矩阵的三角分解
作者:互联网
设实矩阵
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∣Aii∣=0(i=1,2,⋯,n),则可以对
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\boldsymbol{A}=\boldsymbol{LU}
A=LU其中,
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\boldsymbol{L}
L为主对角元素全为1的下三角矩阵(即单位下三角矩阵),
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\boldsymbol{U}
U为上三角矩阵,称这种分解为矩阵的三角分解,也称为
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\boldsymbol{LU}
LU分解。
下面从高斯消去法的基本原理来说明这种分解的具体过程。在高斯消去法中,只是主对角线上的元素不是1。因此,在没有归一化的消去法中,实际上就是通过一系列的初等行变换依次将主对角线以下的元素消成0,而这一系列的初等行变换就相当于用一系列的初等矩阵左乘矩阵
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首先,为了使矩阵
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\boldsymbol{L}_1=\left[\begin{array}{ccccc}1 &&&\\ -l_{21}&1 &&&\\ -l_{31}&0&1&&\\ \vdots& \vdots &\vdots&\ddots&\\ -l_{n1}&0&0 &\cdots &1\end{array}\right]
L1=⎣⎢⎢⎢⎢⎢⎡1−l21−l31⋮−ln110⋮01⋮0⋱⋯1⎦⎥⎥⎥⎥⎥⎤左乘矩阵
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\begin{aligned}\boldsymbol{L}_1\boldsymbol{A}&=\left[\begin{array}{ccccc}1 &&&&\\ -l_{21}&1 &&&\\ -l_{31}&0&1&&\\ \vdots& \vdots &\vdots&\ddots&\\ -l_{n1}&0&0 &\cdots &1\end{array}\right]\left[\begin{array}{ccccc}a_{11} &a_{12}&a_{13}& \cdots&a_{1n}\\ a_{21} &a_{22}&a_{23}& \cdots&a_{2n}\\ a_{31} &a_{32}&a_{33}& \cdots&a_{3n}\\ \vdots& \vdots &\vdots&\vdots&\vdots\\ a_{n1} &a_{n2}&a_{n3}& \cdots&a_{nn}\end{array}\right]\\&=\left[\begin{array}{ccccc}a_{11} &a_{12}&a_{13}& \cdots&a_{1n}\\ 0 &a_{22}^{(1)}&a_{23}^{(1)}& \cdots&a_{2n}^{(1)}\\ 0 &a_{32}^{(1)}&a_{33}^{(1)}& \cdots&a_{3n}^{(1)}\\ \vdots& \vdots &\vdots&\vdots&\vdots\\ 0 &a_{n2}^{(1)}&a_{n3}^{(1)}& \cdots&a_{nn}^{(1)}\end{array}\right]=\boldsymbol{A}_1\end{aligned}
L1A=⎣⎢⎢⎢⎢⎢⎡1−l21−l31⋮−ln110⋮01⋮0⋱⋯1⎦⎥⎥⎥⎥⎥⎤⎣⎢⎢⎢⎢⎢⎡a11a21a31⋮an1a12a22a32⋮an2a13a23a33⋮an3⋯⋯⋯⋮⋯a1na2na3n⋮ann⎦⎥⎥⎥⎥⎥⎤=⎣⎢⎢⎢⎢⎢⎢⎡a1100⋮0a12a22(1)a32(1)⋮an2(1)a13a23(1)a33(1)⋮an3(1)⋯⋯⋯⋮⋯a1na2n(1)a3n(1)⋮ann(1)⎦⎥⎥⎥⎥⎥⎥⎤=A1为了使上式运算成立,初等矩阵
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L1中的元素
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l_{ij}
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a_{i1}-l_{i1}a_{11}=0,\quad i=2,3,\cdots,n
ai1−li1a11=0,i=2,3,⋯,n即
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l_{i1}=a_{i1}/a_{11},\quad i=2,3,\cdots,n
li1=ai1/a11,i=2,3,⋯,n而在新矩阵中,除第一行外,其余各元素相应满足下列关系:
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a_{ij}^{(1)}=a_{ij}-l_{i1}a_{1j},\quad i=2,3,\cdots,n; j=2,3,\cdots,n
aij(1)=aij−li1a1j,i=2,3,⋯,n;j=2,3,⋯,n按以上方法继续往下做。假设已经做了
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k−1步,原矩阵
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\begin{aligned}\boldsymbol{A}_{k-1}&=\boldsymbol{L}_{k-1}\cdots \boldsymbol{L}_{2}\boldsymbol{L}_{1}\boldsymbol{A}\\&=\left[\begin{array}{cccccccc}a_{11} &a_{12}&a_{13}& \cdots&a_{1k}& a_{1,k+1}& \cdots & a_{1n}\\ 0 &a_{22}^{(1)}&a_{23}^{(1)}& \cdots&a_{2k}^{(1)}& a_{2,k+1}^{(1)}& \cdots & a_{2n}^{(1)}\\ 0 & 0 &a_{33}^{(2)}& \cdots&a_{3k}^{(2)}& a_{3,k+1}^{(2)}&\cdots& a_{3n}^{(2)}\\ \vdots& \vdots &\vdots&\vdots&\vdots&\vdots & \vdots &\vdots\\ 0 & 0 & 0 & \cdots &a_{kk}^{(k-1)}& a_{k,k+1}^{(k-1)}& \cdots& a_{kn}^{(k-1)} \\ \vdots& \vdots & \vdots &\cdots & \vdots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 &\cdots& a^{(k-1)}_{n,k}&a_{n,k+1}^{(k-1)}&\cdots &a^{(k-1)}_{nn}\end{array}\right]\end{aligned}
Ak−1=Lk−1⋯L2L1A=⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡a1100⋮0⋮0a12a22(1)0⋮0⋮0a13a23(1)a33(2)⋮0⋮0⋯⋯⋯⋮⋯⋯⋯a1ka2k(1)a3k(2)⋮akk(k−1)⋮an,k(k−1)a1,k+1a2,k+1(1)a3,k+1(2)⋮ak,k+1(k−1)⋮an,k+1(k−1)⋯⋯⋯⋮⋯⋮⋯a1na2n(1)a3n(2)⋮akn(k−1)⋮ann(k−1)⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤现在进行第
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\begin{aligned}\boldsymbol{A}_{k}&=\boldsymbol{L}_{k}\boldsymbol{A}_{k-1}\\&=\left[\begin{array}{cccccccc}a_{11} &a_{12}&a_{13}& \cdots&a_{1k}& a_{1,k+1}& \cdots & a_{1n}\\ 0 &a_{22}^{(1)}&a_{23}^{(1)}& \cdots&a_{2k}^{(1)}& a_{2,k+1}^{(1)}& \cdots & a_{2n}^{(1)}\\ 0 & 0 &a_{33}^{(2)}& \cdots&a_{3k}^{(2)}& a_{3,k+1}^{(2)}&\cdots& a_{3n}^{(2)}\\ \vdots& \vdots &\vdots&\vdots&\vdots&\vdots & \vdots &\vdots\\ 0 & 0 & 0 & \cdots &a_{kk}^{(k-1)}& a_{k,k+1}^{(k-1)}& \cdots& a_{kn}^{(k-1)} \\ \vdots& \vdots & \vdots &\cdots & \vdots & \vdots & \vdots & \vdots\\\ 0 & 0 & 0 &\cdots& 0 &a_{n,k+1}^{(k-1)}&\cdots &a^{(k-1)}_{nn}\end{array}\right]\end{aligned}
Ak=LkAk−1=⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡a1100⋮0⋮ 0a12a22(1)0⋮0⋮0a13a23(1)a33(2)⋮0⋮0⋯⋯⋯⋮⋯⋯⋯a1ka2k(1)a3k(2)⋮akk(k−1)⋮0a1,k+1a2,k+1(1)a3,k+1(2)⋮ak,k+1(k−1)⋮an,k+1(k−1)⋯⋯⋯⋮⋯⋮⋯a1na2n(1)a3n(2)⋮akn(k−1)⋮ann(k−1)⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤其中
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\boldsymbol{L}_k=\left[ \begin{array}{cccccccc}1&&&&&&&\\0&1&&&&&&\\0&0&1&&&&&\\\vdots&\vdots &\vdots &\ddots &&&&\\ 0 & 0 & 0 &\cdots&1&&&\\ 0&0&0&\cdots&-l_{k+1,k}&1&&\\ \vdots&\vdots&\vdots&&\vdots&\vdots&\ddots&\\0&0&0&\cdots&-l_{nk}&0&\cdots&1\end{array}\right]
Lk=⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡100⋮00⋮010⋮00⋮01⋮00⋮0⋱⋯⋯⋯1−lk+1,k⋮−lnk1⋮0⋱⋯1⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤同样,为了使上式运算从成立,初等矩阵
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Lk中元素
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a^{(k)}_{ij}=a_{kj}^{(k-1)}-l_{ik}a_{kj}^{(k-1)},\quad i=k+1,\cdots,n;j=k+1,\cdots,n
aij(k)=akj(k−1)−likakj(k−1),i=k+1,⋯,n;j=k+1,⋯,n如此继续下去,则有
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\boldsymbol{A}_{n-1}=\boldsymbol{L}_{n-1}\cdots \boldsymbol{L}_k\cdots \boldsymbol{L}_2 \boldsymbol{L}_1 \boldsymbol{A}
An−1=Ln−1⋯Lk⋯L2L1A其中,
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U=An−1,则有
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\boldsymbol{U}=\boldsymbol{L}_{n-1}\cdots\boldsymbol{L}_{k}\cdots \boldsymbol{L}_2 \boldsymbol{L}_1 \boldsymbol{A}
U=Ln−1⋯Lk⋯L2L1A最后得到
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\boldsymbol{A}=\boldsymbol{LU}
A=LU由此可以看出,矩阵
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\boldsymbol{A}
A在一定条件下(各阶主子式异于零)可以分解成单位下三角矩阵
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\boldsymbol{A}
A和上三角矩阵
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\boldsymbol{U}
U的乘积。
标签:11,boldsymbol,三角,矩阵,cdots,分解,array,vdots 来源: https://blog.csdn.net/qq_38406029/article/details/120734592