
\begin_inset Formula $\RR^{3}$
\end_inset
中的基
\begin_inset Formula $e_{1}=(1,0,0)^{T}$
\end_inset
,
\begin_inset Formula $e_{2}=(0,1,0)^{T}$
\end_inset
,
\begin_inset Formula $e_{3}=(0,0,1)^{T}$
\end_inset
, 试求
\begin_inset Formula $e_{i}$
\end_inset
与
\begin_inset Formula $e_{j}$
\end_inset
(
\begin_inset Formula $i,j=1,2,3$
\end_inset
) 的内积.
\end_layout
\begin_layout Solution*
直接根据内积的定义计算,
\begin_inset Formula
\begin{align*}
\left[e_{1},e_{2}\right] & =1\times0+0\times1+0\times0=0,\\
\left[e_{2},e_{3}\right] & =0\times0+1\times0+0\times1=0,\\
\left[e_{3},e_{1}\right] & =0\times1+0\times0+1\times0=0.
\end{align*}
\end_inset
同理可得
\begin_inset Formula $\left[e_{i},e_{i}\right]=1$
\end_inset
, (
\begin_inset Formula $i=1,2,3$
\end_inset
).
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
内积的运算性质
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
内积的运算性质 (其中
\begin_inset Formula $x,y,z$
\end_inset
为
\begin_inset Formula $n$
\end_inset
维向量,
\begin_inset Formula $\lambda\in\RR$
\end_inset
):
\end_layout
\begin_layout Enumerate
\begin_inset Formula $[x,y]=[y,x]$
\end_inset
;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $[\lambda x,y]=\lambda[x,y]$
\end_inset
;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $[x+y,z]=[x,z]+[y,z]$
\end_inset
;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $[x,x]\geq0$
\end_inset
; 当且仅当
\begin_inset Formula $x=0$
\end_inset
时,
\begin_inset Formula $[x,x]=0$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
计算内积
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求
\begin_inset Formula $\left[\left([\alpha,\alpha]\beta-\frac{1}{3}[\alpha,\beta]\alpha\right),3\alpha\right]$
\end_inset
.
\end_layout
\begin_layout Solution*
注意使用内积的性质: 对称性.
\begin_inset Formula
\begin{align*}
\left[\left([\alpha,\alpha]\beta-\frac{1}{3}[\alpha,\beta]\alpha\right),3\alpha\right] & =3[\alpha,\alpha][\beta,\alpha]-[\alpha,\beta][\alpha,\alpha]\\
& =\{3[\alpha,\alpha]-[\alpha,\alpha]\}[\alpha,\beta]\\
& =2[\alpha,\alpha][\alpha,\beta].
\end{align*}
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
关于内积运算的说明
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
\begin_inset Formula $\alpha,\beta,\gamma$
\end_inset
是
\begin_inset Formula $n$
\end_inset
维实向量
\begin_inset Formula $(n>1)$
\end_inset
, 判断下列算式有无意义:
\end_layout
\begin_layout Example
(1).
\begin_inset Formula $[\alpha,\beta]\gamma=[\alpha,\alpha][\beta,\gamma]$
\end_inset
;
\end_layout
\begin_layout Example
(2).
\begin_inset Formula $[[\alpha,\beta]\gamma,\gamma]+2\alpha$
\end_inset
.
\end_layout
\begin_layout Solution*
在 (1) 中,
\begin_inset Formula $[\alpha,\beta]$
\end_inset
表示一个数, 因此
\begin_inset Formula $[\alpha,\beta]\gamma$
\end_inset
是一个向量, 而
\begin_inset Formula $[\alpha,\alpha]$
\end_inset
及
\begin_inset Formula $[\beta,\gamma]$
\end_inset
都是数,故
\begin_inset Formula $[\alpha,\alpha][\beta,\gamma]$
\end_inset
也是数.
于是 (1) 式变为一个向量减去一个数, 显然没有意义.
\end_layout
\begin_layout Solution*
在 (2) 中,
\begin_inset Formula $[\alpha,\beta]$
\end_inset
是数,
\begin_inset Formula $[\alpha,\beta]\gamma$
\end_inset
表示
\begin_inset Formula $[\alpha,\beta]$
\end_inset
与
\begin_inset Formula $\gamma$
\end_inset
的数乘,
\begin_inset Formula $[[\alpha,\beta]\gamma,\gamma]$
\end_inset
表示
\begin_inset Formula $[\alpha,\beta]\gamma$
\end_inset
与
\begin_inset Formula $\gamma$
\end_inset
的内积, 事实上
\begin_inset Formula
\[
[[\alpha,\beta]\gamma,\gamma]=[\alpha,\beta][\gamma,\gamma]
\]
\end_inset
因此 (2) 式中第一项是一个数, 而
\begin_inset Formula $2\alpha$
\end_inset
是一个向量, 两者相加无意义.
\end_layout
\end_deeper
\begin_layout Subsection
向量的长度与性质
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
向量的长度与性质
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
令
\begin_inset Formula
\[
\|x\|=\sqrt{[x,x]}=\sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}},
\]
\end_inset
称
\series bold
\begin_inset Formula $\|x\|$
\end_inset
为
\begin_inset Formula $n$
\end_inset
维向量
\begin_inset Formula $x$
\end_inset
的长度 (或范数)
\series default
.
\end_layout
\begin_layout Standard
向量的长度具有下述性质:
\end_layout
\begin_layout Enumerate
\series bold
\begin_inset Argument 1
status open
\begin_layout Plain Layout
+-|alert@+
\end_layout
\end_inset
非负性
\series default
:
\begin_inset Formula $\|x\|\geq0$
\end_inset
; 当且仅当
\begin_inset Formula $x=0$
\end_inset
时,
\begin_inset Formula $\|x\|=0$
\end_inset
;
\end_layout
\begin_layout Enumerate
\series bold
齐次性
\series default
:
\begin_inset Formula $\|\lambda x\|=|\lambda|\|x\|$
\end_inset
;
\end_layout
\begin_layout Enumerate
\series bold
三角不等式
\series default
:
\begin_inset Formula $\|x+y\|\leq\|x\|+\|y\|$
\end_inset
;
\end_layout
\begin_layout Enumerate
对任意
\begin_inset Formula $n$
\end_inset
维向量
\begin_inset Formula $x,y$
\end_inset
, 有
\begin_inset Formula $[x,y]\leq\|x\|\cdot\|y\|$
\end_inset
.
\begin_inset CommandInset label
LatexCommand label
name "enu:prop-4"
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
Cauchy 不等式
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Remark*
若令
\begin_inset Formula $x^{T}=\left(x_{1},x_{2},\cdots,x_{n}\right)$
\end_inset
,
\begin_inset Formula $y^{T}=\left(y_{1},y_{2},\cdots,y_{n}\right)$
\end_inset
, 则性质 (
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:prop-4"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 可表示为
\begin_inset Formula
\[
\left|\sum_{i=1}^{n}x_{i}y_{i}\right|\leq\sqrt{\sum_{i=1}^{n}x_{i}^{2}}\cdot\sqrt{\sum_{i=1}^{n}y_{i}^{2}}
\]
\end_inset
上述不等式称为
\series bold
\color red
柯西–布涅可夫斯基不等式
\series default
\color inherit
, 它说明
\begin_inset Formula $\RR^{n}$
\end_inset
中任意两个向量的内积与它们长度之间的关系.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
单位向量与向量单位化
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
当
\begin_inset Formula $\|x\|=1$
\end_inset
时, 称
\begin_inset Formula $x$
\end_inset
为
\series bold
单位向量
\series default
.
\end_layout
\begin_layout Standard
对
\begin_inset Formula $\RR^{n}$
\end_inset
中的任一非零向量
\begin_inset Formula $\alpha$
\end_inset
, 向量
\begin_inset Formula $\frac{\alpha}{\|\alpha\|}$
\end_inset
是一个单位向量, 这是因为
\begin_inset Formula
\[
\left\Vert \frac{\alpha}{\|\alpha\|}\right\Vert =\frac{1}{\|\alpha\|}\|\alpha\|=1.
\]
\end_inset
\end_layout
\begin_layout Standard
注: 用非零向量
\begin_inset Formula $\alpha$
\end_inset
除以向量
\begin_inset Formula $\alpha$
\end_inset
的模长得到一个单位向量, 这一过程通常称为
\series bold
把向量
\begin_inset Formula $\alpha$
\end_inset
单位化
\series default
.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
向量之间的夹角
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
当
\begin_inset Formula $\|\alpha\|\neq0$
\end_inset
,
\begin_inset Formula $\|\beta\|\neq0$
\end_inset
, 定义
\begin_inset Formula
\[
\theta=\arccos\frac{[\alpha,\beta]}{\|\alpha\|\cdot\|\beta\|},\quad(0\leq\theta\leq\pi),
\]
\end_inset
称
\begin_inset Formula $\theta$
\end_inset
为
\begin_inset Formula $n$
\end_inset
维向量
\begin_inset Formula $\alpha$
\end_inset
与
\begin_inset Formula $\beta$
\end_inset
的夹角.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
向量之间的夹角
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求
\begin_inset Formula $\RR^{3}$
\end_inset
中向量
\begin_inset Formula $\alpha=(4,0,3)^{T}$
\end_inset
,
\begin_inset Formula $\beta=(-\sqrt{3},3,2)^{T}$
\end_inset
之间的夹角
\begin_inset Formula $\theta$
\end_inset
.
\end_layout
\begin_layout Solution*
由
\begin_inset Formula
\begin{align*}
\|\alpha\| & =\sqrt{4^{2}+0^{2}+3^{2}}=5,\\
\|\beta\| & =\sqrt{(-\sqrt{3})^{2}+3^{2}+2^{2}}=4,\\{}
[\alpha,\beta] & =4(-\sqrt{3})+0\times3+3\times2=6-4\sqrt{3},
\end{align*}
\end_inset
所以
\begin_inset Formula
\[
\cos\theta=\frac{[\alpha,\beta]}{\|\alpha\|\cdot\|\beta\|}=\frac{6-4\sqrt{3}}{5\times4}=\frac{3-2\sqrt{3}}{10}\Longrightarrow\theta=\arccos\frac{3-2\sqrt{3}}{10}.
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
向量之间的夹角
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求
\begin_inset Formula $\RR^{5}$
\end_inset
中的向量
\begin_inset Formula $\alpha=(1,0,-1,0,2)^{T}$
\end_inset
,
\begin_inset Formula $\beta=(0,1,2,4,1)^{T}$
\end_inset
的夹角
\begin_inset Formula $\theta$
\end_inset
.
\end_layout
\begin_layout Solution*
因为
\begin_inset Formula
\[
[\alpha,\beta]=1\times0+0\times1+(-1)\times2+0\times4+2\times1=0,
\]
\end_inset
而
\begin_inset Formula $\cos\theta=\frac{[\alpha,\beta]}{\|\alpha\|\cdot\|\beta\|}=0$
\end_inset
, 所以
\begin_inset Formula $\theta=90^{\circ}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Subsection
正交向量组
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
向量的正交
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
若两向量
\begin_inset Formula $\alpha$
\end_inset
与
\begin_inset Formula $\beta$
\end_inset
的内积等于零, 即
\begin_inset Formula
\[
[\alpha,\beta]=0,
\]
\end_inset
则称
\series bold
向量
\begin_inset Formula $\alpha$
\end_inset
与
\begin_inset Formula $\beta$
\end_inset
相互正交
\series default
.
记作
\begin_inset Formula $\alpha\perp\beta$
\end_inset
.
\end_layout
\begin_layout Remark*
显然, 若
\begin_inset Formula $\alpha=0$
\end_inset
, 则
\begin_inset Formula $\alpha$
\end_inset
与任何向量都正交.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
正交向量组
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
若
\begin_inset Formula $n$
\end_inset
维向量
\begin_inset Formula $\alpha_{1},\alpha_{2},\cdots,\alpha_{r}$
\end_inset
是一个非零向量组, 且
\begin_inset Formula $\alpha_{1},\alpha_{2},\cdots,\alpha_{r}$
\end_inset
中的向量两两正交, 则称该
\series bold
向量组为正交向量组
\series default
.
\end_layout
\begin_layout Theorem
若
\begin_inset Formula $n$
\end_inset
维向量
\begin_inset Formula $\alpha_{1},\alpha_{2},\cdots,\alpha_{r}$
\end_inset
是一组正交向量组, 则
\begin_inset Formula $\alpha_{1},\cdots,\alpha_{r}$
\end_inset
线性无关.
\end_layout
\end_deeper
\begin_layout Subsection
规范正交基及其求法
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
规范正交基及其求法
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
设
\begin_inset Formula $V\subset\RR^{n}$
\end_inset
是一个向量空间,
\end_layout
\begin_layout Definition
(1) 若
\begin_inset Formula $\alpha_{1},\alpha_{2},\cdots,\alpha_{r}$
\end_inset
是向量空间
\begin_inset Formula $V$
\end_inset
的一个基, 且是两两正交的向量组, 则称
\series bold
\begin_inset Formula $\alpha_{1},\alpha_{2},\cdots,\alpha_{r}$
\end_inset
是向量空间
\begin_inset Formula $V$
\end_inset
的
\color red
正交基
\series default
\color inherit
.
\end_layout
\begin_layout Definition
(2) 若
\begin_inset Formula $e_{1},e_{2},\cdots,e_{r}$
\end_inset
是向量空间
\begin_inset Formula $V$
\end_inset
的一个基,
\begin_inset Formula $e_{1},\cdots,e_{r}$
\end_inset
两两正交, 且都是单位向量, 则称
\series bold
\begin_inset Formula $e_{1},\cdots,e_{r}$
\end_inset
是向量空间
\begin_inset Formula $V$
\end_inset
的一个
\color red
规范正交基
\series default
\color inherit
.
\end_layout
\begin_layout Standard
若
\begin_inset Formula $e_{1},\cdots,e_{r}$
\end_inset
是
\begin_inset Formula $V$
\end_inset
的一个规范正交基, 则
\begin_inset Formula $V$
\end_inset
中任一向量
\begin_inset Formula $\alpha$
\end_inset
能由
\begin_inset Formula $e_{1},\cdots,e_{r}$
\end_inset
线性表示, 设表示式为
\begin_inset Formula
\[
\alpha=\lambda_{1}e_{1}+\lambda_{2}e_{2}+\cdots+\lambda_{r}e_{r},
\]
\end_inset
为求其中的系数
\begin_inset Formula $\lambda_{i}$
\end_inset
, (
\begin_inset Formula $i=1,2,\cdots,r$
\end_inset
), 可用
\begin_inset Formula $e_{i}^{T}$
\end_inset
左乘上式, 有
\begin_inset Formula
\[
e_{i}^{T}\alpha=\lambda_{i}e_{i}^{T}e_{i}=\lambda_{i},
\]
\end_inset
即
\begin_inset Formula
\[
\lambda_{i}=e_{i}^{T}\alpha=\left[\alpha,e_{i}\right].
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
规范正交基及其求法
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
\begin_inset Formula
\[
\alpha=\lambda_{1}e_{1}+\lambda_{2}e_{2}+\cdots+\lambda_{r}e_{r},\quad\lambda_{i}=e_{i}^{T}\alpha=\left[\alpha,e_{i}\right],\quad i=1,2,\cdots r.
\]
\end_inset
这就是
\uwave on
向量在规范正交基中的坐标的计算公式
\uwave default
.
\end_layout
\begin_layout Standard
利用这个公式能方便地求得向量
\begin_inset Formula $\alpha$
\end_inset
在规范正交基
\begin_inset Formula $e_{1},\cdots,e_{r}$
\end_inset
下的坐标为:
\begin_inset Formula $\left(\lambda_{1},\lambda_{2},\cdots,\lambda_{r}\right)$
\end_inset
.
因此,
\uwave on
我们在给出向量空间的基时常常取规范正交基
\uwave default
.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Subsection
规范正交基的求法
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
规范正交基的求法
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
设
\begin_inset Formula $\alpha_{1},\cdots,\alpha_{r}$
\end_inset
是向量空间
\begin_inset Formula $V$
\end_inset
的一个基, 要
\bar under
求
\begin_inset Formula $V$
\end_inset
的一个规范正交基
\bar default
, 也就是要找一组两两正交的单位向量
\begin_inset Formula $e_{1},\cdots,e_{r}$
\end_inset
, 使
\begin_inset Formula $e_{1},\cdots,e_{r}$
\end_inset
与
\begin_inset Formula $\alpha_{1},\cdots,\alpha_{r}$
\end_inset
等价.
这样一个问题, 称为
\series bold
把基
\begin_inset Formula $\alpha_{1},\cdots,\alpha_{r}$
\end_inset
规范正交化
\series default
, 可按如下两个步骤进行:
\end_layout
\begin_layout Enumerate
\series bold
正交化
\series default
\begin_inset Formula
\[
\begin{aligned}\beta_{1} & =\alpha_{1},\\
\beta_{2} & =\alpha_{2}-\frac{\left[\beta_{1},\alpha_{2}\right]}{\left[\beta_{1},\beta_{1}\right]}\beta_{1},\\
\cdots & \cdots\cdots\cdots\cdots\cdots\\
\beta_{r} & =\alpha_{r}-\frac{\left[\beta_{1},\alpha_{r}\right]}{\left[\beta_{1},\beta_{1}\right]}\beta_{1}-\frac{\left[\beta_{2},\alpha_{r}\right]}{\left[\beta_{2},\beta_{2}\right]}\beta_{2}-\frac{\left[\beta_{r-1},\alpha_{r}\right]}{\left[\beta_{r-1},\beta_{r-1}\right]}\beta_{r-1}.
\end{aligned}
\]
\end_inset
容易验证
\begin_inset Formula $\beta_{1},\cdots,\beta_{r}$
\end_inset
两两正交, 且
\begin_inset Formula $\beta_{1},\cdots,\beta_{r}$
\end_inset
与
\begin_inset Formula $\alpha_{1},\cdots,\alpha_{r}$
\end_inset
等价.
\end_layout
\begin_deeper
\begin_layout Enumerate
注: 上述过程称为
\series bold
施密特 (Schimidt) 正交化过程
\series default
.
它不仅满足
\begin_inset Formula $\beta_{1},\cdots,\beta_{r}$
\end_inset
与
\begin_inset Formula $\alpha_{1},\cdots,\alpha_{r}$
\end_inset
等价, 还满足: 对任何
\begin_inset Formula $k$
\end_inset
, (
\begin_inset Formula $1\leq k\leq r$
\end_inset
), 向量组
\begin_inset Formula $\beta_{1},\cdots,\beta_{k}$
\end_inset
与
\begin_inset Formula $\alpha_{1},\cdots,\alpha_{k}$
\end_inset
等价.
\end_layout
\end_deeper
\begin_layout Enumerate
\series bold
单位化
\series default
: 取
\begin_inset Formula
\[
e_{1}=\frac{\beta_{1}}{\left\Vert \beta_{1}\right\Vert },\quad e_{2}=\frac{\beta_{2}}{\left\Vert \beta_{2}\right\Vert },\cdots,e_{r}=\frac{\beta_{r}}{\left\Vert \beta_{r}\right\Vert },
\]
\end_inset
则
\begin_inset Formula $e_{1},e_{2},\cdots,e_{r}$
\end_inset
是
\begin_inset Formula $V$
\end_inset
的一个规范正交基.
\end_layout
\begin_deeper
\begin_layout Enumerate
注: 施密特 (Schimidt) 正交化过程可将
\begin_inset Formula $\RR^{n}$
\end_inset
中的任一组线性无关的向量组
\begin_inset Formula $\alpha_{1},\cdots,\alpha_{r}$
\end_inset
化为与之等价的正交组
\begin_inset Formula $\beta_{1},\cdots,\beta_{k}$
\end_inset
; 再经过单位化, 得到一组与
\begin_inset Formula $\alpha_{1},\cdots,\alpha_{r}$
\end_inset
等价的规范正交组
\begin_inset Formula $e_{1},e_{2},\cdots,e_{r}$
\end_inset
.
\end_layout
\end_deeper
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
规范正交基及其求法
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $\alpha_{1}=\begin{bmatrix}1\\
2\\
-1
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $\alpha_{2}=\begin{bmatrix}-1\\
3\\
1
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $\alpha_{3}=\begin{bmatrix}4\\
-1\\
0
\end{bmatrix}$
\end_inset
, 试用施密特正交化方法, 将向量组正交规范化.
\end_layout
\begin_deeper
\begin_layout Pause
\end_layout
\end_deeper
\begin_layout Solution*
不难证明
\begin_inset Formula $\alpha_{1},\alpha_{2},\alpha_{3}$
\end_inset
是线性无关的.
取
\begin_inset Formula $\beta_{1}=\alpha_{1}$
\end_inset
;
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\begin_inset Formula
\[
\begin{aligned}\beta_{2} & =\alpha_{2}-\frac{\left[\alpha_{2},\beta_{1}\right]}{\left\Vert \beta_{1}\right\Vert ^{2}}\beta_{1}=\begin{bmatrix}-1\\
3\\
1
\end{bmatrix}-\frac{4}{6}\begin{bmatrix}1\\
2\\
-1
\end{bmatrix}=\frac{5}{3}\begin{bmatrix}-1\\
1\\
1
\end{bmatrix};\\
\beta_{3} & =\alpha_{3}-\frac{\left[\alpha_{3},\beta_{1}\right]}{\left\Vert \beta_{1}\right\Vert ^{2}}\beta_{1}-\frac{\left[\alpha_{3},\beta_{2}\right]}{\left\Vert \beta_{2}\right\Vert ^{2}}\beta_{2}=\begin{bmatrix}4\\
-1\\
0
\end{bmatrix}-\frac{1}{3}\begin{bmatrix}1\\
2\\
-1
\end{bmatrix}+\frac{5}{3}\begin{bmatrix}-1\\
1\\
1
\end{bmatrix}=2\begin{bmatrix}1\\
0\\
1
\end{bmatrix}.
\end{aligned}
\]
\end_inset
再把它们单位化, 取
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
e_{1}=\frac{\beta_{1}}{\left\Vert \beta_{1}\right\Vert }=\frac{1}{\sqrt{6}}\begin{bmatrix}1\\
2\\
-1
\end{bmatrix},\quad e_{2}=\frac{\beta_{2}}{\left\Vert \beta_{2}\right\Vert }=\frac{1}{\sqrt{3}}\begin{bmatrix}-1\\
1\\
1
\end{bmatrix},\quad e_{3}=\frac{\beta_{3}}{\left\Vert \beta_{3}\right\Vert }=\frac{1}{\sqrt{2}}\begin{bmatrix}1\\
0\\
1
\end{bmatrix},
\]
\end_inset
\begin_inset Formula $e,e_{2},e_{3}$
\end_inset
即为所求.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
将向量组正交规范化
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
用施密特正交化方法, 将向量组正交规范化
\begin_inset Formula
\[
\alpha_{1}=(1,1,1,1),\alpha_{2}=(1,-1,0,4),\alpha_{3}=(3,5,1,-1).
\]
\end_inset
\end_layout
\begin_deeper
\begin_layout Pause
\end_layout
\end_deeper
\begin_layout Solution*
显然,
\begin_inset Formula $\alpha_{1},\alpha_{2},\alpha_{3}$
\end_inset
是线性无关的.
先正交化, 取
\begin_inset Formula
\begin{align*}
\beta_{1} & =\alpha_{1}=(1,1,1,1);\\
\beta_{2} & =\alpha_{2}-\frac{\left[\beta_{1},\alpha_{2}\right]}{\left[\beta_{1},\beta_{1}\right]}\beta_{1}=(1,-1,0,4)-\frac{1-1+4}{1+1+1+1}(1,1,1,1)=(0,-2,-1,3);\\
\beta_{3} & =\alpha_{3}-\frac{\left[\beta_{1},\alpha_{3}\right]}{\left[\beta_{1},\beta_{1}\right]}\beta_{1}-\frac{\left[\beta_{2},\alpha_{3}\right]}{\left[\beta_{2},\beta_{2}\right]}\beta_{2}=(1,1,-2,0).
\end{align*}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
再单位化, 得规范正交向量如下:
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
\begin{aligned}e_{1} & =\frac{\beta_{1}}{\left\Vert \beta_{1}\right\Vert }=\frac{1}{2}(1,1,1,1)=\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right);\\
e_{2} & =\frac{\beta_{2}}{\left\Vert \beta_{2}\right\Vert }=\frac{1}{\sqrt{14}}(0,-2,-1,3)=\left(0,\frac{-2}{\sqrt{14}},\frac{-1}{\sqrt{14}},\frac{3}{\sqrt{14}}\right);\\
e_{3} & =\frac{\beta_{3}}{\left\Vert \beta_{3}\right\Vert }=\frac{1}{\sqrt{6}}(1,1,-2,0)=\left(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}},0\right)\text{. }
\end{aligned}
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
求正交基
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
已知三维向量空间中两个向量
\begin_inset Formula $\alpha_{1}=\begin{bmatrix}1\\
1\\
1
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $\alpha_{2}=\begin{bmatrix}1\\
-2\\
1
\end{bmatrix}$
\end_inset
正交, 试求
\begin_inset Formula $\alpha_{3}$
\end_inset
使
\begin_inset Formula $\alpha_{1},\alpha_{2},\alpha_{3}$
\end_inset
构成三维空间的一个正交基.
\end_layout
\begin_layout Solution*
设
\begin_inset Formula $\alpha_{3}=\left(x_{1},x_{2},x_{3}\right)^{T}\neq0$
\end_inset
, 且分别与
\begin_inset Formula $\alpha_{1},\alpha_{2}$
\end_inset
正交.
则
\begin_inset Formula $\left[\alpha_{1},\alpha_{3}\right]=\left[\alpha_{2},\alpha_{3}\right]=0$
\end_inset
, 即
\begin_inset Formula
\[
\begin{cases}
\left[\alpha_{1},\alpha_{3}\right]=x_{1}+x_{2}+x_{3}=0,\\
\left[\alpha_{2},\alpha_{3}\right]=x_{1}-2x_{2}+x_{3}=0,
\end{cases}
\]
\end_inset
解之得
\begin_inset Formula $x_{1}=-x_{3}$
\end_inset
,
\begin_inset Formula $x_{2}=0$
\end_inset
.
令
\begin_inset Formula $x_{3}=1\Longrightarrow\alpha_{3}=\begin{bmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{bmatrix}=\begin{bmatrix}-1\\
0\\
1
\end{bmatrix}$
\end_inset
.
由上可知
\begin_inset Formula $\alpha_{1},\alpha_{2},\alpha_{3}$
\end_inset
构成三维空间的一个正交基.
\end_layout
\end_deeper
\begin_layout Subsection
正交矩阵与正交变换
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
正交矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
若
\begin_inset Formula $n$
\end_inset
阶方阵
\begin_inset Formula $A$
\end_inset
满足
\end_layout
\begin_layout Definition
\begin_inset Formula
\[
A^{T}A=E,\text{ (即 }A^{-1}=A^{T}\text{),}
\]
\end_inset
\end_layout
\begin_layout Definition
则称
\series bold
\begin_inset Formula $A$
\end_inset
为正交矩阵
\series default
, 简称
\series bold
正交阵
\series default
.
\end_layout
\begin_deeper
\begin_layout Pause
\end_layout
\end_deeper
\begin_layout Theorem
\begin_inset Formula $A$
\end_inset
为正交矩阵的充要条件是
\begin_inset Formula $A$
\end_inset
的列向量都是单位正交向量组.
\end_layout
\begin_layout Remark*
由
\begin_inset Formula $A^{T}A=E$
\end_inset
与
\begin_inset Formula $AA^{T}=E$
\end_inset
等价, 定理的结论对行向量也成立.
即
\begin_inset Formula $A$
\end_inset
为正交矩阵的充要条件是
\begin_inset Formula $A$
\end_inset
的行向量都是单位正交向量组.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
正交变换及其性质
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
若
\begin_inset Formula $P$
\end_inset
为正交矩阵, 则
\series bold
线性变换
\begin_inset Formula $y=Px$
\end_inset
\series default
称为
\series bold
正交变换
\series default
.
\end_layout
\begin_layout ColumnsCenterAligned
\end_layout
\begin_deeper
\begin_layout Column
6cm
\end_layout
\begin_layout Standard
\series bold
正交变换的性质
\series default
:
\end_layout
\begin_layout Enumerate
正交变换保持向量的长度不变.
\end_layout
\begin_layout Enumerate
正交变换的例子主要有旋转变换与对称变换.
\end_layout
\begin_layout Enumerate
旋转变换的例子:
\begin_inset Formula
\[
\begin{bmatrix}\cos\theta & \sin\theta\\
-\sin\theta & \cos\theta
\end{bmatrix}.
\]
\end_inset
\end_layout
\begin_layout Column
6cm
\end_layout
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
begin{tikzpicture}[scale=0.8]
\end_layout
\begin_layout Plain Layout
\backslash
tkzInit[xmax=5,ymax=5,xmin=-1,ymin=-1] % limits the size of the axes
\end_layout
\begin_layout Plain Layout
\backslash
tkzDrawX[>=latex]
\end_layout
\begin_layout Plain Layout
\backslash
tkzDrawY[>=latex]
\end_layout
\begin_layout Plain Layout
\backslash
tkzDefPoint(0,0){A}
\end_layout
\begin_layout Plain Layout
\backslash
tkzDefPoint(3,0){B}
\end_layout
\begin_layout Plain Layout
\backslash
tkzDefPoint(0,3){C}
\end_layout
\begin_layout Plain Layout
\backslash
tkzDefPoint(3,3){D}
\end_layout
\begin_layout Plain Layout
\backslash
tkzDefPoint(1.5, 2.59808){E}
\end_layout
\begin_layout Plain Layout
\backslash
tkzDefPoint(-2.59808, 1.5){F}
\end_layout
\begin_layout Plain Layout
\backslash
tkzDefPoint(-1.09808, 4.09808){G}
\end_layout
\begin_layout Plain Layout
\backslash
tkzDrawSegments[vector style](A,B A,C)
\end_layout
\begin_layout Plain Layout
\backslash
tkzDrawSegments[dashed](D,B D,C)
\end_layout
\begin_layout Plain Layout
\backslash
tkzDrawSegments[dashed,red](A,E A,F E,G F,G)
\end_layout
\begin_layout Plain Layout
\backslash
tkzDrawArc[angles,thick,blue,->](A,B)(0,60)
\end_layout
\begin_layout Plain Layout
\backslash
tkzDrawArc[angles,thick,blue,->](A,C)(90,150)
\end_layout
\begin_layout Plain Layout
\backslash
tkzPicAngle["$
\backslash
theta$",draw=orange, <->,angle eccentricity=1.5, angle radius=5mm](B,A,E)
\end_layout
\begin_layout Plain Layout
\backslash
tkzPicAngle["$
\backslash
theta$",draw=orange, <->,angle eccentricity=1.5, angle radius=5mm](C,A,F)
\end_layout
\begin_layout Plain Layout
\backslash
tkzDrawPoints(A,B,C,D)
\end_layout
\begin_layout Plain Layout
\backslash
tkzDrawPoints[red](E,F,G)
\end_layout
\begin_layout Plain Layout
%
\backslash
tkzLabelPoints[below left](A,B,C,D,E,F,G)
\end_layout
\begin_layout Plain Layout
\backslash
end{tikzpicture}
\end_layout
\end_inset
\end_layout
\end_deeper
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
正交矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
判别下列矩形是否为正交阵.
\end_layout
\begin_layout Example
(1).
\begin_inset Formula $\begin{bmatrix}1 & -1/2 & 1/3\\
-1/2 & 1 & 1/2\\
1/3 & 1/2 & -1
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Example
(2).
\begin_inset Formula $\begin{bmatrix}1/9 & -8/9 & -4/9\\
-8/9 & 1/9 & -4/9\\
-4/9 & -4/9 & 7/9
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Pause
\end_layout
\end_deeper
\begin_layout Solution*
(1).
考察矩阵的第一列和第二列, 因为
\begin_inset Formula $1\times\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right)\times1+\frac{1}{3}\times\frac{1}{2}\neq0$
\end_inset
, 因此它不是正交矩阵;
\end_layout
\begin_layout Solution*
(2).
由正交矩阵的定义,
\begin_inset Formula
\[
\begin{bmatrix}1/9 & -8/9 & -4/9\\
-8/9 & 1/9 & -4/9\\
-4/9 & -4/9 & 7/9
\end{bmatrix}\begin{bmatrix}1/9 & -8/9 & -4/9\\
-8/9 & 1/9 & -4/9\\
-4/9 & -4/9 & 7/9
\end{bmatrix}^{T}=\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix},
\]
\end_inset
所以它是正交矩阵.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Subsection
作业
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
作业
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Problem
试将线性无关的向量组正交化
\begin_inset Formula
\[
\alpha_{1}=(1,1,1,1)^{T},\quad\alpha_{2}=(3,3,-1,-1)^{T},\quad\alpha_{3}=(-2,0,6,8)^{T}.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
已知
\begin_inset Formula $\alpha_{1}=\begin{bmatrix}1\\
1\\
1
\end{bmatrix}$
\end_inset
, 求一组非零向量
\begin_inset Formula $\alpha_{2},\alpha_{3}$
\end_inset
, 使
\begin_inset Formula $\alpha_{1},\alpha_{2},\alpha_{3}$
\end_inset
两两正交.
\end_layout
\end_deeper
\begin_layout Frame
\end_layout
\end_body
\end_document
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