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Householder变换 https://www.cnblogs.com/reasno/p/9606224.html Householder transformation https://en.wikipedia.org/wiki/Householder_transformation [矩阵的QR分解系列三] 豪斯霍尔德(Householder)变换 https://blog.csdn.net/honyniu/article/details/110351391 北化 矩阵论 Householder矩阵Householder矩阵: H = I − 2 w w H H = I - 2ww^H H=I−2wwH, w w w为单位列向量 Householder矩阵性质:厄米特矩阵、正交矩阵、对合矩阵、反射矩阵 注:如果一个对合矩阵也是厄米特矩阵,那么它也是正交矩阵 Householder变换Householder变换(初等反射阵或镜象变换阵):将向量
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Hx=x−2wwHx=x−2w x x x和其反射向量 H x Hx Hx可以分解为 u u u与 w w w x = u + α w x = u + \alpha w x=u+αw H x = u − α w Hx = u - \alpha w Hx=u−αw α \alpha α为一个常数 H x = x − 2 w w H x Hx = x - 2ww^H x Hx=x−2wwHx 可写作: H x = y Hx = y Hx=y 定理1定理:H为householder矩阵,则 [ I r 0 0 H ] \begin{bmatrix} I_r & 0 \\ 0 & H\end{bmatrix} [Ir00H]也是为householder矩阵 证明: [ I r 0 0 H ] = [ I r 0 0 I n − 2 w w H ] = I n + r − [ 0 0 0 2 w w H ] = I n + r − 2 [ 0 u ] [ 0 H u H ] = I n + r − 2 u 1 u 1 H \begin{bmatrix} I_r & 0 \\ 0 & H\end{bmatrix} = \begin{bmatrix} I_r & 0 \\ 0 & I_n - 2ww^H \end{bmatrix} = I_{n+r} - \begin{bmatrix} 0 & 0 \\ 0 & 2ww^H \end{bmatrix} = I_{n+r} - 2\begin{bmatrix} 0 \\ u \end{bmatrix} \begin{bmatrix} 0^H & u^H \end{bmatrix} = I_{n+r} - 2u_1 u_1^H [Ir00H]=[Ir00In−2wwH]=In+r−[0002wwH]=In+r−2[0u][0HuH]=In+r−2u1u1H 定理2H = I − 2 w w H H = I - 2ww^H H=I−2wwH 定理:必存在 H x = ± ∣ ∣ x ∣ ∣ z , ∣ ∣ z ∣ ∣ = 1 Hx = \pm ||x|| z, ||z||=1 Hx=±∣∣x∣∣z,∣∣z∣∣=1 证明: x = 0 x=0 x=0,任取 w w w,则 H x = ± ∣ ∣ x ∣ ∣ z Hx=\pm ||x|| z Hx=±∣∣x∣∣z,定理在这种情况下成立 x = ± ∣ ∣ x ∣ ∣ z x = \pm ||x|| z x=±∣∣x∣∣z,取与 x x x正交的 w w w,则 H x = x = ± ∣ ∣ x ∣ ∣ z Hx=x=\pm ||x|| z Hx=x=±∣∣x∣∣z,定理在这种情况下成立 x ≠ ± ∣ ∣ x ∣ ∣ z x \neq \pm ||x|| z x=±∣∣x∣∣z,取 w = x − α z ∣ ∣ x − α z ∣ ∣ w = \frac{x- \alpha z}{||x - \alpha z||} w=∣∣x−αz∣∣x−αz,则 H x = x − 2 ( x − α z ) ( x − α z ) H ∣ ∣ x − α z ∣ ∣ 2 x = α z = ± ∣ ∣ x ∣ ∣ z Hx = x - 2\frac{(x- \alpha z)(x- \alpha z)^H}{||x - \alpha z||^2}x = \alpha z = \pm ||x|| z Hx=x−2∣∣x−αz∣∣2(x−αz)(x−αz)Hx=αz=±∣∣x∣∣z,定理在这种情况下成立 推论: x x x与 e 1 e_1 e1不共线,取 w = x − α e 1 ∣ ∣ x − α e 1 ∣ ∣ w = \frac{x- \alpha e_1}{||x - \alpha e_1||} w=∣∣x−αe1∣∣x−αe1,则 H x = ± ∣ ∣ x ∣ ∣ e 1 Hx = \pm ||x|| e_1 Hx=±∣∣x∣∣e1 Householder矩阵的特征对H H H为正交矩阵,则其有 n n n个特征对 如果 x x x与 w w w垂直,则 H x = x Hx = x Hx=x 如果 x = w x=w x=w垂直,则 H w = − w Hw = -w Hw=−w n n n维空间存在 n − 1 n-1 n−1个向量与 w w w垂直 因此可以看出 H H H的n个特征对: n − 1 n-1 n−1个 ( 1 , x i ) (1,x_i) (1,xi)和1个 ( − 1 , w ) (-1,w) (−1,w) 因此 H H H的行列式为-1 利用Householder变换实现正交分解矩阵 A = [ a 1 , . . . , a n ] A = [a_1, ..., a_n] A=[a1,...,an],实现 A = Q R A = QR A=QR 利用 n − 1 n-1 n−1个Householder矩阵 H i H_i Hi对 A A A进行变换,使得 H n − 1 . . . H 2 H 1 A = R H_{n-1}...H_2 H_1 A = R Hn−1...H2H1A=R, Q = H 1 H 2 . . . H n − 1 Q=H_1H_2...H_{n-1} Q=H1H2...Hn−1 如果 a 1 ≠ 0 a_1 \neq 0 a1=0,有 H 1 a 1 = α 1 e 1 H_1 a_1 = \alpha_1 e_1 H1a1=α1e1 H 1 A n = [ α 1 ∗ 0 B n − 1 ] H_1 A_n = \begin{bmatrix} \alpha_1 & * \\ 0 & B_{n-1} \end{bmatrix} H1An=[α10∗Bn−1] 如果 a 1 = 0 a_1 = 0 a1=0,则不需要进行householder变换,也可以说 H 1 = I H_1=I H1=I B n − 1 = [ b 1 , . . . , b n − 1 ] B_{n-1} = [b_1, ..., b_{n-1}] Bn−1=[b1,...,bn−1] 如果 b 1 ≠ 0 b_1 \neq 0 b1=0,有 H 2 ~ b 1 = α 2 e 1 ~ \tilde{H_2} b_1 = \alpha_2 \tilde{e_1} H2~b1=α2e1~ H 2 ~ B n − 1 = [ α 2 ∗ 0 C n − 2 ] \tilde{H_2} B_{n-1} = \begin{bmatrix} \alpha_2 & * \\ 0 & C_{n-2} \end{bmatrix} H2~Bn−1=[α20∗Cn−2] 如果 b 1 = 0 b_1 = 0 b1=0,则 H 2 ~ = I \tilde{H_2}=I H2~=I H 2 = [ 1 0 T 0 H 2 ~ ] H_2 = \begin{bmatrix} 1 & 0^T \\ 0 & \tilde{H_2} \end{bmatrix} H2=[100TH2~] 也就是说对于n维矩阵,如果其列向量都不为0,用 n − 1 n-1 n−1个Householder矩阵就能实现正交分解 Householder变换的另一个作用便是实现 A A A的正交化,将 A A A变为 Q Q Q 例题例1:通过Householder变换,使得 x = [ − 3 , 0 , 0 , 4 ] T x = [-3,0,0,4]^T x=[−3,0,0,4]T与 e 1 e_1 e1同向 解1: α = ∣ ∣ x ∣ ∣ = 5 \alpha = ||x|| = 5 α=∣∣x∣∣=5 w = x − 5 e 1 ∣ ∣ x − 5 e 1 ∣ ∣ = 1 80 [ − 8 , 0 , 0 , 4 ] T w = \frac{x- 5e_1}{||x - 5 e_1||} = \frac{1}{\sqrt{80}}[-8,0,0,4]^T w=∣∣x−5e1∣∣x−5e1=80 1[−8,0,0,4]T H x = x − 2 w w T x = 5 e 1 Hx = x - 2ww^Tx = 5e_1 Hx=x−2wwTx=5e1 解2: α = ∣ ∣ x ∣ ∣ = − 5 \alpha = ||x|| = -5 α=∣∣x∣∣=−5 w = x + 5 e 1 ∣ ∣ x + 5 e 1 ∣ ∣ = 1 20 [ 2 , 0 , 0 , 4 ] T w = \frac{x + 5e_1}{||x + 5 e_1||} = \frac{1}{\sqrt{20}}[2,0,0,4]^T w=∣∣x+5e1∣∣x+5e1=20 1[2,0,0,4]T H x = x − 2 w w T x = − 5 e 1 Hx = x - 2ww^Tx = -5e_1 Hx=x−2wwTx=−5e1 例2:通过Householder变换,使得 A = [ 0 3 1 0 4 − 2 2 1 2 ] A = \begin{bmatrix} 0 & 3 & 1 \\ 0 & 4 & -2 \\ 2 & 1 & 2 \end{bmatrix} A=⎣⎡0023411−22⎦⎤实现正交分解 解: a 1 = [ 0 0 2 ] a_1 = \begin{bmatrix} 0 \\ 0 \\2 \end{bmatrix} a1=⎣⎡002⎦⎤, α 1 = ∣ ∣ a 1 ∣ ∣ = 2 \alpha_1 = ||a_1|| = 2 α1=∣∣a1∣∣=2, w = a 1 − 2 e 1 ∣ ∣ a 1 − 2 e 1 ∣ ∣ = 1 2 [ − 1 0 1 ] w = \frac{a_1 - 2e_1}{||a_1 - 2 e_1||} = \frac{1}{\sqrt{2}}\begin{bmatrix} -1 \\ 0 \\1 \end{bmatrix} w=∣∣a1−2e1∣∣a1−2e1=2 1⎣⎡−101⎦⎤ H 1 A = [ 2 1 2 0 4 − 2 0 3 1 ] H_1 A = \begin{bmatrix} 2 & 1 & 2 \\ 0 & 4 & -2 \\ 0 & 3 & 1 \end{bmatrix} H1A=⎣⎡2001432−21⎦⎤ b 1 = [ 4 3 ] b_1 = \begin{bmatrix} 4 \\ 3 \end{bmatrix} b1=[43], α 2 = ∣ ∣ b 1 ∣ ∣ = 5 \alpha_2 = ||b_1|| = 5 α2=∣∣b1∣∣=5, w = b 1 − 5 e ~ 1 ∣ ∣ b 1 − 5 e ~ 1 ∣ ∣ = 1 10 [ − 1 3 ] w = \frac{b_1 - 5 \tilde e_1}{||b_1 - 5 \tilde e_1||} = \frac{1}{\sqrt{10}}\begin{bmatrix} -1 \\ 3 \end{bmatrix} w=∣∣b1−5e~1∣∣b1−5e~1=10 1[−13] H 2 H 1 A = [ 2 1 2 0 5 − 1 0 0 − 2 ] H_2H_1 A = \begin{bmatrix} 2 & 1 & 2 \\ 0 & 5 & -1 \\ 0 & 0 & -2 \end{bmatrix} H2H1A=⎣⎡2001502−1−2⎦⎤ A = Q [ 2 1 2 0 5 − 1 0 0 − 2 ] , Q = H 1 H 2 A = Q \begin{bmatrix} 2 & 1 & 2 \\ 0 & 5 & -1 \\ 0 & 0 & -2 \end{bmatrix}, Q = H_1H_2 A=Q⎣⎡2001502−1−2⎦⎤,Q=H1H2 补充知识—正交补https://zh.wikipedia.org/wiki/正交补 W W W为内积空间 V V V的子空间,则 W W W的正交补 W ⊥ W^{\perp} W⊥定义为: W ⊥ = { x ∈ V : ∀ y ∈ W , < x , y > = 0 } W^{\perp} = \{ x \in V: \forall y \in W, =0 \} W⊥={x∈V:∀y∈W,=0} 注: W ⊥ W^{\perp} W⊥也是 V V V的子空间 W ⊥ ⊥ = W ˉ W^{\perp \perp} = \bar{W} W⊥⊥=Wˉ 疑问:线性空间 W W W与线性空间的闭包 W ˉ \bar{W} Wˉ有什么区别 A A A为一个矩阵, R o w A , C o l A , N u l A RowA, ColA, NulA RowA,ColA,NulA分别为其行空间,列空间,零空间 ( R o w A ) ⊥ = N u l A (RowA)^{\perp} = NulA (RowA)⊥=NulA ( C o l A ) ⊥ = N u l A T (ColA)^{\perp} = NulA^T (ColA)⊥=NulAT 注: A A A的右零空间是由 A x = 0 Ax=0 Ax=0的解张成的空间 A A A的左零空间是由 A T x = 0 A^Tx=0 ATx=0的解张成的空间 W W W为巴拿赫空间 V V V的子空间,则 W W W的正交补 W ⊥ W^{\perp} W⊥定义为: W ⊥ = { x ∈ V ∗ : ∀ y ∈ W , x ( y ) = 0 } W^{\perp} = \{ x \in V^*: \forall y \in W, x(y)=0 \} W⊥={x∈V∗:∀y∈W,x(y)=0} 注: W ⊥ W^{\perp} W⊥是 V V V的对偶空间的子空间 对偶空间是行向量与列向量关系的抽象化 疑问: V ∗ ∗ V^{**} V∗∗与 V V V的关系? 设 V V V为 在域 F F F上的向量空间,定义其对偶空间 V ∗ V^∗ V∗ 为由 V → F V \rightarrow F V→F 的所有线性函数的集合 V V V的元素称为反变或逆变contravariant向量; V ∗ V^* V∗的元素称为共变或协变covariant向量,也称余向量或同向量co-vectors,一形或线性型one-form |
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