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\begin_body
\begin_layout Section
矩阵的秩
\end_layout
\begin_layout Subsection
矩阵的秩
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
矩阵秩的概念
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
矩阵秩的概念:
\end_layout
\begin_layout Itemize
讨论向量组的线性相关性;
\end_layout
\begin_layout Itemize
研究线性方程组解的结构;
\end_layout
\begin_layout Itemize
矩阵经
\series bold
有限次初等行变换
\series default
可以化为行阶梯形矩阵;
\end_layout
\begin_layout Itemize
行阶梯形矩阵所含
\series bold
非零行数
\series default
是 (实质上) 就是矩阵的 ``秩'', 它是一个数;
\end_layout
\begin_layout Itemize
利用初等变换求矩阵的秩;
\end_layout
\begin_layout Itemize
``秩'' 的唯一性尚未证明, 唯一性由行列式的性质决定;
\end_layout
\begin_layout Itemize
先讲行列式再讲矩阵的教材, 利用行列式来定义矩阵的秩.
\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
矩阵的
\begin_inset Formula $k$
\end_inset
阶子式
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
在
\begin_inset Formula $m\times n$
\end_inset
矩阵
\begin_inset Formula $A$
\end_inset
中, 任取
\begin_inset Formula $k$
\end_inset
行
\begin_inset Formula $k$
\end_inset
列 (
\begin_inset Formula $1\leq k\leq m,1\leq k\leq n$
\end_inset
), 位于这些行列交叉处的
\begin_inset Formula $k^{2}$
\end_inset
个元素, 不改变它们在
\begin_inset Formula $A$
\end_inset
中所处的位置次序而得到的
\begin_inset Formula $k$
\end_inset
阶行列式, 称为矩阵
\begin_inset Formula $A$
\end_inset
的
\series bold
\begin_inset Formula $k$
\end_inset
阶子式
\series default
.
\end_layout
\begin_layout Remark*
\begin_inset Formula $m\times n$
\end_inset
矩阵
\begin_inset Formula $A$
\end_inset
的
\begin_inset Formula $k$
\end_inset
阶子式共有
\begin_inset Formula $C_{m}^{k}\cdot C_{n}^{k}$
\end_inset
个.
\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
矩阵的子式
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
设
\begin_inset Formula $A$
\end_inset
为
\begin_inset Formula $m\times n$
\end_inset
矩阵
\end_layout
\begin_layout Itemize
当
\begin_inset Formula $A=O$
\end_inset
时, 它的任何子式都为零.
\end_layout
\begin_layout Itemize
当
\begin_inset Formula $A\neq O$
\end_inset
时, 它至少有一个元素不为零, 即它至少有一个一阶子式不为零.
\end_layout
\begin_layout Itemize
再考察二阶子式, 若
\begin_inset Formula $A$
\end_inset
中有一个二阶子式不为零.
\end_layout
\begin_layout Itemize
往下考察三阶子式, 判断
\begin_inset Formula $A$
\end_inset
是否存在非零的三阶子式;
\end_layout
\begin_layout Itemize
如此进行下去, 最后必达到
\begin_inset Formula $A$
\end_inset
中有
\begin_inset Formula $r$
\end_inset
阶子式不为零, 而再没有比
\begin_inset Formula $r$
\end_inset
更高阶的不为零的子式.
\end_layout
\begin_deeper
\begin_layout Itemize
如果
\begin_inset Formula $A$
\end_inset
的所有
\begin_inset Formula $r$
\end_inset
阶子式均为零, 则矩阵
\begin_inset Formula $A$
\end_inset
的所有
\begin_inset Formula $r+1$
\end_inset
阶子式也都为零;
\end_layout
\begin_layout Itemize
如果
\begin_inset Formula $s>r$
\end_inset
, 矩阵
\begin_inset Formula $A$
\end_inset
的所有
\begin_inset Formula $r$
\end_inset
阶子式均为零, 则矩阵
\begin_inset Formula $A$
\end_inset
的所有
\begin_inset Formula $s$
\end_inset
阶子式也都为零;
\end_layout
\end_deeper
\begin_layout Itemize
这个不为零的子式的最高阶数
\begin_inset Formula $r$
\end_inset
反映了矩阵
\begin_inset Formula $A$
\end_inset
内在的重要特征, 在矩阵的理论与应用中都有重要意义.
\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
矩阵的秩 (rank)
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
设
\begin_inset Formula $A$
\end_inset
为
\begin_inset Formula $m\times n$
\end_inset
矩阵, 如果存在
\begin_inset Formula $A$
\end_inset
的
\begin_inset Formula $r$
\end_inset
阶子式不为零, 而任何
\begin_inset Formula $r+1$
\end_inset
阶子式 (
\color orange
如果存在的话
\color inherit
) 皆为零, 则称数
\begin_inset Formula $r$
\end_inset
为
\series bold
矩阵
\begin_inset Formula $A$
\end_inset
的秩
\series default
, 记为
\begin_inset Formula $r(A)$
\end_inset
(或
\begin_inset Formula $R(A)$
\end_inset
).
并规定
\color orange
零矩阵的秩等于零
\color inherit
.
\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
矩阵秩的性质
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
显然, 矩阵的秩具有下列性质:
\end_layout
\begin_layout Enumerate
若矩阵
\begin_inset Formula $A$
\end_inset
中有某个
\begin_inset Formula $s$
\end_inset
阶子式不为
\begin_inset Formula $0$
\end_inset
, 则
\begin_inset Formula $r(A)\geq s$
\end_inset
;
\end_layout
\begin_layout Enumerate
若
\begin_inset Formula $A$
\end_inset
中所有
\begin_inset Formula $t$
\end_inset
阶子式全为
\begin_inset Formula $0$
\end_inset
, 则
\begin_inset Formula $r(A)<t$
\end_inset
;
\end_layout
\begin_layout Enumerate
若
\begin_inset Formula $A$
\end_inset
为
\begin_inset Formula $m\times n$
\end_inset
矩阵, 则
\begin_inset Formula $0\leq r(A)\leq\min\{m,n\}$
\end_inset
;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $r(A)=r\left(A^{T}\right)$
\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
满秩矩阵/降秩矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
当
\begin_inset Formula $r(A)=\min\{m,n\}$
\end_inset
, 称矩阵
\begin_inset Formula $A$
\end_inset
为
\series bold
满秩矩阵
\series default
.
否则称为
\series bold
降秩矩阵
\series default
.
\end_layout
\begin_layout Definition
当
\begin_inset Formula $r(A)=m$
\end_inset
, 称矩阵
\begin_inset Formula $A$
\end_inset
为行满秩矩阵;
\end_layout
\begin_layout Definition
当
\begin_inset Formula $r(A)=n$
\end_inset
, 称矩阵
\begin_inset Formula $A$
\end_inset
为列满秩矩阵.
\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
根据定义计算矩阵的秩
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求矩阵
\begin_inset Formula $A=\begin{bmatrix}1 & 2 & 3\\
2 & 3 & -5\\
4 & 7 & 1
\end{bmatrix}$
\end_inset
的秩.
\end_layout
\begin_deeper
\begin_layout Pause
\end_layout
\end_deeper
\begin_layout Solution*
在
\begin_inset Formula $A$
\end_inset
中,
\begin_inset Formula $\begin{vmatrix}1 & 3\\
2 & -5
\end{vmatrix}\neq0$
\end_inset
.
\end_layout
\begin_layout Solution*
又由于
\begin_inset Formula $A$
\end_inset
的
\begin_inset Formula $3$
\end_inset
阶子式只有一个
\begin_inset Formula $|A|$
\end_inset
, 且
\begin_inset Formula $|A|=\begin{vmatrix}1 & 2 & 3\\
2 & 3 & -5\\
4 & 7 & 1
\end{vmatrix}=\begin{vmatrix}1 & 2 & 3\\
0 & -1 & -11\\
0 & -1 & -11
\end{vmatrix}=0$
\end_inset
, 所以
\begin_inset Formula $r(A)=2$
\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
根据定义计算矩阵的秩
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求矩阵
\begin_inset Formula $B=\begin{bmatrix}2 & -1 & 0 & 3 & -2\\
0 & 3 & 1 & -2 & 5\\
0 & 0 & 0 & 4 & -3\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}$
\end_inset
的秩.
\end_layout
\begin_deeper
\begin_layout Pause
\end_layout
\end_deeper
\begin_layout Solution*
因
\begin_inset Formula $B$
\end_inset
是一个行阶梯形矩阵, 其非零行只有
\begin_inset Formula $3$
\end_inset
行, 所以
\begin_inset Formula $B$
\end_inset
的所有四阶子式全为零.
\end_layout
\begin_layout Solution*
而
\begin_inset Formula $\begin{vmatrix}2 & -1 & 3\\
0 & 3 & -2\\
0 & 0 & 4
\end{vmatrix}\neq0$
\end_inset
, 因此
\begin_inset Formula $r(B)=3$
\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
求矩阵的秩
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
利用定义计算矩阵的秩, 需要由高阶到低阶考虑矩阵的子式, 当矩阵的行数与列数较高时,
\color red
按定义求秩是非常麻烦的.
\end_layout
\begin_layout Standard
由于行阶梯形矩阵的秩很容易判断, 而任意矩阵都可以经过初等变换化为行阶梯形矩阵.
因而可考虑借助初等变换法来求矩阵的秩.
\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
矩阵秩的求法原理
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Theorem
若
\begin_inset Formula $A\sim B$
\end_inset
, 则
\begin_inset Formula $r(A)=r(B)$
\end_inset
.
\end_layout
\begin_layout Standard
定理证明了若
\begin_inset Formula $A$
\end_inset
经一次初等行变换变为
\begin_inset Formula $B$
\end_inset
, 则
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-5mm}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
r(A)\leq r(B).
\]
\end_inset
\end_layout
\begin_layout Standard
由于
\begin_inset Formula $B$
\end_inset
亦可经一次初等行变换为
\begin_inset Formula $A$
\end_inset
, 故也有
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-5mm}
\end_layout
\end_inset
\begin_inset Formula
\[
r(B)\leq r(A).
\]
\end_inset
\end_layout
\begin_layout Standard
因此
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-5mm}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
r(A)=r(B).
\]
\end_inset
\end_layout
\begin_layout Standard
经一次初等行变换矩阵的秩不变, 即经有限次初等行变换矩阵的秩也不变.
\end_layout
\begin_layout Standard
设
\begin_inset Formula $A$
\end_inset
经初等列变换变为
\begin_inset Formula $B$
\end_inset
, 则
\begin_inset Formula $A^{T}$
\end_inset
经初等行变换变为
\begin_inset Formula $B^{T}$
\end_inset
, 由于
\begin_inset Formula $r\left(A^{T}\right)=r\left(B^{T}\right)$
\end_inset
, 又
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-5mm}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
r(A)=r\left(A^{T}\right),\ r(B)=r\left(B^{T}\right),
\]
\end_inset
\end_layout
\begin_layout Standard
因此
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-5mm}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
r(A)=r(B).
\]
\end_inset
\end_layout
\begin_layout Standard
总之, 若
\begin_inset Formula $A$
\end_inset
经过有限次初等变换变为
\begin_inset Formula $B$
\end_inset
(即
\begin_inset Formula $A\sim B)$
\end_inset
, 则
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-5mm}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
r(A)=r(B).
\]
\end_inset
\end_layout
\begin_layout Standard
根据上述定理, 我们得到利用初等变换求矩阵的秩的方法:
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{1mm}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\size small
\uuline on
\color red
把矩阵用初等行变换变成行阶梯形矩阵, 行阶梯形矩阵中非零行的行数就是该矩阵的秩.
\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
矩阵秩的求法
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求矩阵
\begin_inset Formula
\[
A=\begin{bmatrix}1 & 2 & 3 & 4\\
-1 & -1 & -4 & -2\\
3 & 4 & 11 & 8
\end{bmatrix}
\]
\end_inset
的秩.
\end_layout
\begin_deeper
\begin_layout Pause
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
\begin{aligned}\begin{bmatrix}1 & 2 & 3 & 4\\
-1 & -1 & -4 & -2\\
3 & 4 & 11 & 8
\end{bmatrix} & \xrightarrow[r_{3}-3r_{1}]{r_{2}+r_{1}}\begin{bmatrix}1 & 2 & 3 & 4\\
0 & 1 & -1 & 2\\
0 & -2 & 2 & -4
\end{bmatrix}\\
& \xrightarrow{r_{3}+2r_{2}}\begin{bmatrix}1 & 2 & 3 & 4\\
0 & 1 & -1 & 2\\
0 & 0 & 0 & 0
\end{bmatrix}\xrightarrow{r_{1}-2r_{2}}\begin{bmatrix}1 & 0 & 5 & 0\\
0 & 1 & -1 & 2\\
0 & 0 & 0 & 0
\end{bmatrix}\\
& \xrightarrow{c_{3}-5c_{1}}\begin{bmatrix}1 & 0 & 0 & 0\\
0 & 1 & -1 & 2\\
0 & 0 & 0 & 0
\end{bmatrix}\xrightarrow[c_{4}-2c_{2}]{\substack{c_{3}+c_{2}}
}\begin{bmatrix}1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 0
\end{bmatrix},
\end{aligned}
\]
\end_inset
故
\begin_inset Formula $r(A)=2$
\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
矩阵秩的求法
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求矩阵
\begin_inset Formula $A=\begin{bmatrix}1 & 0 & 0 & 1\\
1 & 2 & 0 & -1\\
3 & -1 & 0 & 4\\
1 & 4 & 5 & 1
\end{bmatrix}$
\end_inset
的秩.
\end_layout
\begin_deeper
\begin_layout Pause
\end_layout
\end_deeper
\begin_layout Solution*
\begin_inset Formula
\begin{align*}
A & =\begin{bmatrix}1 & 0 & 0 & 1\\
1 & 2 & 0 & -1\\
3 & -1 & 0 & 4\\
1 & 4 & 5 & 1
\end{bmatrix}\longrightarrow\begin{bmatrix}1 & 0 & 0 & 1\\
0 & 2 & 0 & -2\\
0 & -1 & 0 & 1\\
0 & 4 & 5 & 0
\end{bmatrix}\longrightarrow\begin{bmatrix}1 & 0 & 0 & 1\\
0 & 1 & 0 & -1\\
0 & 0 & 0 & 4\\
0 & 0 & 5 & 0
\end{bmatrix}\\
& \longrightarrow\begin{bmatrix}1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 0
\end{bmatrix}
\end{align*}
\end_inset
最后一矩阵的秩显然等于
\begin_inset Formula $3$
\end_inset
, 故
\begin_inset Formula $r(A)=3$
\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
矩阵秩的求法
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $A=\begin{bmatrix}1 & -2 & 2 & -1\\
2 & -4 & 8 & 0\\
-2 & 4 & -2 & 3\\
3 & -6 & 0 & -6
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $b=\begin{bmatrix}1\\
2\\
3\\
4
\end{bmatrix}$
\end_inset
, 求矩阵
\begin_inset Formula $A$
\end_inset
及矩阵
\begin_inset Formula $\widetilde{A}=(A,b)$
\end_inset
的秩.
\end_layout
\begin_deeper
\begin_layout Pause
\end_layout
\end_deeper
\begin_layout Solution*
\begin_inset Formula
\begin{align*}
\widetilde{A} & =\begin{bmatrix}1 & -2 & 2 & -1 & 1\\
2 & -4 & 8 & 0 & 2\\
-2 & 4 & -2 & 3 & 3\\
3 & -6 & 0 & -6 & 4
\end{bmatrix}\xrightarrow[r_{4}-3r_{1}]{\substack{r_{2}-2r_{1}\\
r_{3}+2r_{1}
}
}\begin{bmatrix}1 & -2 & 2 & -1 & 1\\
0 & 0 & 4 & 2 & 0\\
0 & 0 & -2 & 1 & 5\\
0 & 0 & -6 & -3 & 1
\end{bmatrix}\\
& \xrightarrow[r_{4}+3r_{2}]{r_{2}\div2}\begin{bmatrix}1 & -2 & 2 & -1 & 1\\
0 & 0 & 2 & 1 & 0\\
0 & 0 & 0 & 0 & 5\\
0 & 0 & 0 & 0 & 1
\end{bmatrix}\xrightarrow[r_{4}-r_{3}]{r_{2}\div5}\begin{bmatrix}1 & -2 & 2 & -1 & 1\\
0 & 0 & 2 & 1 & 0\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0
\end{bmatrix},
\end{align*}
\end_inset
所以
\begin_inset Formula $r(A)=2$
\end_inset
,
\begin_inset Formula $r(\widetilde{A})=3$
\end_inset
.
\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
矩阵秩的求法
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $A=\begin{bmatrix}3 & 2 & 0 & 5 & 0\\
3 & -2 & 3 & 6 & -1\\
2 & 0 & 1 & 5 & -3\\
1 & 6 & -4 & -1 & 4
\end{bmatrix}$
\end_inset
, 求矩阵
\begin_inset Formula $A$
\end_inset
的秩, 并求
\begin_inset Formula $A$
\end_inset
的一个最高阶非零子式.
\end_layout
\begin_deeper
\begin_layout Pause
\end_layout
\end_deeper
\begin_layout Solution*
对
\begin_inset Formula $A$
\end_inset
作初等行变换, 变成行阶梯形矩阵.
\begin_inset Formula
\begin{align*}
A & \xrightarrow{r_{1}\leftrightarrow r_{4}}\begin{bmatrix}1 & 6 & -4 & -1 & 4\\
3 & -2 & 3 & 6 & -1\\
2 & 0 & 1 & 5 & -3\\
3 & 2 & 0 & 5 & 0
\end{bmatrix}\xrightarrow{r_{2}-r_{4}}\begin{bmatrix}1 & 6 & -4 & -1 & 4\\
0 & -4 & 3 & 1 & -1\\
2 & 0 & 1 & 5 & -3\\
3 & 2 & 0 & 5 & 0
\end{bmatrix}\\
& \xrightarrow[r_{4}-3r_{1}]{r_{2}-2r_{1}}\begin{bmatrix}1 & 6 & -4 & -1 & 4\\
0 & -4 & 3 & 1 & -1\\
0 & -12 & 9 & 7 & -11\\
0 & -16 & 12 & 8 & -12
\end{bmatrix}\xrightarrow[r_{4}-3r_{1}]{r_{2}-2r_{1}}\begin{bmatrix}1 & 6 & -4 & -1 & 4\\
0 & -4 & 3 & 1 & -1\\
0 & 0 & 0 & 4 & -8\\
0 & 0 & 0 & 4 & -8
\end{bmatrix}\\
& \xrightarrow{r_{2}-r_{4}}\begin{bmatrix}1 & 6 & -4 & -1 & 4\\
0 & -4 & 3 & 1 & -1\\
0 & 0 & 0 & 4 & -8\\
0 & 0 & 0 & 0 & 0
\end{bmatrix},
\end{align*}
\end_inset
由行阶梯形矩阵有三个非零行知
\begin_inset Formula $r(A)=3$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
再求
\begin_inset Formula $A$
\end_inset
的一个最高阶非零子式.
由
\begin_inset Formula $r(A)=3$
\end_inset
知,
\begin_inset Formula $A$
\end_inset
的最高阶非零子式为三阶.
\begin_inset Formula $A$
\end_inset
的三阶子式共有
\begin_inset Formula $C_{4}^{3}\cdot C_{5}^{3}=40$
\end_inset
个.
\end_layout
\begin_layout Solution*
考察
\begin_inset Formula $A$
\end_inset
的行阶梯形矩阵, 记
\begin_inset Formula $A=\left(a_{1},a_{2},a_{3},a_{4},a_{5}\right)$
\end_inset
, 则矩阵
\begin_inset Formula $B=\left(a_{1},a_{2},a_{4}\right)$
\end_inset
的行阶梯形矩阵为
\begin_inset Formula
\[
\begin{bmatrix}1 & 6 & -1\\
0 & -4 & 1\\
0 & 0 & 4\\
0 & 0 & 0
\end{bmatrix},
\]
\end_inset
\begin_inset Formula $r(B)=3$
\end_inset
, 故
\begin_inset Formula $B$
\end_inset
中必有三阶非零子式, 且矩阵
\begin_inset Formula $B$
\end_inset
共有四个三阶子式可能非零.
\end_layout
\begin_layout Solution*
计算
\begin_inset Formula $B$
\end_inset
中前三行构成的子式
\begin_inset Formula
\[
\begin{vmatrix}3 & 2 & 5\\
3 & -2 & 6\\
2 & 0 & 5
\end{vmatrix}=\begin{vmatrix}3 & 2 & 5\\
6 & 0 & 11\\
2 & 0 & 5
\end{vmatrix}=-2\begin{vmatrix}6 & 11\\
2 & 5
\end{vmatrix}=-16\neq0
\]
\end_inset
则这个子式便是
\begin_inset Formula $A$
\end_inset
的一个最高阶非零子式.
\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
矩阵的秩不等式
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $A$
\end_inset
为
\begin_inset Formula $n$
\end_inset
阶非奇异矩阵,
\begin_inset Formula $B$
\end_inset
为
\begin_inset Formula $n\times m$
\end_inset
矩阵.
试证:
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $B$
\end_inset
之积的秩等于
\begin_inset Formula $B$
\end_inset
的秩, 即
\begin_inset Formula $r(AB)=r(B)$
\end_inset
.
\end_layout
\begin_layout Proof
因为
\begin_inset Formula $A$
\end_inset
非奇异, 故可表示成若干初等矩阵之积,
\begin_inset Formula $A=P_{1}P_{2}\cdots P_{s}$
\end_inset
, 其中
\begin_inset Formula $P_{i}$
\end_inset
(
\begin_inset Formula $i=1,2,\cdots,s$
\end_inset
) 皆为初等矩阵.
\begin_inset Formula $AB=P_{1}P_{2}\cdots P_{s}B$
\end_inset
, 即
\begin_inset Formula $AB$
\end_inset
是
\begin_inset Formula $B$
\end_inset
经
\begin_inset Formula $s$
\end_inset
次初等行变换后得出的.
因而
\begin_inset Formula $r(AB)=r(B)$
\end_inset
.
证毕.
\end_layout
\begin_layout Remark*
由矩阵的秩及满秩矩阵的定义, 显然, 若一个
\begin_inset Formula $n$
\end_inset
阶矩阵
\begin_inset Formula $A$
\end_inset
是满秩的, 则
\begin_inset Formula $|A|\neq0$
\end_inset
.
因而
\begin_inset Formula $A$
\end_inset
非奇异; 反之亦然.
\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
矩阵的秩
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
\begin_inset Formula $A=\begin{bmatrix}1 & -1 & 1 & 2\\
3 & \lambda & -1 & 2\\
5 & 3 & \mu & 6
\end{bmatrix}$
\end_inset
, 已知
\begin_inset Formula $r(A)=2$
\end_inset
, 求
\begin_inset Formula $\lambda$
\end_inset
与
\begin_inset Formula $\mu$
\end_inset
的值.
\end_layout
\begin_deeper
\begin_layout Pause
\end_layout
\end_deeper
\begin_layout Solution*
\begin_inset Formula
\[
A\xrightarrow[r_{4}-5r_{1}]{r_{2}-3r_{1}}\begin{bmatrix}1 & -1 & 1 & 2\\
0 & \lambda+3 & -4 & -4\\
0 & 8 & \mu-5 & -4
\end{bmatrix}\xrightarrow{r_{3}-r_{2}}\begin{bmatrix}1 & -1 & 1 & 2\\
0 & \lambda+3 & -4 & -4\\
0 & 5-\lambda & \mu-1 & 0
\end{bmatrix}
\]
\end_inset
因
\begin_inset Formula $r(A)=2$
\end_inset
, 故
\begin_inset Formula
\[
\begin{cases}
5-\lambda=0\\
\mu-1=0
\end{cases}\Longrightarrow\begin{cases}
\lambda=5\\
\mu=1
\end{cases}.
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Subsection
作业
\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
作业
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Problem
已知
\begin_inset Formula $A=\begin{bmatrix}1 & 3 & -2 & 2\\
0 & 2 & -1 & 3\\
-2 & 0 & 1 & 5
\end{bmatrix}$
\end_inset
, 求该矩阵的秩.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
求下列矩阵的秩:
\end_layout
\begin_layout Problem
(1).
\begin_inset Formula $\begin{bmatrix}2 & -1 & 3 & -2 & 4\\
4 & -2 & 5 & 1 & 7\\
2 & -1 & 1 & 8 & 2
\end{bmatrix}$
\end_inset
; (2).
\begin_inset Formula $\begin{bmatrix}1 & 3 & 5 & -1\\
2 & -1 & -3 & 4\\
5 & 1 & -1 & 7\\
7 & 7 & 9 & 1
\end{bmatrix}$
\end_inset
;
\end_layout
\begin_layout Problem
(3).
\begin_inset Formula $\begin{bmatrix}3 & -1 & 3 & 2 & 5\\
5 & -3 & 2 & 3 & 4\\
1 & -3 & -5 & 0 & -7\\
7 & -5 & 1 & 4 & 1
\end{bmatrix}$
\end_inset
; (4).
\begin_inset Formula $\begin{bmatrix}4 & 3 & -5 & 2 & 3\\
8 & 6 & -7 & 4 & 2\\
4 & 3 & -8 & 2 & 7\\
4 & 3 & 1 & 2 & -5\\
8 & 6 & -1 & 4 & -6
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
求
\begin_inset Formula $\lambda$
\end_inset
的值, 使下面的矩阵
\begin_inset Formula $A$
\end_inset
有最小的秩:
\begin_inset Formula $A=\begin{bmatrix}3 & 1 & 1 & 4\\
\lambda & 4 & 10 & 1\\
1 & 7 & 17 & 3\\
2 & 2 & 5 & 3
\end{bmatrix}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Frame
\end_layout
\end_body
\end_document
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