(邱维声)高等代数课程笔记:行列式按k行(列)展开

行列式按 k 行(列)展开

k 阶子式与其余子式:对于 n n n 级矩阵 A = ( a i j ) A=(a_{ij}) A=(aij),任取其 k k k 行:第 i 1 , i 2 , ⋯   , i k i_{1},i_{2},\cdots,i_{k} i1,i2,,ik 行,其中 i 1 < i 2 < ⋯ < i k i_{1}i1<i2<<ik;再任取其 k k k 列:第 j 1 , j 2 , ⋯   , j k j_{1},j_{2},\cdots,j_{k} j1,j2,,jk 列,其中 j 1 < j 2 < ⋯ < j k j_{1}j1<j2<<jk. 则这 k k k k k k 列元素按原来的排法可以构成一个 k k k 阶行列式,称为 A A A 的一个 k k k 阶子式,记作:

A ( i 1 , i 2 , ⋯   , i k j 1 , j 2 , ⋯   , j k ) (1) A\left(\begin{matrix} i_{1},i_{2},\cdots,i_{k}\\ j_{1},j_{2},\cdots,j_{k} \end{matrix}\right) \tag{1} A(i1,i2,,ikj1,j2,,jk)(1)

\quad 若划去上述 k k k k k k 列,则剩下元素按照原来的排法可以形成一个 n − k n-k nk 阶行列式,称为 ( 1 ) (1) (1) 的余子式。

\quad

{ i 1 ′ , i 2 ′ , ⋯   , i n − k ′ } = { 1 , 2 , ⋯   , n } \ { i 1 , i 2 , ⋯   , i k } \{i_{1}',i_{2}',\cdots,i_{n-k}'\} = \{1,2,\cdots,n\} \backslash \{i_{1},i_{2},\cdots,i_{k}\} {i1,i2,,ink}={1,2,,n}\{i1,i2,,ik}

{ j 1 ′ , j 2 ′ , ⋯   , j n − k ′ } = { 1 , 2 , ⋯   , n } \ { j 1 , j 2 , ⋯   , j k } \{j_{1}',j_{2}',\cdots,j_{n-k}'\} = \{1,2,\cdots,n\} \backslash \{j_{1},j_{2},\cdots,j_{k}\} {j1,j2,,jnk}={1,2,,n}\{j1,j2,,jk}

并且约定: i 1 ′ < i 2 ′ < ⋯ < i n − k ′ i_{1}'i1<i2<<ink j 1 ′ < j 2 ′ < ⋯ < j n − k ′ j_{1}'j1<j2<<jnk. 则 ( 1 ) (1) (1) 的余子式也是 A A A 的一个 n − k n-k nk 阶子式,记作:

A ( i 1 ′ , i 2 ′ , ⋯   , i n − k ′ j 1 ′ , j 2 ′ , ⋯   , j n − k ′ ) (2) A\left(\begin{matrix} i_{1}',i_{2}',\cdots,i_{n-k}'\\ j_{1}',j_{2}',\cdots,j_{n-k}' \end{matrix}\right)\tag{2} A(i1,i2,,inkj1,j2,,jnk)(2)

\quad 另外,称

( − 1 ) ( i 1 + i 2 + ⋯ + i k ) + ( j 1 + j 2 + ⋯ + j k ) ⋅ A ( i 1 ′ , i 2 ′ , ⋯   , i n − k ′ j 1 ′ , j 2 ′ , ⋯   , j n − k ′ ) (-1)^{(i_{1}+i_{2}+\cdots +i_{k}) + (j_{1} + j_{2} +\cdots + j_{k})} \cdot A\left(\begin{matrix} i_{1}',i_{2}',\cdots,i_{n-k}'\\ j_{1}',j_{2}',\cdots,j_{n-k}'\end{matrix}\right) (1)(i1+i2++ik)+(j1+j2++jk)A(i1,i2,,inkj1,j2,,jnk)

为子式 ( 1 ) (1) (1) 的代数余子式。


\quad 类比 上一节 n n n 阶行列式按一行(列)展开的理论,容易猜测 定理 1 成立。

定理 1. Laplace 定理:设 A = ( a i j ) A=(a_{ij}) A=(aij) 是一个 n n n 级矩阵,取其第 i 1 , i 2 , ⋯   , i k i_{1},i_{2},\cdots,i_{k} i1,i2,,ik 行( i 1 < i 2 < ⋯ < i k i_{1}i1<i2<<ik),则 ∣ A ∣ |A| A 等于这 k k k 行形成的所有 k k k 阶子式与其代数余子式的乘积之和。即

∣ A ∣ = ∑ 1 ≤ j 1 < j 2 < ⋯ < j k ≤ n A ( i 1 , i 2 , ⋯   , i k j 1 , j 2 , ⋯   , j k ) ⋅ ( − 1 ) ( i 1 + i 2 + ⋯ + i k ) + ( j 1 + j 2 + ⋯ + j k ) ⋅ A ( i 1 ′ , i 2 ′ , ⋯   , i n − k ′ j 1 ′ , j 2 ′ , ⋯   , j n − k ′ ) |A| = \sum_{1\le j_{1}A=1j1<j2<<jknA(i1,i2,,ikj1,j2,,jk)(1)(i1+i2++ik)+(j1+j2++jk)A(i1,i2,,inkj1,j2,,jnk)

证明:此处是在课上邱老师给出的新证明。

\quad 对于指定的第 i 1 , i 2 , ⋯   , i k i_{1},i_{2},\cdots,i_{k} i1,i2,,ik 行,按照 j 1 < j 2 < ⋯ < j k j_{1}j1<j2<<jk 的要求任取列,从而得到 k k k 阶子式。对 k k k 阶子式可以进行“分组”:

  • 对于任取的第 j 1 , j 2 , ⋯   , j k j_{1},j_{2},\cdots,j_{k} j1,j2,,jk 列,令 u 1 u 2 ⋯ u k u_{1}u_{2}\cdots u_{k} u1u2uk 是其一个 k k k 元排列;
  • 对于剩下的第 j 1 ′ < j 2 ′ < ⋯ < j n − k ′ j_{1}'j1<j2<<jnk 列,令 v 1 v 2 ⋯ v n − k v_{1}v_{2}\cdots v_{n-k} v1v2vnk 是其一个 n − k n-k nk 元排列。
  • 这样一来, ∀   u ∈ { u 1 , u 2 , ⋯   , u k } , ∀   v ∈ { v 1 , v 2 , ⋯   , v n − k } : u < v \forall ~ u \in \{u_{1},u_{2},\cdots,u_{k}\},\forall ~ v \in \{v_{1},v_{2},\cdots,v_{n-k}\}:u u{u1,u2,,uk}, v{v1,v2,,vnk}:u<v.

\quad 于是,按照行列式的定义,

∣ A ∣ = ∑ u 1 u 2 ⋯ u k v 1 v 2 ⋯ v n − k ( − 1 ) τ ( i 1 i 2 ⋯ i k i 1 ′ i 2 ′ ⋯ i n − k ′ ) + τ ( u 1 u 2 ⋯ u k v 1 v 2 ⋯ v n − k ) a i 1 u 1 a i 2 u 2 ⋯ a i k u k a i 1 ′ v 1 a i 2 ′ v 2 ⋯ a i n − k ′ v n − k = ∑ 1 ≤ j 1 < j 2 < ⋯ < j n ≤ n ∑ u 1 u 2 ⋯ u k ∑ v 1 v 2 ⋯ v n − k ( − 1 ) τ ( i 1 i 2 ⋯ i k i 1 ′ i 2 ′ ⋯ i n − k ′ ) + τ ( u 1 u 2 ⋯ u k v 1 v 2 ⋯ v n − k ) a i 1 u 1 a i 2 u 2 ⋯ a i k u k a i 1 ′ v 1 a i 2 ′ v 2 ⋯ a i n − k ′ v n − k \begin{aligned} |A| &= \sum_{u_{1}u_{2}\cdots u_{k}v_{1}v_{2}\cdots v_{n-k}}(-1)^{\tau(i_{1}i_{2}\cdots i_{k}i_{1}'i_{2}'\cdots i_{n-k}')+\tau(u_{1}u_{2}\cdots u_{k}v_{1}v_{2}\cdots v_{n-k})} a_{i_{1}u_{1}}a_{i_{2}u_{2}}\cdots a_{i_{k}u_{k}}a_{i_{1}'v_{1}}a_{i_{2}'v_{2}}\cdots a_{i_{n-k}'v_{n-k}} \\ &= \sum_{1\le j_{1} A=u1u2ukv1v2vnk(1)τ(i1i2iki1i2ink)+τ(u1u2ukv1v2vnk)ai1u1ai2u2aikukai1v1ai2v2ainkvnk=1j1<j2<<jnnu1u2ukv1v2vnk(1)τ(i1i2iki1i2ink)+τ(u1u2ukv1v2vnk)ai1u1ai2u2aikukai1v1ai2v2ainkvnk

\quad 要注意到,

a. 对于 n n n 元排列 i 1 i 2 ⋯ i k i 1 ′ i 2 ′ ⋯ i n − k ′ i_{1}i_{2}\cdots i_{k} i_{1}'i_{2}'\cdots i_{n-k}' i1i2iki1i2ink,由于比 i 1 i_{1} i1 小的数有 i 1 − 1 i_{1}-1 i11 个,比 i 2 i_{2} i2 小的数有 i 2 − 1 i_{2}-1 i21 个,再除去 i 1 i_{1} i1,有 i 2 − 2 i_{2}-2 i22 个 ……,比 i k i_{k} ik 小的数有 i k − k i_{k}-k ikk 个,于是

( − 1 ) i 1 i 2 ⋯ i k i 1 ′ i 2 ′ ⋯ i n − k ′ = ( − 1 ) i 1 + i 2 + ⋯ + i k − k ( k + 1 ) 2 (-1)^{i_{1}i_{2}\cdots i_{k} i_{1}'i_{2}'\cdots i_{n-k}'} = (-1)^{i_{1}+i_{2}+\cdots +i_{k} - \frac{k(k+1)}{2}} (1)i1i2iki1i2ink=(1)i1+i2++ik2k(k+1)

b. 假设经过 S S S 次对换,排列 u 1 u 2 ⋯ u k v 1 v 2 ⋯ v n − k u_{1}u_{2}\cdots u_{k}v_{1}v_{2}\cdots v_{n-k} u1u2ukv1v2vnk 化为排列 j 1 j 2 ⋯ j k v 1 v 2 ⋯ v n − k j_{1}j_{2}\cdots j_{k} v_{1}v_{2}\cdots v_{n-k} j1j2jkv1v2vnk. 因为 j 1 < j 2 < ⋯ < j k j_{1}j1<j2<<jk,因此 τ ( j 1 ⋯ j k ) = 0 \tau(j_{1}\cdots j_{k})=0 τ(j1jk)=0,从而

( − 1 ) τ ( u 1 u 2 ⋯ u k ) = ( − 1 ) S (-1)^{\tau(u_{1}u_{2}\cdots u_{k})} = (-1)^{S} (1)τ(u1u2uk)=(1)S

\quad 于是

( − 1 ) τ ( u 1 u 2 ⋯ u k v 1 v 2 ⋯ v n − k ) = ( − 1 ) S ⋅ ( − 1 ) τ ( j 1 j 2 ⋯ j k v 1 v 2 ⋯ v n − k ) = ( − 1 ) τ ( u 1 u 2 ⋯ u k ) ⋅ ( − 1 ) τ ( j 1 j 2 ⋯ j k v 1 v 2 ⋯ v n − k ) = ( − 1 ) j 1 + j 2 + ⋯ + j k − k ( k + 1 ) 2 ⋅ ( − 1 ) τ ( u 1 u 2 ⋯ u k ) + τ ( v 1 v 2 ⋯ v n − k ) \begin{aligned} (-1)^{\tau(u_{1}u_{2}\cdots u_{k}v_{1}v_{2}\cdots v_{n-k})} &= (-1)^{S} \cdot (-1)^{\tau(j_{1}j_{2}\cdots j_{k}v_{1}v_{2}\cdots v_{n-k})}\\ &= (-1)^{\tau(u_{1}u_{2}\cdots u_{k})}\cdot (-1)^{\tau(j_{1}j_{2}\cdots j_{k}v_{1}v_{2}\cdots v_{n-k})}\\ &= (-1)^{j_{1}+j_{2}+\cdots + j_{k} - \frac{k(k+1)}{2}} \cdot (-1)^{\tau(u_{1}u_{2}\cdots u_{k})+ \tau(v_{1}v_{2}\cdots v_{n-k})} \end{aligned} (1)τ(u1u2ukv1v2vnk)=(1)S(1)τ(j1j2jkv1v2vnk)=(1)τ(u1u2uk)(1)τ(j1j2jkv1v2vnk)=(1)j1+j2++jk2k(k+1)(1)τ(u1u2uk)+τ(v1v2vnk)

\quad 从而,

∣ A ∣ = ∑ 1 ≤ j 1 < j 2 < ⋯ < j n ≤ n ∑ u 1 u 2 ⋯ u k ∑ v 1 v 2 ⋯ v n − k ( − 1 ) τ ( i 1 i 2 ⋯ i k i 1 ′ i 2 ′ ⋯ i n − k ′ ) + τ ( u 1 u 2 ⋯ u k v 1 v 2 ⋯ v n − k ) a i 1 u 1 a i 2 u 2 ⋯ a i k u k a i 1 ′ v 1 a i 2 ′ v 2 ⋯ a i n − k ′ v n − k = ∑ 1 ≤ j 1 < j 2 < ⋯ < j n ≤ n ( − 1 ) τ ( i 1 + i 2 + ⋯ + i k ) + τ ( j 1 + j 2 + ⋯ + j k ) ∑ u 1 u 2 ⋯ u k ( − 1 ) τ ( u 1 u 2 ⋯ u k ) a i 1 u 1 a i 2 u 2 ⋯ a i k u k ∑ v 1 v 2 ⋯ v n − k ( − 1 ) τ ( v 1 v 2 ⋯ v n − k ) a i 1 ′ v 1 a i 2 ′ v 2 ⋯ a i n − k ′ v n − k = ∑ 1 ≤ j 1 < j 2 < ⋯ < j k ≤ n ( − 1 ) ( i 1 + i 2 + ⋯ + i k ) + ( j 1 + j 2 + ⋯ + j k ) ⋅ A ( i 1 , i 2 , ⋯   , i k j 1 , j 2 , ⋯   , j k ) ⋅ A ( i 1 ′ , i 2 ′ , ⋯   , i n − k ′ j 1 ′ , j 2 ′ , ⋯   , j n − k ′ ) \begin{aligned} |A| &= \sum_{1\le j_{1} A=1j1<j2<<jnnu1u2ukv1v2vnk(1)τ(i1i2iki1i2ink)+τ(u1u2ukv1v2vnk)ai1u1ai2u2aikukai1v1ai2v2ainkvnk=1j1<j2<<jnn(1)τ(i1+i2++ik)+τ(j1+j2++jk)u1u2uk(1)τ(u1u2uk)ai1u1ai2u2aikukv1v2vnk(1)τ(v1v2vnk)ai1v1ai2v2ainkvnk=1j1<j2<<jkn(1)(i1+i2++ik)+(j1+j2++jk)A(i1,i2,,ikj1,j2,,jk)A(i1,i2,,inkj1,j2,,jnk)

#

\quad 由行列式的行与列的对称性,不难将结论推广到列的情形。


\quad 最后,介绍 Laplace 定理 的一个重要应用。

例题 1:求解下面的行列式。

∣ A ∣ = ∣ a 11 ⋯ a 1 k 0 ⋯ 0 ⋮ ⋮ ⋮ ⋮ a k 1 ⋯ a k k 0 ⋯ 0 c 11 ⋯ c 1 k b 11 ⋯ b 1 t ⋮ ⋮ ⋮ ⋮ c t 1 ⋯ c t k b t 1 ⋯ b t t ∣ |A| = \left|\begin{matrix} 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_{1t}\\ \vdots & &\vdots &\vdots & &\vdots\\ c_{t1} &\cdots &c_{tk} &b_{t1} &\cdots &b_{tt} \end{matrix}\right| A= a11ak1c11ct1a1kakkc1kctk00b11bt100b1tbtt

解:

\quad Laplace 定理,按前 k k k 行展开,则有:

∣ A ∣ = ∣ a 11 ⋯ a 1 k ⋮ ⋮ a k 1 ⋯ a k k ∣ ⋅ ∣ b 11 ⋯ b 1 t ⋮ ⋮ b t 1 ⋯ b t t ∣ |A| = \left|\begin{matrix}a_{11} &\cdots &a_{1k}\\ \vdots & &\vdots \\ a_{k1} &\cdots &a_{kk}\end{matrix}\right| \cdot \left|\begin{matrix}b_{11} &\cdots &b_{1t}\\ \vdots & &\vdots \\ b_{t1} &\cdots &b_{tt}\end{matrix}\right| A= a11ak1a1kakk b11bt1b1tbtt

#


参考

  • 邱维声. 高等代数课程.

你可能感兴趣的:(高等代数,线性代数,高等代数,数学)