2.6 拉普拉斯定理

定义 1：
\(n\)级矩阵\(A\)中任意取定\(k\)行，\(k\)列（\(1 \leq k < n\)），位于这些行和列交叉处的\(k^2\)个元素按原来的排法组成的\(k\)级矩阵的行列式称为\(A\)的一个\(k\)阶子式。取定\(A\)的\(i_1,i_2,\dots ,i_k\)行（\(i_1<i_2<\dots <i_k\)）及\(j_1,j_2,\dots ,j_k\)列（\(j_1<j_2<\dots <j_k\)），所得的\(k\)阶子式记为\(A\tbinom{i_1i_2\dots i_k}{j_1j_2\dots j_k}\) \((1)\)。

划去这个\(k\)阶子式，剩下的元素按原来的排法组成\((n-k)\)级矩阵，称为\((1)\)的余子式。也是\(A\)的一个\((n-k)\)阶子式，令\(\{i_1'i_2'\dots i_{n-k}'\} = \{1,2,\dots ,n\}\) \ \(\{i_1i_2\dots i_k\}\)、\(\{j_1'j_2'\dots j_{n-k}'\} = \{1,2,\dots ,n\}\) \ \(\{j_1j_2\dots j_k\}\)且\(i_1'<i_2'<\dots <i_{n-k}'\)，\(j_1'<j_2'<\dots <j_{n-k}'\)，记余子式为\(A\tbinom{i_1'i_2'\dots i_{n-k}'}{j_1'j_2'\dots j_{n-k}'}\) \((2)\)。
\((-1)^{(i_1 + i_2 + \dots + i_k) + (j_1 + j_2 + \dots + j_k)}A\tbinom{i_1'i_2'\dots i_{n-k}'}{j_1'j_2'\dots j_{n-k}'}\)为\((1)\)的代数余子式。

定理 1：
（Laplace定理）在\(n\)级矩阵\(A = (a_{ij})\)，取定第\(i_1,i_2,\dots ,i_k\)行（\(i_1<i_2<\dots <i_k\)），则这\(k\)阶行元素形成的所有\(k\)阶子式与其代数余子式之和等于\(|A|\)，即

\[|A| = \sum_{1\leq j_1<j_2<\dots <j_k \leq n} A\tbinom{i_1i_2\dots i_k}{j_1j_2\dots j_k}(-1)^{(i_1 + i_2 + \dots + i_k) + (j_1 + j_2 + \dots + j_k)}A\tbinom{i_1'i_2'\dots i_{n-k}'}{j_1'j_2'\dots j_{n-k}'} \]

证明：

\[\begin{aligned} |A| &= \sum_{u_1\dots u_kv_1\dots v_{n-k}}(-1)^{\tau(i_1i_2\dots i_ki_1'i_2'\dots i_{n-k}')+\tau(u_1\dots u_kv_1\dots v_{n-k})}a_{i_1u_1}a_{i_2u_2}\dots a_{i_ku_k}a_{i_1'v_1}a_{i_2'v_2}\dots a_{i_{n-k}'v_{n-k}} \\ &= \sum_{u_1\dots u_kv_1\dots v_{n-k}}(-1)^{(i_1-1)+(i_2-2)+\dots +(i_k-k)+\tau(u_1\dots u_kv_1\dots v_{n-k})}a_{i_1u_1}a_{i_2u_2}\dots a_{i_ku_k}a_{i_1'v_1}a_{i_2'v_2}\dots a_{i_{n-k}'v_{n-k}} \\ &= \sum_{1\leq j_1<j_2<\dots <j_k \leq n}\sum_{u_1\dots u_k}\sum_{v_1\dots v_{n-k}}(-1)^{(i_1+i_2+\dots +i_k)-\frac{k(k+1)}{2}}(-1)^{\tau(u_1\dots u_kv_1\dots v_{n-k})}a_{i_1u_1}a_{i_2u_2}\dots a_{i_ku_k}a_{i_1'v_1}a_{i_2'v_2}\dots a_{i_{n-k}'v_{n-k}} \end{aligned} \]

\[这一步三个连加号分别指的是选定k列、对k元排列求和、对n-k元排列求和 \\ 考察(-1)^{\tau(u_1\dots u_kv_1\dots v_{n-k})}，\\ u_1\dots u_kv_1\dots v_{n-k}经过s次对换变成j_1j_2\dots j_kv_1\dots v_{n-k} \\ 因此(-1)^{\tau(u_1\dots u_kv_1\dots v_{n-k})} = (-1)^s(-1)^{\tau(j_1j_2\dots j_kv_1\dots v_{n-k})} \\ 其中j_1j_2\dots j_k的逆序数为0，则s = \tau(u_1\dots u_k)，故 \]

\[\begin{aligned} (-1)^{\tau(u_1\dots u_kv_1\dots v_{n-k})}&=(-1)^{\tau(u_1\dots u_k)}(-1)^{\tau(j_1j_2\dots j_kv_1\dots v_{n-k})} \\ &=(-1)^{\tau(u_1\dots u_k)}(-1)^{(j_1-1)+(j_2-2)+\dots +(j_k-k)}(-1)^{\tau(v_1\dots v_{n-k})} \end{aligned} \]

\[\begin{aligned} |A|&= \sum_{1\leq j_1<j_2<\dots <j_k \leq n}\sum_{u_1\dots u_k}\sum_{v_1\dots v_{n-k}}(-1)^{(i_1+i_2+\dots +i_k)-\frac{k(k+1)}{2}}(-1)^{\tau(u_1\dots u_k)}(-1)^{(j_1-1)+(j_2-2)+\dots +(j_k-k)}(-1)^{\tau(v_1\dots v_{n-k})}a_{i_1u_1}a_{i_2u_2}\dots a_{i_ku_k}a_{i_1'v_1}a_{i_2'v_2}\dots a_{i_{n-k}'v_{n-k}} \\ &= \sum_{1\leq j_1<j_2<\dots <j_k \leq n}(-1)^{(i_1+i_2+\dots +i_k)+(j_1+j_2+\dots +j_k)}\sum_{u_1\dots u_k}(-1)^{\tau(u_1\dots u_k)}a_{i_1u_1}a_{i_2u_2}\dots a_{i_ku_k}(\sum_{v_1\dots v_{n-k}}(-1)^{\tau(v_1\dots v_{n-k})}a_{i_1'v_1}a_{i_2'v_2}\dots a_{i_{n-k}'v_{n-k}}) \\ &= \sum_{1\leq j_1<j_2<\dots <j_k \leq n} A\tbinom{i_1i_2\dots i_k}{j_1j_2\dots j_k}(-1)^{(i_1 + i_2 + \dots + i_k) + (j_1 + j_2 + \dots + j_k)}A\tbinom{i_1'i_2'\dots i_{n-k}'}{j_1'j_2'\dots j_{n-k}'} \end{aligned} \]

推论：

\[\begin{vmatrix} a_{11} & \cdots & a_{1k} & 0 & \cdots & 0 \\ \vdots & & \vdots & \vdots & & \vdots \\ a_{k1} & \cdots & a_{kk} & 0 & \cdots & 0 \\ c_{11} & \cdots & c_{1k} & b_{11} & \cdots & b_{1r} \\ \vdots & & \vdots & \vdots & & \vdots \\ c_{r1} & \cdots & c_{rk} & b_{r1} & \cdots & b_{rr} \end{vmatrix} = \begin{vmatrix} a_{11} & \cdots & a_{1k} \\ \vdots & & \vdots \\ a_{k1} & \cdots & a_{kk} \end{vmatrix} \cdot \begin{vmatrix} b_{11} & \cdots & b_{1r} \\ \vdots & & \vdots \\ b_{r1} & \cdots & b_{rr} \end{vmatrix} \cdot (-1)^{\frac{k(k+1)}{2}} \cdot (-1)^{\frac{k(k+1)}{2}} \]

即：

\[\begin{vmatrix} A & 0 \\ C & B \end{vmatrix} = |A| \cdot |B| \]

标签：dots,tau,拉普拉斯,定理,vmatrix,cdots,kv,2.6,vdots
来源： https://www.cnblogs.com/miraclepbc/p/14443187.html