(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 nnn 级矩阵 A=(aij)A=(a_{ij})A=(aij),任取其 kkk 行:第 i1,i2,⋯,iki_{1},i_{2},\\cdots,i_{k}i1,i2,⋯,ik 行,其中 i1<i2<⋯<iki_{1}<i_{2}<\\cdots<i_{k}i1<i2<⋯<ik;再任取其 kkk 列:第 j1,j2,⋯,jkj_{1},j_{2},\\cdots,j_{k}j1,j2,⋯,jk 列,其中 j1<j2<⋯<jkj_{1}<j_{2}<\\cdots<j_{k}j1<j2<⋯<jk. 则这 kkk 行 kkk 列元素按原来的排法可以构成一个 kkk 阶行列式,称为 AAA 的一个 kkk 阶子式,记作:
A(i1,i2,⋯,ikj1,j2,⋯,jk)(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 kkk 行 kkk 列,则剩下元素按照原来的排法可以形成一个 n−kn-kn−k 阶行列式,称为 (1)(1)(1) 的余子式。
\\quad 令
{i1′,i2′,⋯,in−k′}={1,2,⋯,n}\\{i1,i2,⋯,ik}\\{i_{1}',i_{2}',\\cdots,i_{n-k}'\\} = \\{1,2,\\cdots,n\\} \\backslash \\{i_{1},i_{2},\\cdots,i_{k}\\} {i1′,i2′,⋯,in−k′}={1,2,⋯,n}\\{i1,i2,⋯,ik}
{j1′,j2′,⋯,jn−k′}={1,2,⋯,n}\\{j1,j2,⋯,jk}\\{j_{1}',j_{2}',\\cdots,j_{n-k}'\\} = \\{1,2,\\cdots,n\\} \\backslash \\{j_{1},j_{2},\\cdots,j_{k}\\} {j1′,j2′,⋯,jn−k′}={1,2,⋯,n}\\{j1,j2,⋯,jk}
并且约定:i1′<i2′<⋯<in−k′i_{1}'<i_{2}'<\\cdots <i_{n-k}'i1′<i2′<⋯<in−k′,j1′<j2′<⋯<jn−k′j_{1}'<j_{2}'<\\cdots <j_{n-k}'j1′<j2′<⋯<jn−k′. 则 (1)(1)(1) 的余子式也是 AAA 的一个 n−kn-kn−k 阶子式,记作:
A(i1′,i2′,⋯,in−k′j1′,j2′,⋯,jn−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′,⋯,in−k′j1′,j2′,⋯,jn−k′)(2)
\\quad 另外,称
(−1)(i1+i2+⋯+ik)+(j1+j2+⋯+jk)⋅A(i1′,i2′,⋯,in−k′j1′,j2′,⋯,jn−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′,⋯,in−k′j1′,j2′,⋯,jn−k′)
为子式 (1)(1)(1) 的代数余子式。
\\quad 类比 上一节
中 nnn 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=(aij)A=(a_{ij})A=(aij) 是一个 nnn 级矩阵,取其第 i1,i2,⋯,iki_{1},i_{2},\\cdots,i_{k}i1,i2,⋯,ik 行(i1<i2<⋯<iki_{1}<i_{2}<\\cdots <i_{k}i1<i2<⋯<ik),则 ∣A∣|A|∣A∣ 等于这 kkk 行形成的所有 kkk 阶子式与其代数余子式的乘积之和。即
∣A∣=∑1≤j1<j2<⋯<jk≤nA(i1,i2,⋯,ikj1,j2,⋯,jk)⋅(−1)(i1+i2+⋯+ik)+(j1+j2+⋯+jk)⋅A(i1′,i2′,⋯,in−k′j1′,j2′,⋯,jn−k′)|A| = \\sum_{1\\le j_{1}<j_{2}<\\cdots<j_{k}\\le n} A\\left(\\begin{matrix}i_{1},i_{2},\\cdots,i_{k}\\\\ j_{1},j_{2},\\cdots,j_{k}\\end{matrix}\\right) \\cdot (-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) ∣A∣=1≤j1<j2<⋯<jk≤n∑A(i1,i2,⋯,ikj1,j2,⋯,jk)⋅(−1)(i1+i2+⋯+ik)+(j1+j2+⋯+jk)⋅A(i1′,i2′,⋯,in−k′j1′,j2′,⋯,jn−k′)
证明:此处是在课上邱老师给出的新证明。
\\quad 对于指定的第 i1,i2,⋯,iki_{1},i_{2},\\cdots,i_{k}i1,i2,⋯,ik 行,按照 j1<j2<⋯<jkj_{1}<j_{2}<\\cdots <j_{k}j1<j2<⋯<jk 的要求任取列,从而得到 kkk 阶子式。对 kkk 阶子式可以进行“分组”:
- 对于任取的第 j1,j2,⋯,jkj_{1},j_{2},\\cdots,j_{k}j1,j2,⋯,jk 列,令 u1u2⋯uku_{1}u_{2}\\cdots u_{k}u1u2⋯uk 是其一个 kkk 元排列;
- 对于剩下的第 j1′<j2′<⋯<jn−k′j_{1}'<j_{2}'<\\cdots<j_{n-k}'j1′<j2′<⋯<jn−k′ 列,令 v1v2⋯vn−kv_{1}v_{2}\\cdots v_{n-k}v1v2⋯vn−k 是其一个 n−kn-kn−k 元排列。
- 这样一来,∀u∈{u1,u2,⋯,uk},∀v∈{v1,v2,⋯,vn−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<v∀ u∈{u1,u2,⋯,uk},∀ v∈{v1,v2,⋯,vn−k}:u<v.
\\quad 于是,按照行列式的定义,
∣A∣=∑u1u2⋯ukv1v2⋯vn−k(−1)τ(i1i2⋯iki1′i2′⋯in−k′)+τ(u1u2⋯ukv1v2⋯vn−k)ai1u1ai2u2⋯aikukai1′v1ai2′v2⋯ain−k′vn−k=∑1≤j1<j2<⋯<jn≤n∑u1u2⋯uk∑v1v2⋯vn−k(−1)τ(i1i2⋯iki1′i2′⋯in−k′)+τ(u1u2⋯ukv1v2⋯vn−k)ai1u1ai2u2⋯aikukai1′v1ai2′v2⋯ain−k′vn−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} <j_{2}<\\cdots<j_{n}\\le n}\\sum_{u_{1}u_{2}\\cdots u_{k}}\\sum_{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}} \\end{aligned} ∣A∣=u1u2⋯ukv1v2⋯vn−k∑(−1)τ(i1i2⋯iki1′i2′⋯in−k′)+τ(u1u2⋯ukv1v2⋯vn−k)ai1u1ai2u2⋯aikukai1′v1ai2′v2⋯ain−k′vn−k=1≤j1<j2<⋯<jn≤n∑u1u2⋯uk∑v1v2⋯vn−k∑(−1)τ(i1i2⋯iki1′i2′⋯in−k′)+τ(u1u2⋯ukv1v2⋯vn−k)ai1u1ai2u2⋯aikukai1′v1ai2′v2⋯ain−k′vn−k
\\quad 要注意到,
a. 对于 nnn 元排列 i1i2⋯iki1′i2′⋯in−k′i_{1}i_{2}\\cdots i_{k} i_{1}'i_{2}'\\cdots i_{n-k}'i1i2⋯iki1′i2′⋯in−k′,由于比 i1i_{1}i1 小的数有 i1−1i_{1}-1i1−1 个,比 i2i_{2}i2 小的数有 i2−1i_{2}-1i2−1 个,再除去 i1i_{1}i1,有 i2−2i_{2}-2i2−2 个 ……,比 iki_{k}ik 小的数有 ik−ki_{k}-kik−k 个,于是
(−1)i1i2⋯iki1′i2′⋯in−k′=(−1)i1+i2+⋯+ik−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)i1i2⋯iki1′i2′⋯in−k′=(−1)i1+i2+⋯+ik−2k(k+1)
b. 假设经过 SSS 次对换,排列 u1u2⋯ukv1v2⋯vn−ku_{1}u_{2}\\cdots u_{k}v_{1}v_{2}\\cdots v_{n-k}u1u2⋯ukv1v2⋯vn−k 化为排列 j1j2⋯jkv1v2⋯vn−kj_{1}j_{2}\\cdots j_{k} v_{1}v_{2}\\cdots v_{n-k}j1j2⋯jkv1v2⋯vn−k. 因为 j1<j2<⋯<jkj_{1}<j_{2}<\\cdots<j_{k}j1<j2<⋯<jk,因此 τ(j1⋯jk)=0\\tau(j_{1}\\cdots j_{k})=0τ(j1⋯jk)=0,从而
(−1)τ(u1u2⋯uk)=(−1)S(-1)^{\\tau(u_{1}u_{2}\\cdots u_{k})} = (-1)^{S} (−1)τ(u1u2⋯uk)=(−1)S
\\quad 于是
(−1)τ(u1u2⋯ukv1v2⋯vn−k)=(−1)S⋅(−1)τ(j1j2⋯jkv1v2⋯vn−k)=(−1)τ(u1u2⋯uk)⋅(−1)τ(j1j2⋯jkv1v2⋯vn−k)=(−1)j1+j2+⋯+jk−k(k+1)2⋅(−1)τ(u1u2⋯uk)+τ(v1v2⋯vn−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(u1u2⋯ukv1v2⋯vn−k)=(−1)S⋅(−1)τ(j1j2⋯jkv1v2⋯vn−k)=(−1)τ(u1u2⋯uk)⋅(−1)τ(j1j2⋯jkv1v2⋯vn−k)=(−1)j1+j2+⋯+jk−2k(k+1)⋅(−1)τ(u1u2⋯uk)+τ(v1v2⋯vn−k)
\\quad 从而,
∣A∣=∑1≤j1<j2<⋯<jn≤n∑u1u2⋯uk∑v1v2⋯vn−k(−1)τ(i1i2⋯iki1′i2′⋯in−k′)+τ(u1u2⋯ukv1v2⋯vn−k)ai1u1ai2u2⋯aikukai1′v1ai2′v2⋯ain−k′vn−k=∑1≤j1<j2<⋯<jn≤n(−1)τ(i1+i2+⋯+ik)+τ(j1+j2+⋯+jk)∑u1u2⋯uk(−1)τ(u1u2⋯uk)ai1u1ai2u2⋯aikuk∑v1v2⋯vn−k(−1)τ(v1v2⋯vn−k)ai1′v1ai2′v2⋯ain−k′vn−k=∑1≤j1<j2<⋯<jk≤n(−1)(i1+i2+⋯+ik)+(j1+j2+⋯+jk)⋅A(i1,i2,⋯,ikj1,j2,⋯,jk)⋅A(i1′,i2′,⋯,in−k′j1′,j2′,⋯,jn−k′)\\begin{aligned} |A| &= \\sum_{1\\le j_{1} <j_{2}<\\cdots<j_{n}\\le n}\\sum_{u_{1}u_{2}\\cdots u_{k}}\\sum_{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}<j_{2}<\\cdots<j_{n}\\le n}(-1)^{\\tau(i_{1}+i_{2} +\\cdots + i_{k}) + \\tau(j_{1} + j_{2} + \\cdots + j_{k})} \\sum_{u_{1}u_{2}\\cdots u_{k}} (-1)^{\\tau(u_{1}u_{2}\\cdots u_{k})} a_{i_{1}u_{1}}a_{i_{2}u_{2}}\\cdots a_{i_{k}u_{k}} \\sum_{v_{1}v_{2}\\cdots v_{n-k}}(-1)^{\\tau(v_{1}v_{2}\\cdots v_{n-k})}a_{i_{1}'v_{1}}a_{i_{2}'v_{2}}\\cdots a_{i_{n-k}'v_{n-k}}\\\\ &= \\sum_{1\\le j_{1}<j_{2}<\\cdots<j_{k}\\le n} (-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_{k}\\\\ j_{1},j_{2},\\cdots,j_{k}\\end{matrix}\\right) \\cdot A\\left(\\begin{matrix} i_{1}',i_{2}',\\cdots,i_{n-k}'\\\\ j_{1}',j_{2}',\\cdots,j_{n-k}'\\end{matrix}\\right) \\end{aligned} ∣A∣=1≤j1<j2<⋯<jn≤n∑u1u2⋯uk∑v1v2⋯vn−k∑(−1)τ(i1i2⋯iki1′i2′⋯in−k′)+τ(u1u2⋯ukv1v2⋯vn−k)ai1u1ai2u2⋯aikukai1′v1ai2′v2⋯ain−k′vn−k=1≤j1<j2<⋯<jn≤n∑(−1)τ(i1+i2+⋯+ik)+τ(j1+j2+⋯+jk)u1u2⋯uk∑(−1)τ(u1u2⋯uk)ai1u1ai2u2⋯aikukv1v2⋯vn−k∑(−1)τ(v1v2⋯vn−k)ai1′v1ai2′v2⋯ain−k′vn−k=1≤j1<j2<⋯<jk≤n∑(−1)(i1+i2+⋯+ik)+(j1+j2+⋯+jk)⋅A(i1,i2,⋯,ikj1,j2,⋯,jk)⋅A(i1′,i2′,⋯,in−k′j1′,j2′,⋯,jn−k′)
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\\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∣=∣a11⋯a1k0⋯0⋮⋮⋮⋮ak1⋯akk0⋯0c11⋯c1kb11⋯b1t⋮⋮⋮⋮ct1⋯ctkbt1⋯btt∣|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∣=a11⋮ak1c11⋮ct1⋯⋯⋯⋯a1k⋮akkc1k⋮ctk0⋮0b11⋮bt1⋯⋯⋯⋯0⋮0b1t⋮btt
解:
\\quad 由 Laplace 定理
,按前 kkk 行展开,则有:
∣A∣=∣a11⋯a1k⋮⋮ak1⋯akk∣⋅∣b11⋯b1t⋮⋮bt1⋯btt∣|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∣=a11⋮ak1⋯⋯a1k⋮akk⋅b11⋮bt1⋯⋯b1t⋮btt
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参考:
- 邱维声. 高等代数课程.