k阶子式及其余子式
定义1 n阶行列式|A|中任意取定k行、k列(1≤k<n),位于这些行和列的交叉处的\(k^2\)个元素按原来的排法组成的k阶行列式,称为|A|的一个k阶子式. 选取|A|的第\(i_1,i_2,...,i_k\)行\((i_1<i_2<...<i_k)\),第\(j_1,j_2,...,j_k\)列\((j_1<j_2<...<j_k)\),所得的k阶子式记作:
\[\tag{1}
A\begin{pmatrix}
i_1,i_2,...,i_k\\
j_1,j_2,...,j_k
\end{pmatrix}
\]
划去该k阶子式所在的行和列,剩下元素按原来的排法组成的(n-k)阶行列式,称为子式(1)的余子式,它乘以\((-1)^{(i_1+i_2+...+i_k)+(j_1+j_2+...+j_k)}\),则称为子式(1)的代数余子式.
剩下的行、列:令
\[\tag{2}
\begin{aligned}
\{i_1',i_2',...,i_{n-k}'\}&=第\{1,2,...,n\}行去掉第\{i_1,i_2,...,i_k\}行\\
\{j_1',j_2',...,j_{n-k}'\}&=第\{1,2,...,n\}列去掉第\{j_j,j_2,...,j_k\}列
\end{aligned}
\]
其中,\(i_1'<i_2'<...<i_{n-k}',j_1'<j_2'<...<j_{n-k}'\)
取定k行、k列的k阶子式后,剩下的,也是|A|的n-k阶子式:
\[A\begin{pmatrix}
i_1',i_2',...,i_{n-k}'\\
j_1',j_2',...,j_{n-k}'
\end{pmatrix}
\]
按k行(列)展开
Laplace定理
在行列式按行(列)展开中,已知每个元素跟自己的代数余子式乘积之和,等于行列式.
对于k阶子式,是否有同样结论?
这就是Laplace定理.
定理(Laplace) n阶行列式\(A=(a_{ij})\),取定第\(i_1,i_2,...,i_k\)行(其中\(i_1<i_2<...<i_n\)),则|A|等于这k行形成的所有k阶子式与它自己的代数余子式的乘积之和. 即
\[\tag{3}
|A|=\sum_{1\le j_1<j_2<...<j_k\le n}A\begin{pmatrix}
i_1,i_2,...,i_k\\
j_1,j_2,...,j_k
\end{pmatrix}
(-1)^{(i_1+...+i_k)+(j_1+...+j_k)}A\begin{pmatrix}
i_1',i_2',...,i_k'\\
j_1',j_2',...,j_k'
\end{pmatrix}
\]
证明:
\[左边=\sum_{j_1j_2...j_n}a_{1j_1}a_{2j_2}...a_{nj_n}
\]
(3)式左边是n!项的代数和,右边是多少项代数和?
因为每个求和项对应一个\(j_1,j_2,...,j_k\)的排列,一共\(C_n^k\)项. 而每个乘积项,k阶子式(是一个k阶行列式)有k!项,余子式有\((n-k)!\)项,所以k阶子式+余子式有\(k!(n-k)!\)项.
∴(3)式右边共有项数:
\[C_n^kk!(n-k)!=\frac{n!}{k!(n-k)!}k!(n-k)!=n!
\]
即左边、右边项数相同. 如果能证明右边每项都是|A|的一项, 那么右边的n!项和正好为|A|.
右边任取一项:
\[\begin{aligned}
&(-1)^{\tau(i_1i_2...i_k)+\tau(\mu_1\mu_2...\mu_k)}a_{i_1\mu_1}a_{i_2\mu_2}...a_{i_k\mu_k}(-1)^{(i_1+...+i_k)+(j_1+...+j_k)}(-1)^{\tau(i_1'i_2'...i_{n-k}')+\tau(v_1v_2...v_{n-k})}a_{i_1'v_1}a_{i_2'v_2}...a_{i_{n-k}'v_{n-k}}\\
=&(-1)^{\tau(\mu_1\mu_2...\mu_k)}(-1)^{(i_1+...+i_k)+(j_1+...+j_k)}(-1)^{\tau(v_1v_2...v_{n-k})}a_{i_1\mu_1}a_{i_2\mu_2}...a_{i_k\mu_k}a_{i_1'v_1}a_{i_2'v_2}...a_{i_{n-k}'v_{n-k}}
\end{aligned}
\]
其中,\(\mu_1\mu_2...\mu_k\)是\(j_1,j_2,...,k_k\)的一个k元排列,\(v_1v_2...v_{n-k}\)是\(j_1',j_2',...,j_{n-k}'\)的一个n-k元排列
tips:
∵\(i_1<i_2<...<i_k, i_1'<i_2'<...<i_{n-k}\)'
∴逆序数\(\tau(i_1i_2...i_k)=0, \tau(i_1'i_2'...i_{n-k}')=0\)
左边通项可写成:
\[\begin{aligned}
&(-1)^{\tau(i_1i_2...i_ki_1'i_2'...i_{n-k}')+\tau(\mu_1\mu_2...\mu_kv_1...v_{n-k})}a_{i_1\mu_1}a_{i_2\mu_2}...a_{i_k\mu_k}a_{i_1'v_1}...a_{i_{n-k}'v_{n-k}}\\
=&[(-1)^{\tau(i_1i_2...i_k)+\tau(i_1'i_2'...i_k')}(-1)^{i_1+i_2+...+i_k}(-1)^{\frac{k(k+1)}{2}}][(-1)^{\tau(\mu_1\mu_2...\mu_k)+\tau(v_1...v_{n-k})}(-1)^{\mu_1+\mu_2+...+\mu_k}(-1)^{\frac{k(k+1)}{2}}]\\
&\cdot a_{i_1\mu_1}a_{i_2\mu_2}...a_{i_k\mu_k}a_{i_1'v_1}...a_{i_{n-k}'v_{n-k}}(由下面引理)\\
=&[(-1)^{0+0}(-1)^{\sum_{r=1}^ki_r+\frac{k(k+1)}{2}}][(-1)^{\tau(\mu_1\mu_2...\mu_k)+\tau(v_1...v_{n-k})}(-1)^{\sum_{r=1}^k\mu_r+\frac{k(k+1)}{2}}]\\
&\cdot a_{i_1\mu_1}a_{i_2\mu_2}...a_{i_k\mu_k}a_{i_1'v_1}...a_{i_{n-k}'v_{n-k}}\\
=&(-1)^{\sum_{r=1}^ki_r+\frac{k(k+1)}{2}}(-1)^{\tau(\mu_1\mu_2...\mu_k)+\tau(v_1...v_{n-k})}(-1)^{\sum_{r=1}^kj_r+\frac{k(k+1)}{2}}\\
&\cdot a_{i_1\mu_1}a_{i_2\mu_2}...a_{i_k\mu_k}a_{i_1'v_1}...a_{i_{n-k}'v_{n-k}}\\
=&(-1)^{\sum_{r=1}^ki_r+\sum_{r=1}^kj_r+k(k+1)}(-1)^{\tau(\mu_1\mu_2...\mu_k)+\tau(v_1...v_{n-k})} \cdot a_{i_1\mu_1}a_{i_2\mu_2}...a_{i_k\mu_k}a_{i_1'v_1}...a_{i_{n-k}'v_{n-k}}\\
=&(-1)^{(i_1+i_2+...+i_k)+(j_1+j_2+...+j_k)}(-1)^{\tau(\mu_1\mu_2...\mu_k)+\tau(v_1...v_{n-k})} \cdot a_{i_1\mu_1}a_{i_2\mu_2}...a_{i_k\mu_k}a_{i_1'v_1}...a_{i_{n-k}'v_{n-k}}
\end{aligned}
\]
其中,\(i_1i_2...i_ki_1'i_{n-k}'\)和\(\mu_1\mu_2...\mu_kv_1...v_{n-k}\)都是\(12...n\)的一个n元排列;而\(\mu_1\mu_2...\mu_k\)取自\(j_1j_2...j_k\),\(v_1...v_{n-k}\)取自\(j_1'j_2'...j_{n-k}'\).
∴\(\sum_{r=1}^k \mu_r = \sum_{r=1}^kj_r\)
可以看出,左边通项与右边通项是相同的
故得证
引理 设\(c_1c_2...c_kd_1d_2...d_{n-k}\)是由1,2,...,n形成的一个n元排列, 则:
\[(-1)^{\tau(c_1c_2...c_kd_1d_2...d_{n-k})}
=(-1)^{\tau(c_1c_2...c_k)+\tau(d_1d_2...d_{n-k})}\cdot (-1)^{c_1+c_2+...c_k}\cdot (-1)^{\frac{k(k+1)}{2}}
\]
证明:
\[c_1c_2...c_k\xrightarrow{s次对换}a_1a_2...a_k, a_1<a_2<...<a_k
\]
∵\(τ(a_1a_2...a_k)=0\)
∴\(a_1a_2...a_k\)是偶排列
∴\(c_1c_2...c_k\)奇偶性同s,即
\[(-1)^{\tau(c_1c_2...c_k)}=(-1)^s
\]
在上面s次对换下,
\[c_1c_2...c_kd_1d_2...d_{n-k}\xrightarrow{s次对换}a_1a_2...a_kd_1d_2...d_{n-k}
\]
对换改变排列的奇偶性,有,
\[\begin{aligned}
(-1)^{\tau(c_1c_2...c_kd_1d_2...d_{n-k})}&=(-1)^s(-1)^{\tau (a_1a_2...a_kd_1d_2...d_{n-k})}\\
&=(-1)^{\tau(c_1c_2...c_k)}(-1)^{(a_1-1)+(a_2-1)+...+(a_k-1)+\tau(d_1d_2...d_{n-k})}\\
&=(-1)^{\tau(c_1c_2...c_k)+\tau(d_1d_2...d_{n-k})}(-1)^{a_1+a_2+...+a_k}(-1)^{-\frac{k(k+1)}{2}}\\
&=(-1)^{\tau(c_1c_2...c_k)+\tau(d_1d_2...d_{n-k})}(-1)^{a_1+a_2+...+a_k}(-1)^{\frac{k(k+1)}{2}}
\end{aligned}
\]
补充说明:
1)
∵\(a_1a_2...a_kd_1d_2...d_{n-k}\)是数1,2,...,n的一个排列
∴小于\(a_1\)的数必有\(a_1-1\)个,小于\(a_2\)的数必有\(a_2-1\)个,依次类推,小于\(a_k\)的数有\(a_k-1\)个.
∵计算\(a_1,a_2,...,a_k\)逆序数的时候,已经算上了\(d_1,d_2,...,d_{n-k}\)
∴计算\(d_1,d_2,...,d_{n-k}\)逆序数时,需要排除\(a_1,a_2,...,a_k\)
∴\(\tau (a_1a_2...a_kd_1d_2...d_{n-k})=(a_1-1)+(a_2-1)+...+(a_k-1)+\tau(d_1d_2...d_{n-k})\)
2)
\[(-1)^{-\frac{k(k+1)}{2}}=(-1)^{\frac{k(k+1)}{2}}
\iff 1=(-1)^{\frac{k(k+1)}{2}+\frac{k(k+1)}{2}}
\iff 1 = (-1)^{k(k+1)}
\]
不论k为奇数还是偶数,k(k+1)一定是偶数
∴\(1 = (-1)^{k(k+1)}\)成立
Laplace定理的推论
推论 下式成立:
\[\begin{vmatrix}
a_{11} & ... & a_{1k} & 0 & ... & 0\\
... & & ... & ... & & ...\\
a_{k1} & ... & a_{kk} & 0 & ... & 0\\
c_{11} & ... & c_{1k} & b_{11} & ... & b_{1r}\\
... & & ... & ... & & ...\\
c_{r1} & ... & c_{rk} & b_{r1} & ... & b_{rr}
\end{vmatrix}
=\begin{vmatrix}
a_{11} & ... & a_{1k}\\
... & & ...\\
a_{1k} & ... & a_{kk}
\end{vmatrix}
\cdot \begin{vmatrix}
b_{11} & ... & b_{1r}\\
... & & ...\\
b_{r1} & ... & b_{rr}
\end{vmatrix}
\]
证明:
左边行列式按前k行展开. 前k行元素形成的k阶子式中,只有左上角的k阶子式值不为0,如果不是从左上角开始的任意k阶子式必包含0列,从而行列式值为0.
左上角的k阶子式的余子式是右下角的r阶子式.
于是,
\[\begin{aligned}
左边&=\begin{vmatrix}
a_{11} & ... & a_{1k}\\
... & & ...\\
a_{k1} & ... & a_{kk}
\end{vmatrix}
(-1)^{(1+2+...+k)+(1+2+...+k)}
\begin{vmatrix}
b_{11} & ... & b_{1r}\\
... & & ...\\
b_{r1} & ... & b_{rr}
\end{vmatrix}\\
&=\begin{vmatrix}
a_{11} & ... & a_{1k}\\
... & & ...\\
a_{k1} & ... & a_{kk}
\end{vmatrix}
\cdot \begin{vmatrix}
b_{11} & ... & b_{1r}\\
... & & ...\\
b_{r1} & ... & b_{rr}
\end{vmatrix}\\
&=右边
\end{aligned}
\]
简记:
\[\begin{aligned}
A&=\begin{vmatrix}
a_{11} & ... & a_{1k}\\
... & & ...\\
a_{k1} & ... & a_{kk}
\end{vmatrix},
B=\begin{vmatrix}
b_{11} & ... & b_{1r}\\
... & & ...\\
b_{r1} & ... & b_{rr}
\end{vmatrix},\\
C&=\begin{vmatrix}
c_{11} & ... & c_{1k}\\
... & & ...\\
c_{r1} & ... & c_{rk}
\end{vmatrix},
0=\begin{vmatrix}
0 & ... & 0\\
... & & ...\\
0 & ... & 0
\end{vmatrix}
\end{aligned}
\]
推论可写为:
\[\begin{vmatrix}
A & 0\\
C & B
\end{vmatrix}
=|A||B|
\]
小结
1)0行较多时,行列式可分块求值
\[\begin{vmatrix}
A & 0\\
C & B
\end{vmatrix}
=|A||B|
\]
其中,A,C,B都是分块矩阵