LA@三角行列式
三角行列式🎈
概念
-
主对角线(main diagonal)
-
In linear algebra, the main diagonal (sometimes :principal diagonal, primary diagonal, leading diagonal, major diagonal, or good diagonal) of a matrix A is the list of entries a i j a_{ij} aijwhere i = j i=j i=j.
-
All off-diagonal elements are zero in a diagonal matrix.
-
-
副对角线(antidiagonal)
- The antidiagonal (sometimes counter diagonal, secondary diagonal, trailing diagonal, minor diagonal, off diagonal, or bad diagonal) of an order N square matrix B(N阶方阵B) is the collection of entries b i j b_{ij} bij,such that i + j = N + 1 i+j=N+1 i+j=N+1 for all 1 ⩽ i , j ⩽ N 1\leqslant{i,j}\leqslant{N} 1⩽i,j⩽N.
- That is ,it runs from the top right corner to the bottom left corner.
主对角线三角行列式👺
-
- 记为 A T D A_{TD} ATD(triangular determinant)
-
主对角线三角行列式的值等于主对角线元素的乘积,(包括上三角和下三角行列式)
-
∣ A T D ∣ = ∏ i = 1 n a i j |A_{TD}|=\prod\limits_{i=1}^{n}a_{ij} ∣ATD∣=i=1∏naij
-
∣ a 11 a 12 ⋯ a 1 n a 22 ⋯ a 2 n ⋱ ⋮ a n n ∣ = ∣ a 11 a 21 a 22 ⋮ ⋮ ⋱ a n 1 a n 2 ⋯ a n n ∣ = ( − 1 ) τ ( 12 ⋯ n ) a 11 a 22 ⋯ a n n = a 11 a 22 ⋯ a n n = ∏ i = 1 n a i i \\ \begin{vmatrix} a_{11}& a_{12}& \cdots & a_{1n} \\ & a_{22}& \cdots & a_{2n} \\ & & \ddots & \vdots \\ & & & a_{nn} \end{vmatrix} =\begin{vmatrix} a_{11}& & & \\ a_{21}& a_{22}& & \\ \vdots & \vdots & \ddots & \\ a_{n1}& a_{n2}& \cdots & a_{nn} \end{vmatrix} \\=(-1)^{\tau{(12\cdots{n})}}a_{11}a_{22}\cdots{a_{nn}} =a_{11}a_{22}\cdots{a_{nn}} \\ =\prod_{i=1}^{n}a_{ii} a11a12a22⋯⋯⋱a1na2n⋮ann = a11a21⋮an1a22⋮an2⋱⋯ann =(−1)τ(12⋯n)a11a22⋯ann=a11a22⋯ann=i=1∏naii
解法1:
- 由行列式公式 ∣ A ∣ = ∑ k = 1 n ! s ( p k ) θ ( p k ) |A|=\sum_{k=1}^{n!}s(p_k)\theta(p_k) ∣A∣=∑k=1n!s(pk)θ(pk)
- 其中:
p
k
p_k
pk是
1
,
⋯
,
n
1,\cdots,n
1,⋯,n的某个排列;
s
(
p
k
)
s(p_k)
s(pk)排列
p
k
p_k
pk的逆序数;
θ
(
p
k
)
\theta(p_k)
θ(pk)是来自不同行不同列的n个元素的乘积
- 容易观察到,三角行列式的第一列只有第一个元素可能非0,其余全为0
- 如果
p
k
=
j
1
j
2
,
⋯
j
n
p_k=j_1j_2,\cdots{j_n}
pk=j1j2,⋯jn中
j
1
≠
1
j_1\neq{1}
j1=1那么
j
2
j
3
,
⋯
,
j
n
j_2j_3,\cdots,j_n
j2j3,⋯,jn中有且只有一个会等于1,而这些元素分别来自第2至第n行,因此一定有某个元素
a
i
,
j
i
=
0
,
i
≠
1
\large{a_{i,j_i}}=0,i\neq{1}
ai,ji=0,i=1
- 这种情况下, θ ( p k ) \theta(p_k) θ(pk)一定会取0
- 只有当
j
1
=
1
j_1=1
j1=1,
θ
=
θ
(
p
k
)
≠
0
\theta=\theta(p_k)\neq{0}
θ=θ(pk)=0才可能发生(
j
i
=
i
j_i=i
ji=i,
i
=
1
,
2
,
⋯
,
n
i=1,2,\cdots,n
i=1,2,⋯,n)
- 这种情况下, a 2 , j 2 a_{2,j_2} a2,j2也必须是 a 22 a_{22} a22才可能使 θ ≠ 0 \theta\neq{0} θ=0,否则 j 3 , ⋯ , j n j_3,\cdots,j_n j3,⋯,jn中一定会有一个元素取2,使得矩阵A中某个元素 a i , 2 = 0 , i ≠ 1 , 2 a_{i,2}=0,i\neq{1,2} ai,2=0,i=1,2,导致 θ = 0 \theta=0 θ=0
- 类似的,只有排列 p δ = 12 ⋯ n p_{\delta}=12\cdots n pδ=12⋯n,使得 θ = θ ( p δ ) = a 11 a 22 ⋯ a n n = ∏ i = 1 n a i i \theta=\theta(p_\delta)=a_{11}a_{22}\cdots{a_{nn}}=\prod_{i=1}^{n}a_{ii} θ=θ(pδ)=a11a22⋯ann=∏i=1naii可能不为0( p k = 1 , 2 , ⋯ , n p_k=1,2,\cdots,{n} pk=1,2,⋯,n)
- 容易球得 τ ( p δ ) = 0 \tau(p_{\delta})=0 τ(pδ)=0, τ \tau τ表示求逆序数
- 从而 ∣ A ∣ = ( − 1 ) 0 ∏ i = 1 n a i i = ∏ i = 1 n a i i |A|=(-1)^{0}\prod_{i=1}^{n}a_{ii}=\prod_{i=1}^{n}a_{ii} ∣A∣=(−1)0∏i=1naii=∏i=1naii
解法2:
- 利用拉普拉斯降阶展开,对第一列进行展开可知,只有第一个元素可能产生非零项:
∣
A
∣
=
a
11
(
−
1
)
1
+
1
M
11
=
a
11
M
11
|A|=a_{11}(-1)^{1+1}M_{11}=a_{11}M_{11}
∣A∣=a11(−1)1+1M11=a11M11
- 由于 M 11 M_{11} M11依然是一个上三角行列式,因此类似的有: M 11 = a 22 M 22 M_{11}=a_{22}M_{22} M11=a22M22
- 类似的:
M
i
i
=
a
i
+
1
,
i
+
1
M
i
+
1
,
i
+
1
M_{ii}=a_{i+1,i+1}M_{i+1,i+1}
Mii=ai+1,i+1Mi+1,i+1,
i
=
0
,
1
,
⋯
,
n
−
1
i=0,1,\cdots,n-1
i=0,1,⋯,n−1
- M i + 1 , i + 1 = a i + 2 , i + 2 M i + 2 , i + 2 M_{i+1,i+1}=a_{i+2,i+2}M_{i+2,i+2} Mi+1,i+1=ai+2,i+2Mi+2,i+2
- ⋯ \cdots ⋯
- 从而 ∣ A ∣ = a 11 a 22 ⋯ a n − 1 , n − 1 a n n = ∏ i = 1 n a i i |A|=a_{11}a_{22}\cdots{a_{n-1,n-1}a_{nn}}=\prod_{i=1}^{n}a_{ii} ∣A∣=a11a22⋯an−1,n−1ann=∏i=1naii
副对角线三角行列式
-
∣ A A T D ∣ = ( − 1 ) 1 2 n ( n − 1 ) ∏ i = 1 n a i j |A_{ATD}|=(-1)^{\frac{1}{2}n(n-1)}\prod\limits_{i=1}^{n}a_{ij} ∣AATD∣=(−1)21n(n−1)i=1∏naij
-
∣ a 11 a 12 ⋯ a 1 , n − 1 a 1 n a 21 a 22 ⋯ a 2 , n − 1 0 ⋮ ⋮ ⋮ ⋮ a n − 1 , 1 a n − 1 , 2 ⋯ 0 0 a n 1 0 ⋯ 0 0 ∣ = ∣ 0 0 ⋯ 0 a 1 n 0 0 ⋯ a 2 , n − 1 a 2 n ⋮ ⋮ ⋮ ⋮ 0 a n − 2 , 2 ⋯ a n − 2 , n − 2 a n − 2 , n a n 1 a n 2 ⋯ a n , n − 1 a n , n ∣ = ( − 1 ) τ ( n ⋯ 21 ) a 1 n a 2 , n − 1 ⋯ a n 1 = ( − 1 ) 1 2 n ( n − 1 ) a 1 n a 2 , n − 1 ⋯ a n 1 \\ \begin{vmatrix} a_{11}& a_{12}& \cdots &a_{1,n-1} & a_{1n} \\ a_{21}& a_{22}& \cdots &a_{2,n-1} & 0 \\ \vdots & \vdots & &\vdots & \vdots \\ a_{n-1,1}&a_{n-1,2}&\cdots&0&0\\ a_{n1}& 0& \cdots &0 &0 \end{vmatrix} =\begin{vmatrix} 0&0& \cdots &0 & a_{1n} \\ 0& 0&\cdots &a_{2,n-1} & a_{2n} \\ \vdots & \vdots & &\vdots & \vdots \\ 0&a_{n-2,2}&\cdots&a_{n-2,n-2}&a_{n-2,n}\\ a_{n1}& a_{n2}& \cdots &a_{n,n-1} &a_{n,n} \end{vmatrix} \\ =(-1)^{\tau(n\cdots21)}a_{1n}a_{2,n-1}\cdots{a_{n1}} \\ =(-1)^{\frac{1}{2}n(n-1)}a_{1n}a_{2,n-1}\cdots{a_{n1}} a11a21⋮an−1,1an1a12a22⋮an−1,20⋯⋯⋯⋯a1,n−1a2,n−1⋮00a1n0⋮00 = 00⋮0an100⋮an−2,2an2⋯⋯⋯⋯0a2,n−1⋮an−2,n−2an,n−1a1na2n⋮an−2,nan,n =(−1)τ(n⋯21)a1na2,n−1⋯an1=(−1)21n(n−1)a1na2,n−1⋯an1
- 其中逆序数的计算:
τ ( n ⋯ 21 ) = ∑ i = 1 n − 1 i = 1 2 n ( n − 1 ) \tau(n\cdots{21})=\sum\limits_{i=1}^{n-1}i=\frac{1}{2}n(n-1) τ(n⋯21)=i=1∑n−1i=21n(n−1)
- 其中逆序数的计算:
-
可以通过行列式性质将副对角线三角行列式转换为主对角线行列式.
- 只需执行若干次行互换即可
对角行列式
-
特别的,非0元素仅出现在对角线上的行列式称为对角行列式
-
∣ λ 1 λ 2 ⋱ λ n ∣ = λ 1 λ 2 ⋯ λ n = ∏ i = 1 n λ i \begin{vmatrix} {{\lambda _1}} & {} & {} & {} \cr {} & {{\lambda _2}} & {} & {} \cr {} & {} & \ddots & {} \cr {} & {} & {} & {{\lambda _n}} \cr \end{vmatrix} =\lambda_1\lambda_2\cdots\lambda_n =\prod_{i=1}^{n}\lambda_i λ1λ2⋱λn =λ1λ2⋯λn=i=1∏nλi
特殊的拉普拉斯行列式展开🎈
形式1
-
准主三角行列式:
-
设方阵A是 m + n m+n m+n阶的矩阵,且 A A A可以被划分为如下形式
-
∣ A m R m × n O n × m B n ∣ = ∣ A m O m × n C n × m B n ∣ = ∣ A m ∣ ⋅ ∣ B n ∣ \begin{vmatrix} A_m&R_{m\times{n}} \\ O_{n\times{m}}&B_n \end{vmatrix} = \begin{vmatrix} A_m& O_{m\times{n}}\\ C_{n\times{m}}&B_n \end{vmatrix} =|A_m|\cdot|B_n| AmOn×mRm×nBn = AmCn×mOm×nBn =∣Am∣⋅∣Bn∣
- O O O分布在副对角线上( A A A的有上角或左下角有一个零方阵)
-
-
以"下三角"的情况为例:
-
∣ Q ∣ = ∣ a 11 ⋯ a 1 m ⋮ ⋮ 0 a m 1 ⋯ a m m c 11 ⋯ c 1 m b 11 ⋯ b 1 n ⋮ ⋮ ⋮ ⋮ c n 1 ⋯ c n m b n 1 ⋯ b n n ∣ |Q|=\begin{vmatrix} a_{11}&\cdots&a_{1m}&&&\\ \vdots&\quad&\vdots&&\mathcal{\Huge{0}}&\\ a_{m1}&\cdots&a_{mm}&&&\\ c_{11}&\cdots&c_{1m}&b_{11}&\cdots&b_{1n}\\ \vdots&\quad&\vdots&\vdots&&\vdots\\ c_{n1}&\cdots&c_{nm}&b_{n1}&\cdots&b_{nn}\\ \end{vmatrix} ∣Q∣= a11⋮am1c11⋮cn1⋯⋯⋯⋯a1m⋮ammc1m⋮cnmb11⋮bn10⋯⋯b1n⋮bnn
-
分块矩阵中的方阵A:可以通过若干次行倍增操作( r i + k r j r_i+kr_j ri+krj)将A化为行列式等值的小下三角阵.
- ∣ A ∣ = ∣ a 11 ⋯ a 1 m ⋮ ⋮ a m 1 ⋯ a m m ∣ = ∣ a 11 ′ ⋮ ⋱ a m 1 ′ ⋯ a m m ′ ∣ = ∏ i = 1 m a i i ′ |A|=\begin{vmatrix} a_{11}&\cdots&a_{1m}\\ \vdots&\quad&\vdots\\ a_{m1}&\cdots&a_{mm} \end{vmatrix} =\begin{vmatrix} a'_{11}&&\\ \vdots&\ddots&\\ a'_{m1}&\cdots&a'_{mm} \end{vmatrix} =\prod_{i=1}^{m}a'_{ii} ∣A∣= a11⋮am1⋯⋯a1m⋮amm = a11′⋮am1′⋱⋯amm′ =i=1∏maii′
-
分块矩阵中的方阵B:可以通过若干次列倍增操作( c i + k c j c_i+kc_j ci+kcj)将B化为小下三角阵.(逐行的将B的右上角元素化为0)
- ∣ B ∣ = ∣ b 11 ⋯ b 1 n ⋮ ⋮ b n 1 ⋯ b n n ∣ = ∣ b 11 ′ ⋮ ⋱ b n 1 ′ ⋯ b n n ′ ∣ = ∏ i = 1 n b i i ′ |B|=\begin{vmatrix} b_{11}&\cdots&b_{1n}\\ \vdots&\quad&\vdots\\ b_{n1}&\cdots&b_{nn} \end{vmatrix} =\begin{vmatrix} b'_{11}&& \\ \vdots&\ddots&\\ b'_{n1}&\cdots&b'_{nn} \end{vmatrix} =\prod_{i=1}^{n}b'_{ii} ∣B∣= b11⋮bn1⋯⋯b1n⋮bnn = b11′⋮bn1′⋱⋯bnn′ =i=1∏nbii′
-
将上述2组变换对 ∣ Q ∣ |Q| ∣Q∣执行一遍,得到
-
∣ Q ∣ = ∣ a 11 ′ ⋮ ⋱ 0 a m 1 ′ ⋯ a m m ′ c 11 ⋯ c 1 m b 11 ′ ⋮ ⋮ ⋮ ⋱ c n 1 ⋯ c n m b n 1 ′ ⋯ b n n ′ ∣ = ∏ i = 1 m a i i ′ ∏ i = 1 n b i i ′ |Q|=\begin{vmatrix} a'_{11}&&&&&\\ \vdots&\ddots&&&\mathcal{\Huge{0}}&\\ a'_{m1}&\cdots&a'_{mm}&&&\\ c_{11}&\cdots&c_{1m}&b'_{11}&&\\ \vdots&\quad&\vdots&\vdots&\ddots&\\ c_{n1}&\cdots&c_{nm}&b'_{n1}&\cdots&b'_{nn}\\ \end{vmatrix} =\prod_{i=1}^{m}a'_{ii} \prod_{i=1}^{n}b'_{ii} ∣Q∣= a11′⋮am1′c11⋮cn1⋱⋯⋯⋯amm′c1m⋮cnmb11′⋮bn1′0⋱⋯bnn′ =i=1∏maii′i=1∏nbii′
-
Note:分块 C C C在 Q Q Q经过两组变换后并不受影响(将任意行列式化为等值的三角行列式只需要行倍增或列倍增中的一种即可实现)
-
-
可见 ∣ Q ∣ = ∣ A ∣ ∣ B ∣ |Q|=|A||B| ∣Q∣=∣A∣∣B∣
-
形式2
-
准副三角行列式
-
∣ T ∣ = ∣ O A m B n R ∣ = ∣ R A m B n O ∣ = ( − 1 ) m n ∣ A m ∣ ⋅ ∣ B n ∣ |T|=\begin{vmatrix} O&A_m \\ B_n&R \end{vmatrix} = \begin{vmatrix} R&A_m \\ B_n&O \end{vmatrix} =(-1)^{mn}|A_m|\cdot|B_n| ∣T∣= OBnAmR = RBnAmO =(−1)mn∣Am∣⋅∣Bn∣
-
其证明原理和形式1中的相仿,根据都是行列式的等值变换
-
类似于上述情况,通过若干行变换和列变换,转换为副对角线行列式:
-
这里方阵 T T T是 m + n m+n m+n阶的(为了计算或理解方便,可以令 t = m + n t=m+n t=m+n)
-
∣ A ∣ = ( − 1 ) 1 2 m ( m − 1 ) ∏ i = 1 m a i i ′ ∣ B ∣ = ( − 1 ) 1 2 n ( n − 1 ) ∏ i = 1 n b i i ′ ∣ T ∣ = ( − 1 ) 1 2 ( n + m ) ( n + m − 1 ) ∏ i = 1 m a i i ′ ∏ i = 1 n b i i ′ |A|=(-1)^{\frac{1}{2}m(m-1)}\prod_{i=1}^{m}a'_{ii} \\ |B|=(-1)^{\frac{1}{2}n(n-1)}\prod_{i=1}^{n}b'_{ii} \\ |T|=(-1)^{\frac{1}{2}{(n+m)(n+m-1)}} \prod_{i=1}^{m}a'_{ii}\prod_{i=1}^{n}b'_{ii} ∣A∣=(−1)21m(m−1)i=1∏maii′∣B∣=(−1)21n(n−1)i=1∏nbii′∣T∣=(−1)21(n+m)(n+m−1)i=1∏maii′i=1∏nbii′
-
该公式中的指数部分涉及表达式: τ = ( n + m ) ( n + m − 1 ) \tau=(n+m)(n+m-1) τ=(n+m)(n+m−1),由于我们仅关心 τ \tau τ的奇偶性,可对其形式进行变形
-
τ \tau τ= ( n + m ) 2 − ( n + m ) = n 2 + 2 m n + m 2 − n − m (n+m)^2-(n+m)=n^2+2mn+m^2-n-m (n+m)2−(n+m)=n2+2mn+m2−n−m
- = n ( n − 1 ) + m ( m − 1 ) + 2 m n =n(n-1)+m(m-1)+2mn =n(n−1)+m(m−1)+2mn
-
1 2 ( n + m ) ( n + m − 1 ) \frac{1}{2}{(n+m)(n+m-1)} 21(n+m)(n+m−1)= 1 2 n ( n − 1 ) + 1 2 m ( m − 1 ) + m n \frac{1}{2}n(n-1)+\frac{1}{2}m(m-1)+mn 21n(n−1)+21m(m−1)+mn
-
∣ T ∣ = ( − 1 ) 1 2 n ( n − 1 ) + 1 2 m ( m − 1 ) + m n ∏ i = 1 m a i i ′ ∏ i = 1 n b i i ′ = ( − 1 ) m n [ ( − 1 ) 1 2 m ( m − 1 ) ∏ i = 1 m a i i ′ ] [ ( − 1 ) 1 2 n ( n − 1 ) ∏ i = 1 n b i i ′ ] = ( − 1 ) m n ∣ A ∣ ∣ B ∣ \begin{aligned} |T|&=(-1)^{\frac{1}{2}n(n-1)+\frac{1}{2}m(m-1)+mn} \prod_{i=1}^{m}a'_{ii}\prod_{i=1}^{n}b'_{ii} \\ &=(-1)^{mn}[(-1)^{\frac{1}{2}m(m-1)}\prod_{i=1}^{m}a'_{ii}] [(-1)^{\frac{1}{2}n(n-1)}\prod_{i=1}^{n}b'_{ii}] \\&=(-1)^{mn}|A||B| \end{aligned} ∣T∣=(−1)21n(n−1)+21m(m−1)+mni=1∏maii′i=1∏nbii′=(−1)mn[(−1)21m(m−1)i=1∏maii′][(−1)21n(n−1)i=1∏nbii′]=(−1)mn∣A∣∣B∣
-
-
-
-
浙公网安备 33010602011771号