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\end_header

\begin_body

\begin_layout Section
向量空间*
\end_layout

\begin_layout Subsection
向量空间与子空间
\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
向量空间与子空间
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Definition
设 
\begin_inset Formula $V$
\end_inset

 为 
\begin_inset Formula $n$
\end_inset

 维向量的集合, 若集合 
\begin_inset Formula $V$
\end_inset

 非空, 且集合 
\begin_inset Formula $V$
\end_inset

 对于 
\begin_inset Formula $n$
\end_inset

 维向量的加法及数乘两种运算封闭, 即
\end_layout

\begin_deeper
\begin_layout Enumerate
若 
\begin_inset Formula $\alpha\in V$
\end_inset

, 
\begin_inset Formula $\beta\in V$
\end_inset

, 则 
\begin_inset Formula $\alpha+\beta\in V$
\end_inset

;
\end_layout

\begin_layout Enumerate
若 
\begin_inset Formula $\alpha\in V$
\end_inset

, 
\begin_inset Formula $\lambda\in\RR$
\end_inset

, 则 
\begin_inset Formula $\lambda\alpha\in V$
\end_inset

.
 
\end_layout

\end_deeper
\begin_layout Definition
则称
\series bold
集合 
\begin_inset Formula $V$
\end_inset

 为 
\begin_inset Formula $\RR$
\end_inset

 上的向量空间
\series default
.
 
\end_layout

\begin_layout Definition
记
\series bold
所有 
\begin_inset Formula $n$
\end_inset

 维向量的集合
\series default
为 
\begin_inset Formula $\RR^{n}$
\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
向量空间的例子
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
由 
\begin_inset Formula $n$
\end_inset

 维向量的线性运算规律, 容易验证集合 
\begin_inset Formula $\RR^{n}$
\end_inset

 对于加法及数乘两种运算封闭.
 因而集合 
\begin_inset Formula $\RR^{n}$
\end_inset

 构成一向量空间, 称 
\begin_inset Formula $\RR^{n}$
\end_inset

 为 
\begin_inset Formula $n$
\end_inset

 维向量空间.
\end_layout

\begin_layout Remark*
\begin_inset Formula $n=3$
\end_inset

 时, 三维向量空间 
\begin_inset Formula $\RR^{3}$
\end_inset

 表示实空间;
\end_layout

\begin_layout Remark*
\begin_inset Formula $n=2$
\end_inset

 时, 二维向量空间 
\begin_inset Formula $\RR^{2}$
\end_inset

 表示平面;
\end_layout

\begin_layout Remark*
\begin_inset Formula $n=1$
\end_inset

 时, 一维向量空间 
\begin_inset Formula $\RR^{1}$
\end_inset

 表示数轴.
\end_layout

\begin_layout Remark*
\begin_inset Formula $n>3$
\end_inset

 时, 
\begin_inset Formula $\RR^{n}$
\end_inset

 没有直观的几何形象.
\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
(向量空间的) 子空间
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Definition
设有向量空间 
\begin_inset Formula $V_{1}$
\end_inset

 和 
\begin_inset Formula $V_{2}$
\end_inset

, 若向量空间 
\begin_inset Formula $V_{1}\subset V_{2}$
\end_inset

, 则称 
\series bold

\begin_inset Formula $V_{1}$
\end_inset

 是 
\begin_inset Formula $V_{2}$
\end_inset

 的子空间
\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
向量空间的例子
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
判别下列集合是否为向量空间
\end_layout

\begin_layout Example
\begin_inset Formula 
\[
V_{1}=\left\{ x=\left(0,x_{2},\cdots,x_{n}\right)^{T}\mid x_{2},\cdots,x_{n}\in\RR\right\} .
\]

\end_inset


\end_layout

\begin_deeper
\begin_layout Pause

\end_layout

\end_deeper
\begin_layout Solution*
\begin_inset Formula $V_{1}$
\end_inset

 是向量空间.
 因为对于 
\begin_inset Formula $V_{1}$
\end_inset

 的任意两个元素
\begin_inset Formula 
\[
\alpha=\left(0,a_{2},\cdots,a_{n}\right)^{T},\ \beta=\left(0,b_{2},\cdots,b_{n}\right)^{T}\in V_{1},
\]

\end_inset

有 
\begin_inset Formula $\alpha+\beta=\left(0,a_{2}+b_{2},\cdots,a_{n}+b_{n}\right)^{T}\in V_{1}$
\end_inset

, 
\begin_inset Formula $\lambda\alpha=\left(0,\lambda a_{2},\cdots,\lambda a_{n}\right)^{T}\in V_{1}$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Frame

\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
不是向量空间的例子
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
判别下列集合是否为向量空间
\begin_inset Formula 
\[
V_{2}=\left\{ x=\left(1,x_{2},\cdots,x_{n}\right)^{T}\mid x_{2},\cdots,x_{n}\in\RR\right\} .
\]

\end_inset


\end_layout

\begin_deeper
\begin_layout Pause

\end_layout

\end_deeper
\begin_layout Solution*
集合 
\begin_inset Formula $V_{2}$
\end_inset

 不是向量空间.
 因为若 
\begin_inset Formula $\alpha=\left(1,a_{2},\cdots,a_{n},\right)^{T}\in V_{2}$
\end_inset

, 则 
\begin_inset Formula $2\alpha=\left(2,2a_{2},\cdots,2a_{n},\right)^{T}\notin V_{2}$
\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
由两个向量生成的向量空间
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
设 
\begin_inset Formula $\alpha,\beta$
\end_inset

 为两个已知的 
\begin_inset Formula $n$
\end_inset

 维向量, 集合
\begin_inset Formula 
\[
V=\{\xi=\lambda\alpha+\mu\beta\mid\lambda,\mu\in\RR\},
\]

\end_inset

试判断集合 
\begin_inset Formula $V$
\end_inset

 是否为向量空间.
 
\end_layout

\begin_deeper
\begin_layout Pause

\end_layout

\end_deeper
\begin_layout Solution*
\begin_inset Formula $V$
\end_inset

 是一个向量空间.
 因为若 
\begin_inset Formula $\xi_{1}=\lambda_{1}\alpha+\mu_{1}\beta$
\end_inset

, 
\begin_inset Formula $\xi_{2}=\lambda_{2}\alpha+\mu_{2}\beta$
\end_inset

, 则有 
\begin_inset Formula 
\[
\xi_{1}+\xi_{2}=\left(\lambda_{1}+\lambda_{2}\right)\alpha+\left(\mu_{1}+\mu_{2}\right)\beta\in V,\ k\xi_{1}=\left(k\lambda_{1}\right)\alpha+\left(k\mu_{1}\right)\beta\in V.
\]

\end_inset

这个向量空间称为
\series bold
由向量 
\begin_inset Formula $\alpha,\beta$
\end_inset

 所生成的向量空间
\series default
.
 通常记 
\begin_inset Formula $V$
\end_inset

 为 
\begin_inset Formula $\mathrm{span}\left\{ \alpha,\beta\right\} $
\end_inset

.
\end_layout

\begin_layout Pause

\end_layout

\begin_layout Remark*
通常由向量组 
\begin_inset Formula $a_{1},a_{2},\cdots,a_{m}$
\end_inset

 所生成的向量空间记为
\begin_inset Formula 
\[
V=\mathrm{span}\{a_{1},\cdots,a_{m}\}=\left\{ \xi=\lambda_{1}a_{1}+\lambda_{2}a_{2}+\cdots+\lambda_{m}a_{m}\mid\lambda_{1},\lambda_{2},\cdots,\lambda_{m}\in\RR\right\} .
\]

\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
等价向量组生成的向量空间相等
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
设向量组 
\begin_inset Formula $\alpha_{1},\cdots,\alpha_{m}$
\end_inset

 与向量组 
\begin_inset Formula $\beta_{1},\cdots,\beta_{s}$
\end_inset

 等价, 记
\begin_inset Formula 
\[
\begin{aligned}V_{1} & =\left\{ \xi=\lambda_{1}\alpha_{1}+\lambda_{2}\alpha_{2}+\cdots+\lambda_{m}\alpha_{m}\mid\lambda_{1},\lambda_{2},\cdots,\lambda_{m}\in\RR\right\} ,\\
V_{2} & =\left\{ \xi=\mu_{1}\beta_{1}+\mu_{2}\beta_{2}+\cdots+\mu_{s}\beta_{s}\mid\mu_{1},\mu_{2},\cdots,\mu_{s}\in\RR\right\} ,
\end{aligned}
\]

\end_inset

试证: 
\begin_inset Formula $V_{1}=V_{2}$
\end_inset

.
\end_layout

\begin_deeper
\begin_layout Pause

\end_layout

\end_deeper
\begin_layout Proof
设 
\begin_inset Formula $\xi\in V_{1}$
\end_inset

, 则 
\begin_inset Formula $\xi$
\end_inset

 可由 
\begin_inset Formula $\alpha_{1},\cdots,\alpha_{m}$
\end_inset

 线性表示.
 因 
\begin_inset Formula $\alpha_{1},\cdots,\alpha_{m}$
\end_inset

 可由 
\begin_inset Formula $\beta_{1},\cdots,\beta_{s}$
\end_inset

 线性表示, 故 
\begin_inset Formula $\xi$
\end_inset

 可由 
\begin_inset Formula $\beta_{1},\cdots,\beta_{s}$
\end_inset

 线性表示, 因此 
\begin_inset Formula $\xi\in V_{2}$
\end_inset

.
 即, 若 
\begin_inset Formula $\xi\in V_{1}$
\end_inset

, 则 
\begin_inset Formula $\xi\in V_{2}\Longrightarrow V_{1}\subset V_{2}$
\end_inset

.
\end_layout

\begin_layout Proof
类似地可证: 若 
\begin_inset Formula $\xi\in V_{2}$
\end_inset

, 则 
\begin_inset Formula $\xi\in V_{1}\Longrightarrow V_{2}\subset V_{1}$
\end_inset

.
\end_layout

\begin_layout Proof
因为 
\begin_inset Formula $V_{1}\subset V_{2},V_{2}\subset V_{1}$
\end_inset

, 所以 
\begin_inset Formula $V_{1}=V_{2}$
\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
齐次线性方程组的解集是向量空间
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
考虑齐次线性方程组 
\begin_inset Formula $Ax=0$
\end_inset

, 全体解的集合为
\begin_inset Formula 
\[
S=\{\alpha\mid A\alpha=0\}.
\]

\end_inset

显然, 
\begin_inset Formula $S$
\end_inset

 非空 (因为 
\begin_inset Formula $0\in S$
\end_inset

).
 任取 
\begin_inset Formula $\alpha,\beta\in S$
\end_inset

, 
\begin_inset Formula $k$
\end_inset

 为任一常数, 则
\begin_inset Formula 
\[
\begin{aligned}A(\alpha+\beta) & =A\alpha+A\beta=0\Longrightarrow\alpha+\beta\in S,\\
A(k\alpha) & =kA\alpha=k0=0\Longrightarrow k\alpha\in S,
\end{aligned}
\]

\end_inset

故 
\begin_inset Formula $S$
\end_inset

 是一向量空间.
 称 
\begin_inset Formula $S$
\end_inset

 为齐次线性方程组 
\begin_inset Formula $Ax=0$
\end_inset

 的解空间.
\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
三维空间 
\begin_inset Formula $\RR^{3}$
\end_inset

 的子空间
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
\begin_inset Formula $\RR^{3}$
\end_inset

 中过原点的平面是 
\begin_inset Formula $\RR^{3}$
\end_inset

 的子空间.
\end_layout

\begin_deeper
\begin_layout Pause

\end_layout

\end_deeper
\begin_layout Proof
\begin_inset Formula $\RR^{3}$
\end_inset

 中过原点的平面可以看作集合
\begin_inset Formula 
\[
V=\left\{ (x,y,z)\in\RR^{3}\mid\alpha x+\beta y+\gamma z=0,\text{ 其中 }(\alpha,\beta,\gamma)\in\RR^{3}\right\} .
\]

\end_inset

若 
\begin_inset Formula $\left(x_{1},y_{1},z_{1}\right)\in V$
\end_inset

, 
\begin_inset Formula $\left(x_{2},y_{2},z_{2}\right)\in V$
\end_inset

, 即
\begin_inset Formula 
\[
\alpha x_{1}+\beta y_{1}+\gamma z_{1}=0,\ \alpha x_{2}+\beta y_{2}+\gamma z_{2}=0,
\]

\end_inset

则有
\begin_inset Formula 
\[
\alpha\left(x_{1}+x_{2}\right)+\beta\left(y_{1}+y_{2}\right)+\gamma\left(z_{1}+z_{2}\right)=0,\ k\alpha x_{1}+k\beta y_{1}+k\gamma z_{1}=0,
\]

\end_inset

即
\begin_inset Formula 
\[
\left(x_{1},y_{1},z_{1}\right)+\left(x_{2},y_{2},z_{2}\right)\in V,\ k\left(x_{1},y_{1},z_{1}\right)\in V,
\]

\end_inset

故 
\begin_inset Formula $\RR^{3}$
\end_inset

 中过原点的平面是 
\begin_inset Formula $\RR^{3}$
\end_inset

 的子空间.
\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
\begin_inset Formula $\RR^{2}$
\end_inset

 不是 
\begin_inset Formula $\RR^{3}$
\end_inset

 的子空间
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
向量空间 
\begin_inset Formula $\RR^{2}$
\end_inset

 不是 
\begin_inset Formula $\RR^{3}$
\end_inset

 的子空间, 因为 
\begin_inset Formula $\RR^{2}$
\end_inset

 根本不是 
\begin_inset Formula $\RR^{3}$
\end_inset

 的子集 (
\begin_inset Formula $\RR^{3}$
\end_inset

 中的向量有三个分量, 但 
\begin_inset Formula $\RR^{2}$
\end_inset

 中的分量却只有两个).
 集合
\begin_inset Formula 
\[
H=\{(s,t,0)\mid s,t\in\RR\}
\]

\end_inset

是 
\begin_inset Formula $\RR^{3}$
\end_inset

 的子集
\begin_inset Foot
status open

\begin_layout Plain Layout
与 
\begin_inset Formula $\RR^{2}$
\end_inset

 有相同的性态, 尽管严格意义上 
\begin_inset Formula $H$
\end_inset

 不同于 
\begin_inset Formula $\RR^{2}$
\end_inset


\end_layout

\end_inset

.
 证明 
\begin_inset Formula $H$
\end_inset

 是 
\begin_inset Formula $\RR^{3}$
\end_inset

 的子空间.
\end_layout

\begin_deeper
\begin_layout Pause

\end_layout

\end_deeper
\begin_layout Proof
任取 
\begin_inset Formula $\left(s_{1},t_{1},0\right),\left(s_{2},t_{2},0\right)\in H$
\end_inset

, 
\begin_inset Formula $k$
\end_inset

 为任意常数, 则
\end_layout

\begin_layout Proof
\begin_inset Formula 
\[
\left(s_{1},t_{1},0\right)+\left(s_{2},t_{2},0\right)\in H,\ k\left(s_{1},t_{1},0\right)\in H,
\]

\end_inset

因此 
\begin_inset Formula $H$
\end_inset

 是 
\begin_inset Formula $\RR^{3}$
\end_inset

 的子空间.
\end_layout

\end_deeper
\begin_layout Subsection
向量空间的基与维数
\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
向量空间的基与维数
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Definition
设 
\begin_inset Formula $V$
\end_inset

 是向量空间, 若有 
\begin_inset Formula $r$
\end_inset

 个向量 
\begin_inset Formula $\alpha_{1},\alpha_{2},\cdots,\alpha_{r}\in V$
\end_inset

, 且满足
\end_layout

\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\alpha_{1},\cdots,\alpha_{r}$
\end_inset

 线性无关;
\end_layout

\begin_layout Enumerate
\begin_inset Formula $V$
\end_inset

 中任一向量都可由 
\begin_inset Formula $\alpha_{1},\cdots,\alpha_{r}$
\end_inset

 线性表示.
\end_layout

\end_deeper
\begin_layout Definition
则称
\series bold
向量组 
\begin_inset Formula $\alpha_{1},\cdots,\alpha_{r}$
\end_inset

 为向量空间 
\begin_inset Formula $V$
\end_inset

 的一个基
\series default
, 数 
\begin_inset Formula $r$
\end_inset

 称为
\series bold
向量空间 
\begin_inset Formula $V$
\end_inset

 的维数
\series default
, 记为 
\begin_inset Formula $\mathrm{dim}V=r$
\end_inset

 并称 
\series bold

\begin_inset Formula $V$
\end_inset

 为 
\begin_inset Formula $r$
\end_inset

 维向量空间
\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

\series bold
向量空间的注记
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Remark*
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
vspace{1mm}
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Enumerate
只含零向量的向量空间称为 
\series bold

\begin_inset Formula $0$
\end_inset

 维向量空间
\series default
, 它没有基;
\end_layout

\begin_layout Enumerate
若把向量空间 
\begin_inset Formula $V$
\end_inset

 看作向量组, 则 
\begin_inset Formula $V$
\end_inset

 的基就是向量组的极大无关组, 
\begin_inset Formula $V$
\end_inset

 的维数就是向量组的秩;
\end_layout

\begin_layout Enumerate
若向量组 
\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, 则 
\begin_inset Formula $V$
\end_inset

 可表示为
\begin_inset Formula 
\[
V=\left\{ x=\lambda_{1}\alpha_{1}+\cdots+\lambda_{r}\alpha_{r}\mid\lambda_{1},\lambda_{2},\cdots,\lambda_{r}\in\RR\right\} .
\]

\end_inset

此时, 
\begin_inset Formula $V$
\end_inset

 又称为
\series bold
由基 
\begin_inset Formula $\alpha_{1},\cdots,\alpha_{r}$
\end_inset

 所生成的向量空间
\series default
, 记作 
\begin_inset Formula $V=\mathrm{span}\left\{ \alpha_{1},\cdots,\alpha_{r}\right\} $
\end_inset

.
 故数组 
\begin_inset Formula $\lambda_{1},\cdots,\lambda_{r}$
\end_inset

 称为
\series bold
向量 
\begin_inset Formula $x$
\end_inset

 在基 
\begin_inset Formula $\alpha_{1},\cdots,\alpha_{r}$
\end_inset

 中的坐标
\series default
.
\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
向量空间中的基与坐标
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Remark*
如果在向量空间 
\begin_inset Formula $V$
\end_inset

 中取定一个基 
\begin_inset Formula $a_{1},a_{2},\cdots,a_{r}$
\end_inset

, 那么 
\begin_inset Formula $V$
\end_inset

 中任一向量 
\begin_inset Formula $x$
\end_inset

 可惟一地表示为
\begin_inset Formula 
\[
x=\lambda_{1}a_{1}+\lambda_{2}a_{2}+\cdots+\lambda_{r}a_{r},
\]

\end_inset

数组 
\begin_inset Formula $\lambda_{1},\lambda_{2},\cdots,\lambda_{r}$
\end_inset

 称为
\series bold
向量 
\begin_inset Formula $x$
\end_inset

 在基 
\begin_inset Formula $a_{1},a_{2},\cdots,a_{r}$
\end_inset

 中的坐标
\series default
.
 特别地, 在 
\begin_inset Formula $n$
\end_inset

 维向量空间 
\begin_inset Formula $\RR^{n}$
\end_inset

 中取单位坐标向量组 
\begin_inset Formula $e_{1},e_{2},\cdots,e_{n}$
\end_inset

 为基, 则以 
\begin_inset Formula $x_{1},x_{2},\cdots,x_{n}$
\end_inset

 为分量的向量 
\begin_inset Formula $x$
\end_inset

, 可表示为
\begin_inset Formula 
\[
x=x_{1}e_{1}+x_{2}e_{2}+\cdots+x_{n}e_{n},
\]

\end_inset

可见向量 
\begin_inset Formula $x$
\end_inset

 在基 
\begin_inset Formula $e_{1},e_{2},\cdots,e_{n}$
\end_inset

 中的坐标就是该向量的分量.
 因此 
\begin_inset Formula $e_{1},e_{2},\cdots,e_{n}$
\end_inset

 叫做 
\series bold

\begin_inset Formula $\RR^{n}$
\end_inset

 中的自然基
\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
向量空间中基的例子
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
证明单位向量组
\begin_inset Formula 
\[
e_{1}=(1,0,0,\cdots,0)^{T},\quad e_{2}=(0,1,0,\cdots,0)^{T},\cdots,\quad e_{n}=(0,0,0,\cdots,1)^{T},
\]

\end_inset

是 
\begin_inset Formula $n$
\end_inset

 维向量空间 
\begin_inset Formula $\RR^{n}$
\end_inset

 的一个基.
\end_layout

\begin_deeper
\begin_layout Pause

\end_layout

\end_deeper
\begin_layout Proof

\end_layout

\begin_deeper
\begin_layout Enumerate
易见 
\begin_inset Formula $n$
\end_inset

 维向量组 
\begin_inset Formula $e_{1},e_{2},\cdots,e_{n}$
\end_inset

 线性无关;
\end_layout

\begin_layout Enumerate
对 
\begin_inset Formula $n$
\end_inset

 维向量空间 
\begin_inset Formula $\RR^{n}$
\end_inset

 中的任意一向量 
\begin_inset Formula $\alpha=\left(a_{1},a_{2},\cdots,a_{n}\right)^{T}$
\end_inset

, 有 
\begin_inset Formula $\alpha=a_{1}e_{1}+a_{2}e_{2}+\cdots+a_{n}e_{n}$
\end_inset

, 即 
\begin_inset Formula $\RR^{n}$
\end_inset

 中的任意一向量都可由向量组 
\begin_inset Formula $e_{1},e_{2},\cdots,e_{n}$
\end_inset

 线性表出.
 因此, 向量组 
\begin_inset Formula $e_{1},e_{2},\cdots,e_{n}$
\end_inset

 是 
\begin_inset Formula $n$
\end_inset

 维向量空间 
\begin_inset Formula $\RR^{n}$
\end_inset

 的一个基.
\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
向量空间中的基
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
\begin_inset Argument 1
status open

\begin_layout Plain Layout
\begin_inset Formula $\star\star\star\star\star$
\end_inset


\end_layout

\end_inset

给定向量
\begin_inset Formula 
\[
\alpha_{1}=(-2,4,1)^{T},\ \alpha_{2}=(-1,3,5)^{T},\ \alpha_{3}=(2,-3,1)^{T},\ \beta=(1,1,3)^{T},
\]

\end_inset

试证明: 向量组 
\begin_inset Formula $\alpha_{1},\alpha_{2},\alpha_{3}$
\end_inset

 是三维向量空间 
\begin_inset Formula $\RR^{3}$
\end_inset

 的一个基, 并将向量 
\begin_inset Formula $\beta$
\end_inset

 用这个基线性表示.
\end_layout

\begin_deeper
\begin_layout Pause

\end_layout

\end_deeper
\begin_layout Proof
令矩阵 
\begin_inset Formula $A=\left(\alpha_{1},\alpha_{2},\alpha_{3}\right)$
\end_inset

, 要证明 
\begin_inset Formula $\alpha_{1},\alpha_{2},\alpha_{3}$
\end_inset

 是 
\begin_inset Formula $\RR^{3}$
\end_inset

 的一个基, 只需证明矩阵 
\begin_inset Formula $A$
\end_inset

 的初等行变换过程: 
\begin_inset Formula $A\rightarrow\cdots\rightarrow E$
\end_inset


\begin_inset Foot
status collapsed

\begin_layout Plain Layout
这说明向量组构成是满秩矩阵, 从而线性无关, 向量空间中的任何向量都可以由向量组线性表出.
\end_layout

\end_inset

;
\end_layout

\begin_layout Proof
又设 
\begin_inset Formula $\beta=x_{1}\alpha_{1}+x_{2}\alpha_{2}+x_{3}\alpha_{3}$
\end_inset

 或 
\begin_inset Formula $Ax=\beta$
\end_inset

, 解得 
\begin_inset Formula $x=A^{-1}\beta$
\end_inset

.
 若对 
\begin_inset Formula $\begin{bmatrix}A & \beta\end{bmatrix}$
\end_inset

 进行初等行变换, 当将 
\begin_inset Formula $A$
\end_inset

 化为单位矩阵 
\begin_inset Formula $E$
\end_inset

 时, 可同时将向量 
\begin_inset Formula $\beta$
\end_inset

 化为 
\begin_inset Formula $X=A^{-1}\beta$
\end_inset

.
\begin_inset Formula 
\[
\begin{bmatrix}A & \beta\end{bmatrix}=\begin{bmatrix}-2 & -1 & 2 & 1\\
4 & 3 & -3 & 1\\
1 & 5 & 1 & 3
\end{bmatrix}\xrightarrow{\text{ 行变换 }}\begin{bmatrix}1 & 0 & 0 & 4\\
0 & 1 & 0 & -1\\
0 & 0 & 1 & 4
\end{bmatrix},
\]

\end_inset

可见 
\begin_inset Formula $A\rightarrow E$
\end_inset


\begin_inset Foot
status collapsed

\begin_layout Plain Layout
不要忘了, 
\begin_inset Formula $A\rightarrow E$
\end_inset

 意味着矩阵 
\begin_inset Formula $A$
\end_inset

 与 
\begin_inset Formula $E$
\end_inset

 等价, (回忆矩阵等价的定义)
\end_layout

\end_inset

, 故 
\begin_inset Formula $\alpha_{1},\alpha_{2},\alpha_{3}$
\end_inset

 是 
\begin_inset Formula $\RR^{3}$
\end_inset

 的一个基, 且 
\begin_inset Formula $\beta=4\alpha_{1}-\alpha_{2}+4\alpha_{3}$
\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
向量在给定基下的坐标
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
考虑 
\begin_inset Formula $\RR^{2}$
\end_inset

 的一个基 
\begin_inset Formula $\alpha_{1},\alpha_{2}$
\end_inset

, 其中 
\begin_inset Formula $\alpha_{1}=(1,0)^{T}$
\end_inset

, 
\begin_inset Formula $\alpha_{2}=(1,2)^{T}$
\end_inset

, 若 
\begin_inset Formula $\RR^{2}$
\end_inset

 的一向量 
\begin_inset Formula $x$
\end_inset

 在基 
\begin_inset Formula $\alpha_{1},\alpha_{2}$
\end_inset

 的坐标为 
\begin_inset Formula $(-2,3)^{T}$
\end_inset

, 求 
\begin_inset Formula $x$
\end_inset

.
 
\end_layout

\begin_layout Example
又若 
\begin_inset Formula $y=(4,5)^{T}$
\end_inset

, 试确定向量 
\begin_inset Formula $y$
\end_inset

 在基 
\begin_inset Formula $\alpha_{1},\alpha_{2}$
\end_inset

 的坐标.
\end_layout

\begin_deeper
\begin_layout Pause

\end_layout

\end_deeper
\begin_layout Solution*
根据 
\begin_inset Formula $x$
\end_inset

 在基 
\begin_inset Formula $\alpha_{1},\alpha_{2}$
\end_inset

 下的坐标, 计算 
\begin_inset Formula 
\[
x=(-2)\begin{bmatrix}1\\
0
\end{bmatrix}+3\begin{bmatrix}1\\
2
\end{bmatrix}=\begin{bmatrix}1\\
6
\end{bmatrix}.
\]

\end_inset

设 
\begin_inset Formula $y$
\end_inset

 在基 
\begin_inset Formula $\alpha_{1},\alpha_{2}$
\end_inset

 下的坐标为 
\begin_inset Formula $\left(\lambda_{1},\lambda_{2}\right)^{T}$
\end_inset

, 则
\begin_inset Formula 
\[
\lambda_{1}\begin{bmatrix}1\\
0
\end{bmatrix}+\lambda_{2}\begin{bmatrix}1\\
2
\end{bmatrix}=\begin{bmatrix}4\\
5
\end{bmatrix}\quad\text{ 或 }\begin{bmatrix}1 & 1\\
0 & 2
\end{bmatrix}\begin{bmatrix}\lambda_{1}\\
\lambda_{2}
\end{bmatrix}=\begin{bmatrix}4\\
5
\end{bmatrix}.
\]

\end_inset

该方程可以通过利用等号左边矩阵的逆来求解.
 得到方程的解 
\begin_inset Formula $\lambda_{1}=\frac{3}{2}$
\end_inset

, 
\begin_inset Formula $\lambda_{2}=\frac{5}{2}$
\end_inset

.
 因此 
\begin_inset Formula $y=\frac{3}{2}\alpha_{1}+\frac{5}{2}\alpha_{2}$
\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
向量在给定基下的坐标
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
设 
\begin_inset Formula $v_{1}=(3,6,2)^{T}$
\end_inset

, 
\begin_inset Formula $v_{2}=(-1,0,1)^{T}$
\end_inset

, 
\begin_inset Formula $x=(3,12,7)^{T}$
\end_inset

.
 判断 
\begin_inset Formula $x$
\end_inset

 是否属于由 
\begin_inset Formula $v_{1},v_{2}$
\end_inset

 生成的向量空间.
 如果是, 求出 
\begin_inset Formula $x$
\end_inset

 在 
\begin_inset Formula $v_{1},v_{2}$
\end_inset

 中的坐标.
\end_layout

\begin_layout Solution*
如果 
\begin_inset Formula $x$
\end_inset

 属于由 
\begin_inset Formula $v_{1},v_{2}$
\end_inset

 生成的向量空间, 则下列向量方程是有解的:
\begin_inset Formula 
\[
\lambda_{1}\begin{bmatrix}3\\
6\\
2
\end{bmatrix}+\lambda_{2}\begin{bmatrix}-1\\
0\\
1
\end{bmatrix}=\begin{bmatrix}3\\
12\\
7
\end{bmatrix}.
\]

\end_inset

若满足上式的 
\begin_inset Formula $\lambda_{1},\lambda_{2}$
\end_inset

 存在, 它们应该是 
\begin_inset Formula $x$
\end_inset

 在 
\begin_inset Formula $v_{1},v_{2}$
\end_inset

 中的坐标.
 利用行变换可得
\end_layout

\begin_layout Solution*
\begin_inset Formula 
\[
\begin{bmatrix}3 & -1 & 3\\
6 & 0 & 12\\
2 & 1 & 7
\end{bmatrix}\rightarrow\begin{bmatrix}1 & 0 & 2\\
0 & 1 & 3\\
0 & 0 & 0
\end{bmatrix}.
\]

\end_inset

因此 
\begin_inset Formula $\lambda_{1}=2$
\end_inset

, 
\begin_inset Formula $\lambda_{2}=3$
\end_inset

.
\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
设向量组 
\begin_inset Formula $A:\alpha_{1}=(1,0,1)^{T}$
\end_inset

, 
\begin_inset Formula $\alpha_{2}=(2,2,0)^{T}$
\end_inset

, 
\begin_inset Formula $\alpha_{3}=(2,4-1)^{T}$
\end_inset

, 向量组 
\begin_inset Formula $B:\beta_{1}=(-1,2,4)^{T}$
\end_inset

, 
\begin_inset Formula $\beta_{2}=(2,4,-4)^{T}$
\end_inset

.
\end_layout

\begin_layout Problem
试证明向量组 
\begin_inset Formula $A$
\end_inset

 是三维向量空间 
\begin_inset Formula $\RR^{3}$
\end_inset

 的一个基, 并将向量组 
\begin_inset Formula $B$
\end_inset

 用这个基线性表示.
\end_layout

\end_deeper
\begin_layout Frame

\end_layout

\end_body
\end_document