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(邱维声)高等代数课程笔记:行列式按k行(列)展开

(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 kkkkkk 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 kkkkkk 列,则剩下元素按照原来的排法可以形成一个 n−kn-knk 阶行列式,称为 (1)(1)(1) 的余子式。

\\quad

{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,,ink}={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,,jnk}={1,2,,n}\\{j1,j2,,jk}

并且约定:i1′<i2′<⋯<in−k′i_{1}'<i_{2}'<\\cdots <i_{n-k}'i1<i2<<inkj1′<j2′<⋯<jn−k′j_{1}'<j_{2}'<\\cdots <j_{n-k}'j1<j2<<jnk. 则 (1)(1)(1) 的余子式也是 AAA 的一个 n−kn-knk 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,,inkj1,j2,,jnk)(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,,inkj1,j2,,jnk)

为子式 (1)(1)(1) 的代数余子式。


\\quad 类比 上一节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=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 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}u1u2uk 是其一个 kkk排列
  • 对于剩下的第 j1′<j2′<⋯<jn−k′j_{1}'<j_{2}'<\\cdots<j_{n-k}'j1<j2<<jnk 列,令 v1v2⋯vn−kv_{1}v_{2}\\cdots v_{n-k}v1v2vnk 是其一个 n−kn-knk 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,,vnk}: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=u1u2ukv1v2vnk(1)τ(i1i2iki1i2ink)+τ(u1u2ukv1v2vnk)ai1u1ai2u2aikukai1v1ai2v2ainkvnk=1j1<j2<<jnnu1u2ukv1v2vnk(1)τ(i1i2iki1i2ink)+τ(u1u2ukv1v2vnk)ai1u1ai2u2aikukai1v1ai2v2ainkvnk

\\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}'i1i2iki1i2ink,由于比 i1i_{1}i1 小的数有 i1−1i_{1}-1i11 个,比 i2i_{2}i2 小的数有 i2−1i_{2}-1i21 个,再除去 i1i_{1}i1,有 i2−2i_{2}-2i22 个 ……,比 iki_{k}ik 小的数有 ik−ki_{k}-kikk 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)i1i2iki1i2ink=(1)i1+i2++ik2k(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}u1u2ukv1v2vnk 化为排列 j1j2⋯jkv1v2⋯vn−kj_{1}j_{2}\\cdots j_{k} v_{1}v_{2}\\cdots v_{n-k}j1j2jkv1v2vnk. 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(j1jk)=0,从而

(−1)τ(u1u2⋯uk)=(−1)S(-1)^{\\tau(u_{1}u_{2}\\cdots u_{k})} = (-1)^{S} (1)τ(u1u2uk)=(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(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≤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=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)

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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=a11ak1c11ct1a1kakkc1kctk00b11bt100b1tbtt

解:

\\quadLaplace 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=a11ak1a1kakkb11bt1b1tbtt

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参考

  • 邱维声. 高等代数课程.